Class 9 Science

Class 9 Science Chapter 8 Force And Laws Of Motion Very Short Answer Type Questions

Question 1. Give two examples of the effects of force.
Answer:

A toy car starts moving when pushed.

The shape of the dough ball changes when rolled.

Question 2. Apart from changing the magnitude of velocity of an object or changing the direction of motion of an object, what other changes can force bring on an object?
Answer:

Force can change the state of rest to the state of motion and vice-versa. It can also change the shape and position of the body.

Question 3. Why friction does not pull a ball backward if no force is pushing or pulling it?
Answer:

Friction comes into action only in the presence of an applied force.

Question 4. Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the frictional force that will be exerted on the cabinet?
Answer:

The frictional force is 200 N but in the opposite direction.

Question 5. What do you mean by resultant force?
Answer:

When two or more forces act on a body simultaneously then the single force which produces the same effect as produced by all the other forces acting together is known as the resultant force.

Question 6. While riding on the bicycle, if we stop peddling, why does the bicycle begin to slow down?
Answer:

The bicycle begins to slow down because of the force of friction acting in the opposite direction.

Question 7. While performing a practical, student A kept the wooden block on a polished wooden surface and measured the minimum force required to pull it as F₁ while student B kept the wooden block on a rough surface, covered with sand, and measured the minimum force required to pull it as F₂. They repeated the experiment five times. Which student applied more force to pull the block?
Answer:

Student B applied more force to pull the block because the friction force acting on the block placed on a rough surface was of large magnitude and opposite to the direction of force applied.

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Question 8. What did Galileo conclude on the basis of his experiments on the motion of objects?
Solution:

A body continues to move with the same velocity if no unbalanced force acts on it.

Question 9. Velocity-time graph of a moving particle of mass 1 kg

Is any force acting on the body? Justify your answer.

Answer:

The velocity of the body is uniform, thus acceleration is zero. Hence, no unbalanced force acts on the body.

Question 10. Why should we wear safety seat belts in a car?
Answer:

Due to our inertia, we may get hurt when the car suddenly starts or stops. Safety belts protect us by slowing down the motion in such situations.

Question 11. A water tanker filled up to — of its height is moving with a uniform speed. On sudden application of the brake, in which direction the water in the tanker would flow?
Answer:

On the sudden application of brake, the tanker will come in the state of rest but the water remains in the state of motion, so the water will move forward.

Question 12. Why do we tend to fall outwards when a car takes a steep turn?
Answer:

We tend to fall outwards when a car takes a steep turn due to inertia of direction.

Question 13. Name the agency which when applied to a body is directly proportional to the rate of change of momentum which it produces in the body.
Answer:

The force applied to a body is directly proportional to the rate of change of momentum.

Question 14. Two similar vehicles are moving with the same velocity on the roads such that one of them is loaded and the other is empty. Which of the vehicles will require a larger force to stop it? Give reasons.
Answer:

Both have the same velocity but one of them is loaded, so its momentum is greater than the other. Thus, for the loaded vehicle greater change of momentum, so larger force will be required to stop it.

Question 15. Athletes in pole jump events fall on a cushioned surface and not on the floor. Why?
Answer:

The change in momentum occurs over a long duration of time which reduces the force acting on the athlete and he does not hurt.

Question 16. A passenger in a moving train tosses a coin which falls behind him. From this incident, what can you predict about the motion of the train?
Answer:

If the coin falls behind the passenger that means the train is accelerated. When the coin is tossed, then it has the same velocity as that of the train but during the time it is in the air its velocity becomes less than that of the train (because the train is accelerated), so it falls behind the passenger.

Question 17. An athlete always runs for some distance before taking a jump. Why?
Answer:

When an athlete runs for some distance, then he increases his inertia of motion. At the jumping time, when he takes a jump, his increased inertia of motion carries him over a long distance or height.

Question 18. Name the physical quantity which corresponds to the rate of change of momentum.
Answer:

Force (according to Newton’s second law rate of change of momentum is known as force).

Question 19. A body accelerates when a balanced force acts on it. Is it true or false? Why?
Answer:

It is false because according to Newton’s second law, the momentum of a body changes only when an unbalanced force acts on it.

Question 20. A goalkeeper in a game of football pulls his hands backward after holding the ball shot at the goal. Explain.
Answer:

The goalkeeper pulls his hands backward after holding the ball to decrease the rate of change of momentum by increasing the time.

By doing this, less force is exerted on his hands. Since force is directly proportional to the rate of change of momentum.

Question 21. If a man jumps out from a boat to the bank of the river, the boat moves backward. Identify the action and reaction in this situation.
Answer:

The push of the man on the boat will be action while the boat exerts an equal force on the man, which will be a reaction.

Question 22. A runner presses the ground with his feet before he starts his run. Identify action and reaction in this situation.
Answer:

When a runner presses the ground, action is done by his feet and a reaction is produced by the ground, as a result runner moves forward.

Question 23. According to the third law of motion, do action and reaction forces act on the same body or different bodies?
Answer:

According to the third law of motion, action and reaction force always act on different bodies in opposite directions.

Question 24. Does Newton’s third law apply to a system where bodies do not touch each other?
Answer:

Yes. For example, in the case of attraction or repulsion between two magnets, etc. Newton’s third law is applicable.

Question 25. A balloon is inflated and released. Why does it fly forward as air escapes out of it?
Answer:

The reaction force of air coming out of the balloon pushes the balloon in the forward direction.

Question 26. Shyam throws a heavy stone out of his boat. As a result, the boat moves in the opposite direction, why?
Answer:

According to Newton’s third law of motion, action and reaction are always equal and opposite and act on two different bodies.

As the heavy stone is thrown in the forward direction, the boat moves backward.

Question 27.  If F and F’ are balanced forces, then what will be the magnitude of \(F_2\)?
Answer:

As per the question, F and F’ are balanced forces.

So, F=\(F^{\prime}\)

15+F_2=25 N

⇒ \(F_2=10 \mathrm{~N}\)

Class 9 Science Chapter 8 Force And Laws OF Motion Short Answer Type Questions

Question 1. A car is moving horizontally on a surface along a straight line at a constant speed. Is any force acting on it? If not, why is the fuel required to run it?
Answer:

The car moves with constant velocity. So, no unbalanced force acts on it.

The fuel is used to move the car with a force that balances with the force of friction exerted by the surface due to the motion of the car.

Question 2. Make a pile of similar carom coins on a table. Then, remove the lower coin without touching the other coins. With your fingers, you may give a sharp horizontal hit at the bottom of the pile using a striker.

1.  What will happen, if the hit is strong enough?
2. What will happen to the remaining coins once the lower coin is removed?

Answer:

1.  In case, the hit is strong enough, the bottom coin moves quickly, while the pile above remains at rest due to inertia.
2.  Once the lower coin is removed, the inertia of the remaining coins makes them fall vertically on the table.

Question  3. Why does the pillion rider fall forward when brakes are applied?
Answer:

During the ride, the pillion rider and driver are in a state of motion. But when the driver applies brakes, then the body of the pillion rider continues moving forward on account of the inertia of motion. Therefore, the pillion rider falls forward.

Question  4. There are three solids made up of aluminum, steel, and wood of the same shape and volume. Which of them would have the highest inertia?
Answer:

So Steel. As the mass is a measure of inertia, the ball having more mass than other balls will have the highest inertia. Since steel has the greatest density among these. So, it has the highest mass and inertia.

Question 5. A bullet fired against a glass window pane makes a hole in it and the glass pane is not cracked. But on the other hand, when a stone strikes the same glass pane, then it gets smashed. Why is it so?
Answer:

When the bullet strikes the glass pane, then the part of the glass pane that comes in contact with the bullet immediately shares the large velocity of the bullet and makes a hole, while the remaining part of the glass remains at rest and is therefore not smashed due to inertia of rest.

But when a slow-moving stone strikes the same glass pane, then the various parts of the glass pane get enough time to share the velocity of the stone, and the glass is smashed.

Question 6. Why can a small mass such as a bullet kill a person when fired from a gun?
Answer:

It is so because even, if the mass of the bullet is small, it moves out of the gun with a very high velocity, due to which its momentum is high (p =mv). This high momentum of the bullet exerts a large force and kills a person.

Question 7. Why does a boat tend to leave the shore, when passengers are alighting from it?
Answer:

When the passengers are alight from the boat, the boat moves in a backward direction. This is an example of Newton’s third law. This difficulty is usually overcome by the boatman by binding the boat to some rigid support.

Question 8. How are action and reaction forces related to magnitude and direction?
Answer:

Action and reaction forces have the same magnitude but opposite directions.

Question 9. Describe our walking in terms of Newton’s third law of motion.
Answer:

When we walk on the ground or road, our foot pushes the ground backward (action) and the ground pushes our foot forward (reaction). Thus, the forward reaction force exerted by the ground on our foot makes us walk forward.

Question 10. Two identical bullets are fired one by one by a light rifle and another by a heavy rifle with the same force. Which rifle will hurt the shoulder more and why?
Answer:

Both the bullets are identical and are fired with the same force. So, according to Newton’s third law of motion, the same force will be applied to both rifles. With the application of the same force, the light rifle will move more quickly in the forward direction, so it will hurt more to the shoulder.

Question 11. Which Newton’s law of motion is applied in the flight of a bird?
Answer:

For flight, a bird applies force on air in a backward direction by flapping its wings (action). Air exerts (a force) on the bird in the forward direction. Therefore Newton’s third law of motion is applied in the flight of a bird.

Question 12. (1) A heavy and a light object has the same momentum. Which of these is traveling faster?

(2) State the law of conservation of momentum. Give examples.

Answer:

1. If two objects are moving with the same momentum, then the object having lightweight moves faster.
2. According to the law of conservation of momentum, if an external force is not present, then the total momentum before the collision between the two bodies is equal to the total momentum after the collision.

Question 13. Two children stand on wheel carts facing each other. One student throws a heavy mass towards the other who catches it. What will be the direction of motion of both the children? Give a reason for your answer.
Answer:

The child who throws the mass towards the other and the one who catches it, both move backwards.

Reason: The Thrower moves due to a backward reaction of mass pushed forward.

The catcher moves due to the forward reaction of mass pushed and held backward.

Question 14. Take a big rubber balloon and inflate it fully. Tie its neck using a thread. Also, using adhesive tape, fix a straw on the surface of this balloon. Pass a thread through the straw and hold one end of the thread in your hand or fix it on the wall.

Ask your friend to hold the other end of the thread or fix it on a wall at some distance.

This arrangement is shown in the figure. Now, remove the thread tied to the neck of the balloon. Let the air escape from the mouth of the balloon. Observe the direction in which the straw moves. Give reason.

Answer:

The straw moves in the opposite direction of air because when the air is released out of the balloon, it imparts momentum to the straw from Newton’s third law of motion.

Question 15. Two balls of the same size but of different materials, rubber, and iron are kept on the smooth floor of a moving train. The brakes are applied suddenly to stop the train. Will the balls start rolling? If, so in which direction? Will they move at the same speed? Give reasons for your answer.
Answer:

When the train is stopped suddenly, then it comes in the state of rest but the balls remain in the state of motion. So, due to the inertia of motion, the balls move in the forward direction.

As the balls are of the same size but of different materials that means their masses will be different. So, both the balls will move at different speeds.

Question 16. Suppose a ball of mass m is thrown vertically upward with an initial speed of v, its speed decreases continuously till it becomes zero. Thereafter, the ball begins to fall downward and attain the speed v again before striking the ground. It implies that the magnitude of the initial and final momentum of the ball are the same. Yet, it is not an example of conservation of momentum. Explain, why.
Answer:

The momentum of a system remains conserved if no external force acts on the system. In the given example, there is a gravitational force acting on the ball, so it is not an example of conservation of momentum.

Question 17. Two friends on roller skates are standing 5 m apart facing each other. One of them throws a ball of 2 kg towards the other, who catches it? How will this activity affect the position of the two? Explain your answer.
Answer:

Separation between them will increase. Initially, the momentum of both of them is zero as they are at rest. In order to conserve the momentum, the one who throws the ball would move backward.

The second will experience a net force after catching the ball and therefore will move backwards that is in the direction of the force.

Question 18. On what factors do the following physical quantities depend?

1. Inertia
2. Momentum
3. Force
4. Impulse

Answer:

1. Inertia depends on the mass of a body.
2. Momentum depends on the mass and velocity of the body.
3.  The force depends on the mass and acceleration of the body.
4.  Impulse depends on force and the short time in which force acts.

Question 19. Give a reason for the following questions:

1. Road accidents occurring due to high speeds are much worse than accidents due to low speeds of vehicles.
2. When a motorcar makes a sharp turn at a high speed passengers tend to get thrown to one side.

Answer:

1. Road accidents occurring due to high speeds are much worse than accidents due to low speeds of vehicles. This is because the momentum of high-speed vehicles is more than that of the low speeds of vehicles.
2. When a motorcar makes a sharp turn left or right at a high speed. The lower portion of their passengers turns suddenly along with the motorcar but the upper portion does not change its direction due to inertia.

• • So, this portion of a passenger moves forward and the passenger tends to get thrown to one side or another side.

Question 20. Glasswares are wrapped in straw during their transportation. Justify giving a reason.
Answer:

During transportation, the glasswares may break, if they collide with each other in the event of jerks. When they are wrapped in straw, then the force of the jerk is transmitted to them through the pieces of straw for a longer period of time.

Thus, the change in momentum of the glasswares takes place over a longer period of time. Therefore, a very small force is experienced by them in the event of jerks and hence they do not break.

Question 21. What would happen if a fielder stops the fast-moving ball suddenly? Justify your answer.
Answer:

The high velocity of the ball decreases to zero in a short time. It means that in a short time, there is a large change in the momentum of the ball.

Therefore, to stop the fast-moving ball, the fielder will have to apply a large force and in the process, he may hurt his palm.

Question 22. Give a reason for the following:

1. The water sprinkler used for grass lawns begins to rotate as soon as water is supplied.
2. Water drops are removed from wet clothes by giving a tight jerk to the cloth.

Answer:

1. When water is pushed out of the sprinkler with a force, it exerts reaction force on the sprinkler causing it to rotate.
2. Water drops have the inertia of rest. They do not move with the cloth as it is jerked and comes out.

Question 23. What do you understand by momentum? A vehicle is moving with a velocity of 5 m/s. If the momentum of the vehicle is 5000 kg-m/s, then what is its mass?
Answer:

Momentum is the physical quantity which is the product of mass and velocity of an object. Momentum has both masses as well as the direction of motion. It is denoted by p. Its SI unit is kg-m/s.

Momentum, p = mv

Given, p – 5000 kg-m/s, v = 5 m/s, m -?

5000 = m x 5 => m = \(\frac{5000}{5}\) = 1000 kg

Question 24. A man Throws a ball weighing 500 g vertically upwards with a speed of 20 m/s.

1.  What will be its initial momentum?
2. What would be its momentum at the highest point of its flight?

Answer:

Given, initial velocity, u = 20 m/s

Mass, m=500 \(\mathrm{~g}=\frac{1}{2} \mathrm{~kg}\)

Initial momentum, p_1=m u

⇒ \(p_1=\frac{1}{2} \times 20=10 \mathrm{~kg}-\mathrm{m} / \mathrm{s}\)

At the highest point, v=0

Final momentum,\(p_2\)=m v

∴ \(p_2=\frac{1}{2} \times 0=0\)

Question 25. A force of 30 N produces an acceleration of 2 m/s² in a body of mass m. Find the mass of the body. Also, find the acceleration, if force is doubled in magnitude along the same direction.
Answer:

Acceleration, a=2 \(\mathrm{~m} / \mathrm{s}^2\)

Mass of a body, m=\(\frac{F}{a}=\frac{30}{2}=15 \mathrm{~kg}\) [F=m a]

If F is doubled, a is also doubled as

F \(\propto\) a [since, m is constant]

So, a=4 \(\mathrm{~m} / \mathrm{s}^2\)

Thus, acceleration when force is doubled is 4 m/s.

Question 26. For how much time should a force of 200 N act on an object having a mass 5 kg, so as to increase its velocity from 50 m/s to 100 m/s? Given, force, F = 200 N Mass of an object, m = 5 kg Initial velocity, u – 50 m/s Final velocity, v = 100 m/s
Answer:

Given, force, F = 200 N

Mass of an object, m =5l.g

Initial velocity u -50 m/s

Final velociry u -100 m/s

From Newton’s second law of motion, we get

F=m a = \(m(\frac{v-u}{t}) [a=\frac{v-u}{t}]\)

200 =5\(\left(\frac{100-50}{t}\right)\)

t =5\(\left(\frac{100-50}{200}\right)\)

= 1.25 s

Thus, the time taken is 1.25 s.

Question 27. A cricket ball of mass 70 g moving with a velocity of 0.5 m/s is stopped by a player in 0.5 s. What is the force exerted by the player to stop it?
Answer:

Given, the mass of the cricket ball, m = 70 g = 0.07 kg

Initial velocity, u =0.5 m/s, final velocity, v = 0 Time, t =0.5 s

From Newton’s second law of motion, we get

F =m a=m\((\frac{v-u}{t}) [ a=\frac{v-u}{t}]\)

=0.07 \(\times\left(\frac{0-0.5}{0.5}\right)=-0.07 \mathrm{~N}\)

A negative sign shows that force is in the opposite direction to the motion of the ball. So, the magnitude of the force is 0.07 N.

Question 28. The velocity-time graph of a ball of mass 30 g moving along a straight line on a long table is given in the figure below.

How much force does the table exert on the ball to bring it to rest?

Answer:

Given, the mass of the ball, m= 30 g = 0.03 kg

From the given graph, we note that u =20 cm/s = 0.20 m/s

v =0, t =10 -0 =10 s

As we know, the force exerted on the ball is given by

F = \(m\left(\frac{v-u}{t}\right)=0.03\left(\frac{0-0.20}{10}\right)\)

=-0.0006 N

Thus, the force exerted by the table on the ball is -0.0006 N.

Question  29. Two billiard balls each of mass 0.05 kg moving in opposite directions each with a speed of 6 m/s collide and rebound with the same speed. Find the impulse imported to each ball due to the other.
Answer:

The initial momentum of the ball

A = 0.05 x 6 = 03kg m/s and final momentum of the same ball

A = 0.05 x (-6) = -03 kg m/s /.

Impulse imported to ball A = Change in momentum = -03 -03 = -0.6kg m/s

An equal impulse is imported to ball B by ball A.

Question  30. A bullet of mass 20 g is fired from a pistol of mass 2 kg with a horizontal velocity of 150 m/s. Calculate the recoil velocity of the pistol.
Answer:

Given, mass of bullet, m = 20 g = 0.02 kg

Mass of pistol, M = 2 kg

Velocity of bullet, =150m/s

From the law of conservation of momentum, momentum of pistol = momentum of bullet MV = -mv

[Here, a negative sign shows that the direction of motion of the bullet and pistol are opposite to each other]

Velocity of pistol, V =\(-\frac{m v}{M}\)

= \(\frac{-0.02 \times 150}{2}\)

V =-1.5 m/s

Thus, the recoil velocity of the pistol is 1.5 m/s.

Velocity of pistol, V =-\(\frac{m v}{M}\)

= \(\frac{-0.02 \times 150}{2}\)

V =-1.5 m/s

Question 31. Find the acceleration and force acting on a body of mass 4 kg using the following v-t graph.

Answer:

Acceleration, a= Slope of the v-t graph

=\(\frac{Y \text {-intercept }}{X \text {-intercept }}=\frac{10 \mathrm{~m} / \mathrm{s}}{4 \mathrm{~s}}=2.5 \mathrm{~m} / \mathrm{s}^2\)

From Newton’s second law of motion, we get

F=m a=4 \(\times\) 2.5=10 \(\mathrm{~N}\)

Thus, acceleration and force on the body are 2.5 \(\mathrm{~m} / \mathrm{s}^2\) and 10 N, respectively.

Question 32. The force of ION applied to a mass mx produces an acceleration of 5 m/s² and when applied to mass produces an acceleration of 15 m/s² in the mass. How much acceleration will the same force produce, if the two masses are tied together?
Answer:

We know that, F = ma

10 = \(m_1 \times 5 or m_1=\frac{10}{5}=2 \mathrm{~kg}\)

Also, \(m_2=\frac{10}{15}=\frac{2}{3} \mathrm{~kg}\)

Required acceleration produced when the masses are tied together (i.e. m = \(m_1+m_2\) ),

a = \(\frac{F}{m}=\frac{10}{m_1+m_2}=\frac{10}{2+\frac{2}{3}}\)

=\(\frac{30}{8}=3.75 \mathrm{~m} / \mathrm{s}^2\)

Question 33. A bullet of mass 4 g when fired with a velocity of 50 m/s can enter a wall up to a depth of 10 cm. How much will be the average resistance offered by the wall?
Answer:

The hindrance offered by the wall to the motion of a bullet is called the resistance offered by the wall. Given, Mass of the bullet, m = 4g = 4 x 10^-3 kg

Initial velocity, u=50 \(\mathrm{~m} / \mathrm{s}\),

Depth, s=10 \(\mathrm{~cm}=\frac{1}{10} \mathrm{~m}\)

Final velocity, v=0, Force, F=?

Using the second equation of motion,

⇒ \(v^2 = u^2+2 a s\)

⇒ \(v^2-u^2 =2 a s \)

⇒ \(0-(50)^2  =2 a \times \frac{1}{10}\)

⇒ \(-2500 =\frac{a}{5} \Rightarrow a=-12500 \mathrm{~m} / \mathrm{s}^2\)

Force, \(F=m a=4 \times 10^{-3} \times(-12500)=-50 \mathrm{~N}\)

Thus, the average resistance offered is 50 N.

The negative sign indicates that the force is acting opposite to the motion.

Question 34. A heavy car A of mass 2000 kg traveling at 30 m/s has a head-on collision with a sport car B of mass 1000 kg. If both cars stop dead on colliding, then what was the velocity of car B?
Answer:

Given, mass of car  A, \(m_1 =2000 \mathrm{~kg}\)

Mass of car B, \(m_2\) =1000 kg

Velocity of car A, \(v_1 =30 \mathrm{~m} / \mathrm{s}\)

and velocity of car B, \(v_2\)=?

Now, according to the law of conservation of momentum, \(m_1 v_1=m_2 v_2\)

2000 \(\times\)30 =1000 \(\times v_2\)

⇒ \(v_2 =2 \times 30\)

=60 \(\mathrm{~m} / \mathrm{s}\)

Question 35. The motion of a body of mass 5 kg is shown in the velocity-time graph.

Find from the graph.

1.  Its acceleration.
2. The force acting on the body.
3. The change in momentum of a body in 2 seconds after the start.

Answer:

(1) Acceleration = Slope of the line of the velocity-time graph,

a =\(\frac{v_2-v_1}{t-t_1}=\frac{5-0}{2-0}=\frac{5}{2}=\frac{10}{4}\)

= \(\frac{15}{6}=2.5 \mathrm{~m} / \mathrm{s}^2\)

(2) The force acting on the body is given by

F=m a=5 \(\times\) 2.5=12.5 N

Change in momentum =m v-m u  [ u=0 and v=5 m/s]

=5 \(\times\) 5-5 \(\times\) 0=25 kg- m/s

Question 36. The velocity-time graph of a ball moving on the surface of the floor is shown in the figure. Calculate the force acting on the ball, if the mass of the ball is 100 g.

Answer:

The velocity-time graph shows that the velocity of the ball at =0 is zero.

So, the initial velocity of the ball, u = 0.

Velocity of the ball at t = 4 s is 20 m/s

i.e. final velocity, v =20 m/s; Time, t = 4 s

Acceleration of the ball, a =\(\frac{v-u}{t}=\frac{20 \mathrm{~m} / \mathrm{s}-0}{4 \mathrm{~s}}\)

=5 \(\mathrm{~m} / \mathrm{s}^2\)

Also, the mass of the ball,

m=100 \(\mathrm{~g}=\frac{100}{1000} \mathrm{~kg}=\frac{1}{10} \mathrm{~kg}\)

Force acting on the ball, F=m a=\(\frac{1}{10} \mathrm{~kg} \times 5 \mathrm{~m} / \mathrm{s}^2\)

=0.5 \(\mathrm{~kg}-\mathrm{m} / \mathrm{s}^2=0.5 \mathrm{~N}[1 \mathrm{~kg}-\mathrm{m} / \mathrm{s}^2=1 \mathrm{~N}]\)

Question 37. The speed-time graph of a car is given. The car weighs 1000 kg.

1. What is the distance traveled by the car in the first 2s?
2.  What is the braking force applied at the end of 5 s to bring the car to stop within one second?

Answer:

(1) Distance traveled by the car in first 2 s= Area of \(\triangle\) O A D=\(\frac{1}{2} \times 2 \times 15=15 \mathrm{~m}\)

(2) Braking force, F=m \(\times\) a

Given, mass of the car, m=1000 kg, initial velocity, u=15 m/s,

final velocity, v=0, time, t = 1s

On applying, a=\(\frac{v-u}{t} \Rightarrow a=\frac{0-15}{1}=-15 \mathrm{~m} / \mathrm{s}^2\)

F = ma =1000 \(\times(-15)\)=-15000 N

Question 38. A constant force of friction of 50 N is acting on a body of mass 200 kg moving initially with a speed of 15 m/s. How long does the body take to stop? What distance will it cover before coming to rest?
Answer:

Given, F = 50 N, m = 200 kg, u= 15 m/s

The acceleration of the body is obtained from F = ma

or a=\(\frac{F}{m}=\frac{50}{200}=-0.25 \mathrm{~m} / \mathrm{s}^2\)

Now, from the first equation of motion

v=u+a t

t=\(\left(\frac{v-u}{a}\right)=\left(\frac{0-15}{-0.25}\right)=60 \mathrm{~s}\)

Also, the distance traveled is obtained from

s =u t+\(\frac{1}{2} a t^2\)

=50 \(\times 60+1 / 2 \times(-0.25) \times(60)^2=450 \mathrm{~m}\)

Question 39. Look at the diagram below and answer the following questions:

1.  When a force is applied through the free end of the spring balance A, then the reading on the spring balance A is 15 g-wt. What will be the measure of the reading shown by spring balance B?
2.  Write reasons for your answer.
3.  Name the force that balance A exerts on balance B and the force of balance B on balance A.

Answer:

1. 15 g-wt.
2. From Newton’s third law, the force exerted by B on A and the force exerted by A on B is equal.
3.  Force of reaction balance A exerts on balance B and force of action balance B exerts on balance A.

Question 40. An iron sphere of mass 1 kg is dropped from a height of 10 m. If the acceleration of the sphere is 9.8 m/s², then calculate the momentum transferred to the ground by the ball.
Answer:

Here, initial velocity of sphere, u = 0 Distance travelled, s = 10 m

Acceleration of sphere, a = 9.8 m/s²

Final velocity, v =?

The final velocity of the sphere when it just reaches the ground can be calculated using the,

v² – u² = 2 as

⇒ v² -0 = 2 x 9.8 x 10 =196 m/s

v = √196 m/s = 14 m/s

Momentum of the sphere just before it touches the ground = mv

= 1 kg x 14 m/s =14 kg-m/s

On reaching the ground, the iron sphere comes to rest, so its final momentum is equal to zero.

According to the law of conservation of momentum, momentum transferred to the ground = momentum of the sphere just before it comes to rest = 14 kg-m/s

Class 9 Science Chapter 8 Long Answer Type Questions

Question 1. What is momentum? Write its SI unit. Interpret force in terms of momentum. Represent the following graphically:

1. Momentum versus velocity when mass is fixed.
2. Momentum versus mass when velocity is constant.

Answer:

Momentum The quantity of motion possessed by a moving body is known as the momentum of the body.

It is the product of the mass and velocity of the body. Momentum, p = mu. Its SI unit is kg-m/s

Force applied on an object of mass m moving with acceleration a.

F=m a=m \(\frac{\Delta v}{\Delta t}\)

[acceleration = rate of change of velocity =\(\frac{\Delta v}{\Delta t}]\)}

=\(\frac{\Delta p}{\Delta t}\)

Force applied on an object is equal to the rate of change of momentum of the object.

(1) Momentum versus velocity graph when mass is fixed, p=m v. If m is fixed, then p \(\propto v\)

The momentum versus velocity graph will be a straight line passing through the origin (if v=0, then p=0 ).

(2) Momentum versus mass graph when velocity is constant, p — mv.

If velocity is constant, then p m.

So, the momentum versus mass graph will be a straight line passing through the origin (if m = 0, then p = 0).

Question 2. (1) When a carpet is beaten with a stick dust comes out of it. Explain.

(2) Calculate the force required to impart a car with a velocity of 30 m/s in 10 s starting from rest. The mass of the car is 15000 kg.

Answer:

(1) When we beat the carpet with a strick, then it comes into motion. However, the dust particles continue to be at rest due to inertia and get detached from the carpet.

(2) Given, initial velocity, u = 30 m/s

Time, t = 10 s, final velocity, v = 0

Mass, m = 15000 kg

From first equation of motion, a=\(\frac{v-u}{t}=\frac{-30}{10}=-3 \mathrm{~m} / \mathrm{s}^2\)

So, retarding force required = ma

= 15000 x (-3) = – 45000 N

Thus, the force required by the car is 45000 N.

Question 3. An object of mass 5 kg is accelerated uniformly from a velocity of 4 m/s to 8 m/s in 8 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.

Answer:

Given, mass, m=5 kg, initial velocity, u=4 m/s

Final velocity, v=8 m/s, time, t=8 s

Initial momentum, \(p_1\)=m u=5 \(\times\) 4=20 kg- m/s and final momentum, \(p_2=m v=5 \times 8\)

=40 kg – m/s

Now, force =\(\frac{\Delta p}{t}=\frac{p_2-p_1}{t}=\frac{40-20}{8}=\frac{20}{8} N\) = 2.5 N

Question 4. Using the second law of motion, derive the relation between force and acceleration. A bullet of 10 g strikes a sandbag at a speed of \(10^3 \mathrm{~m} / \mathrm{s}\) and gets embedded after traveling 5 cm. Calculate

1.  The resistive force exerted by sand on the bullet.
2.  The time is taken by a bullet to come to rest.

Answer:

Given, m=10 g=0.01kg, u=\(10^3 \mathrm{~m} / \mathrm{s}\), v=0,

s=5 cm=0.05 m, t=?, a=?, F=?

(1) From the third equation of motion,

a=\(\frac{v^2-u^2}{2 s}=\frac{0-\left(10^3\right)^2}{2 \times 0.05}=-10^7 \mathrm{~m} / \mathrm{s}^2\)

The force applied by the bullet,

F=m a=0.01 \(\times\left(-10^7\right)=-10^5 \mathrm{~N}\)

[negative sign shows that force is against the direction of motion]

The resistive force exerted by the sand on the bullet is \(10^5\) N.

(2) Time taken by a bullet to come to rest,

t=\(\frac{v-u}{a}=\frac{0-10^3}{-10^7}=10^{-4} \mathrm{~s}\)

Question 5. Derive the unit of force using the second law of motion. A force of 5 N produces an acceleration of 8 \(m/s^2\) on a mass \(m_1\) and an acceleration of 24 \(\mathrm{~m} / \mathrm{s}^2\) on a mass \(m_2\). What acceleration would the same provide, if both the masses are tied together?
Answer:

⇒ \(F_1=5 \mathrm{~N}, a_1=8 \mathrm{~m} / \mathrm{s}^2, m_1\)= ?

⇒ \(F_2=5 \mathrm{~N}, a_2=24 \mathrm{~m} / \mathrm{s}^2, m_2\)=?

⇒ From , F=m a, 5 = \(m_1 \times 8\)

⇒ \(m_1=\frac{5}{8} \mathrm{~kg}\)

Similarly, \(5=m_2 \times 24\)

⇒ \(m_2=\frac{5}{24} \mathrm{~kg}\)

⇒ \(m_1+m_2=\frac{5}{8}+\frac{5}{24}\)

= \(\frac{15+5}{24}=\frac{20}{24}=\frac{5}{6}\)

Given that acceleration produced by the same force provided, if both the masses are tied together is

a =\(\frac{F}{\left(m_1+m_2\right)}\)

= \(\frac{5}{5 / 6}=6 \mathrm{~m} / \mathrm{s}^2\)

Question 6. If the engine of a car provides an acceleration of 2 m/s² to start it from rest, then assuming the mass to be roughly 1000 kg. Calculate

1.  force provided by the engine.
2.  velocity after 10 s.
3.  time after which the car comes to rest, if the engine is turned off after 15 s. (take, frictional force =15 N)

Answer:

Given, initial velocity, u=0 m/s

Acceleration, a=2 \(m / s^2\),

Mass, m=1000 kg

(1) From Newton’s second law of motion, force = mass \(\times\) acceleration

F=m a=1000 \(\times\) 2=2000 N

(2) Velocity after 10 s,

from the first equation of motion,

v=u+a t

v=0+2 \(\times\) 10=20 m/s

(3) Final velocity of the car when the engine is on after 15 s.

v=0+2 \(\times\) 15=30 m/s [t=15 s]

So, p=m v

=1000 \(\times\) 20=20000 \(\mathrm{~kg}-\mathrm{m} / \mathrm{s}\)

After 15 s, net force = force exerted by engine + friction force

=2000-15=1985 N [friction force=15 N]

[Here, friction force is subtracted because it opposes the motion]

Acceleration =\(\frac{\text { Net force }}{\text { Mass }} \Rightarrow a=\frac{1985}{1000}\)

a=1.985 \(\mathrm{~m} / \mathrm{s}^2 \approx 2 \mathrm{~m} / \mathrm{s}^2\)

Time taken by the car (comes to rest)

v=u-a t [first equation of motion][final velocity, v=0]}

0=30-2 \(\times\) t

2 t=30 [from Eq. (1)]

t=15 s

Force And Laws Of Motion Multiple Choice Questions

Question 1. The resultant force acting on a body is zero, then a

1. The body is in unequilibrium
2. The body is in equilibrium
3. The body moves with constant acceleration
4. Body moves with retardation

Answer: 2. Body is in equilibrium

When the resultant force acting on a body is zero, then the body will be in equilibrium.

Question 2. If a ball thrown up in a moving car comes back to the boy’s hands, then it explains

1. Newton’s first law of motion
2. Newton’s second law of motion
3. Newton’s third law of motion
4. law of conservation of momentum

Answer: 1. The balls come back to the boy’s hand due to the law of inertia (Newton’s first law of motion). During the period, when the ball is in the air, the ball covers the same horizontal distance as the train and so it comes back to the boy’s hand.

Question 3. A thrust of 200 N is applied on the surface of the wall, then the normal reaction on the wall is

1. 200 N
2. 100 N
3. 400 N
4. 300 N

Answer: 1. 200 N

Every action has an equal and opposite reaction, therefore the normal reaction is 200 N.

Question 4. A bus of mass 500 kg is moving with a velocity of 5 m/s and is acted upon by a forward force of 500 N due to the engine and retarding force of 200 N due to friction velocity of the bus after 20 s will be

1. 15 m/s
2. 17 m/s
3. 19 m/s
4. 21 m/s

Answer: 2. 17 m/s

Given, m = 500 kg,

u=5 m/s and t=20 s

Resultant force, F=500-200=300 N

Acceleration, a=\(\frac{F}{m}=\frac{300}{500}=0.6 \mathrm{~m} / \mathrm{s}^2\)

By equation of motion,

v=u+a t=5+0.6 \(\times 20=17 \mathrm{~m} / \mathrm{s}\)

Question 5. 1 dyne is equal to

1. \(10^5 \mathrm{~N}\)
2. \(10^7 \mathrm{~N}\)
3. \(10^{-5} \mathrm{~N}\)
4. \(10^{-7} \mathrm{~N}\)

Answer: 3. \(10^{-5} \mathrm{~N}\)

Dyne is the unit of force in the CGS system.

1 N = \(\frac{1 \mathrm{~kg}-\mathrm{m}}{\mathrm{s}^2}=\frac{1000 \mathrm{~g} \times 100 \mathrm{~cm}}{1 \mathrm{~s}^2}\)

1 dyne =\(\frac{1}{10^5} \mathrm{~N}=10^{-5} \mathrm{~N}\)

Question 6.  1 kg-wt is equal to

1. 9.8 kg
2. 1 kg
3. 9.8 N
4. 98 N

Answer: 3. 9.8 N

1 kg-wt = mg N = 1 x 9.8 N = 9.8 N

Question 7. A truck moving with a speed of 54 km/h. The truck driver applied brakes suddenly and brings the truck to rest in 5 s, then the average retarding force on the truck, if the mass of the truck and driver is 400 kg, will be

1. 1200 N
2. 600 N
3. 800 N
4. 500 N

Answer: 1. 1200 N

Given, u=54 \(\mathrm{~km} / \mathrm{h}=54 \times \frac{5}{18} \mathrm{~m} / \mathrm{s}\)

=15 m/s, v=0 and t=5 s

Retardation, a=\(\frac{v-u}{t}=\frac{0-15}{5}=-3 \mathrm{~m} / \mathrm{s}^2\)

Average retarding force, F=m a=400 \(\times\) 3=1200 N

Question 8. Two forces acting on a body in different directions, then acceleration produced in the body is due to

1. Resultant of both forces
2. The sum of both forces
3. Difference between both forces
4. None of the above

Answer: 1. Resultant of both forces

According to Newton’s second law of motion, the resultant force on a body is responsible for its acceleration.

Question 9. Two forces F₁ and F2 are acting on a body as shown in the figure, then acceleration in the body is

1. 23 \(\mathrm{~m} / \mathrm{s}^2 \)
2. 3 \(\mathrm{~m} / \mathrm{s}^2 \)
3. 2 \(\mathrm{~m} / \mathrm{s}^2 \)
4. 22 \(\mathrm{~m} / \mathrm{s}^2 \)

Answer: 2. 3 \(\mathrm{~m} / \mathrm{s}^2 \)

The resultant force on a body,

F=\(F_2-F_1=65-50=15 \mathrm{~N}\)

By Newton’s second law of motion,

F=m a \(\Rightarrow\) 15=5 a

a=3 \(\mathrm{~m} / \mathrm{s}^2\)

Question 10. The dirty Blanket is beaten with a stick to remove dust particles. Which law holds good for this?

1. Law of conservation of momentum
2. Law of inertia
3. Law of impulse
4. Law of conservation of energy

Answer: 2. Law of inertia

When we beat a blanket with the help of a stick, then it comes into motion. However, the dust particles continue to rest due to inertia and get detached from the blanket.

Question 11. A bullet of 20 g strikes a sandbag at a speed of 200 m/s and gets embedded after traveling 2 cm, then the resistive force exerted by sand on the bullet is

1. 2 \(\times 10^3 \mathrm{~N}\)
2. 2 \(\times 10^4 \mathrm{~N}\)
3. 2 \(\times 10^6 \mathrm{~N}\)
4. 2 \(\times 10^5 \mathrm{~N}\)

Answer: 2. 2 \(\times 10^4 \mathrm{~N}\)

Given, u=200 \(\mathrm{~m} / \mathrm{s}\), v=0, s=0.02 m

and m=20 \(\mathrm{~g}=20 \times 10^{-3} \mathrm{~kg}\)

From the third equation of motion,

⇒ \(v^2 =u^2+2 a s\)

a =\(\frac{v^2-u^2}{2 s}\)

=\(\frac{(0)^2-(200)^2}{2 \times 0.02}\)

=-\(10^6 \mathrm{~m} / \mathrm{s}^2\)

Negative sign indicates retardation.

Resistive force, F =m a=20 \(\times 10^{-3} \times 10^6\)

=2 \(\times 10^4 \mathrm{~N}\)

Question 12. Impulse is equal to

1. Rate of change in momentum
2. Rate of change in force
3. Change in reaction force
4. Change in momentum

Answer: 4. Change in momentum

Impulse, I =F \(\times \Delta t\)

= \(\frac{\Delta p}{\Delta t} \cdot \Delta t=\Delta p\) = Change in momentum.

Question 13. Jetplanes and rockets work on the principle of conservation of

1. Energy
2. Momentum
3. Mass
4. Heat

Solution: 2. Momentum

Initially, the total momentum of the rocket and its fuel is zero, when the fuel is exploded. The burnt gases are allowed to escape through a nozzle with very high downward velocity and with large momentum in a downward direction.

To conserve this momentum, the rocket acquires an equal momentum in the upward direction and hence starts moving upwards.

Question 14. A Bullet Of Mass 30g is fired horizontally with a velocity of 120 m/s from a pistol of mass 2 kg, the recoil velocity of the pistol is

1. 1.6 m/s
2. 2.8 m/s
3. 2.4 m/s
4. 1.8 m/s

Answer: 4. 1.8 m/s

By the law of conservation of momentum,

Initial momentum = Final momentum 0=0.03×120 + 2xv

where v is recoil velocity.

0 = 3.6 + 2v

v=\(\frac{-3.6}{2}=-1.8 \mathrm{~m} / \mathrm{s}\)

Question 15. An object of mass 4 kg moves with a velocity of 4 m/s, then its momentum will be

1. 16 m/s
2. 4 m/s
3. 16 kg-m/s
4. 4 kg-m/s

Solution: 3. 16 kg-m/s

Momentum,p = mv = 4x 4 = 16kg-m/s

Question 16. A force acts on a body of mass 5 kg and changes its velocity from 8 m/s to 12 m/s in 4s, then the magnitude of the force is

1. 8 N
2. 4 N
3. 5 N
4. 6 N

Answer: 3. 5 N

According to Newton’s second law of motion, Force = Rate of change in momentum

F=\(\frac{m(v-u)}{t}=\frac{5(12-8)}{4}=5 \mathrm{~N}\)

Question 17. When a person jumps down from a tower into a stretched tarpaulin, then receives

1. Greater injury
2. Less injury
3. No injury
4. None of the above

Answer:

When a person jumps, the tarpaulin gets depressed at the place of impact, therefore impact time interval increases, As a result, a person experiences a very small force, and hence he receives no injury.

Question 18. A cricket ball of mass 0.25 kg moving with a velocity of 10 m/s is brought to rest by a player in 0.1 seconds, and then the impulse exerted by the player is

1. 2.5 N-s
2. 1.5 N-s
3. 25 N-s
4. 30 N-s

Answer: 1. 2.5 N-s

Impulse = Change in momentum -m{v -u)

= 0.23 (0-10) = -2.3 N-s

A negative sign indicates the direction of the impulse is opposite to the motion of the ball.

Class 9 Science Notes For Chapter 8 Force And Laws Of Motion

In our everyday life, we observe that some efforts are required to put a stationary object into motion or to stop a moving object. We ordinarily experience this as a muscular effort and say that we must push, hit, or pull on an object to change its state of motion. The concept of force is based on this push, hit, or pull.

Force

Any action which causes pull, hit, or push on a body is called force. Force cannot be seen but it can be judged only by the effects which it produces in various bodies around us. Many effects of force are given below:

1. A force can move a stationary body.
2. A force can stop a moving body.
3. A force can change the direction and speed of a moving body.
4. A force can change the shape and size of a body.
5. Balanced and Unbalanced Forces
6. Forces are of two types such as balanced forces and unbalanced forces which are as given below:

Balanced Forces

When the net effect produced by several forces acting on a body is zero, then the forces are said to be balanced forces. Balanced forces can only bring a change in the shape of the body. A block of wood is placed on a horizontal surface and two strings A and B are connected to it. The block is in a state of rest.

Read and Learn  More Class 9 Science Notes

If we pull A and B strings with equal magnitude of forces, then the block does not change its state of rest. Such types of forces are known as balanced forces.

Unbalanced Forces

When the net effect produced by several forces acting on a body is non-zero, then the forces are said to be unbalanced forces. A boy wants to relocate the refrigerator in his house as shown

He pushes the refrigerator with a small force, the refrigerator does not move due to frictional force acting in a direction opposite to the push. If he pushes the refrigerator harder, then the pushing force becomes more than the friction, and due to this, the refrigerator starts moving in the direction of push as shown in the figure.

 

In the above example, there is an unbalanced force that causes motion in the refrigerator. The unbalanced forces cause a change in the state of rest or of uniform motion of a body.

For Example, In a tug of war when one of the teams suddenly releases the rope, then an unbalanced force acts on the other team due to which it falls backward.

If an unbalanced force is applied to a moving object, there will be a change either in its speed or in the direction of its motion. Thus, to accelerate the motion of an object, an unbalanced force is required

Newton’s Laws Of Motion

Newton studied the ideas of Galileo regarding the motion of an object. He formulated three fundamental laws that govern the motion of objects. These three laws are known as Newton’s laws of motion, which are as given below:

Newton’s First Law of Motion

It states that an object will continue to remain in its state of rest or of uniform motion along a straight line path unless an external force acts on it. This means all objects resist change in their state.

The state of any object can be changed by applying external forces only.

(1) A person standing on a bus falls backward when the bus starts moving suddenly. This happens because the person and bus both are at rest while the person is not moving as the bus starts moving. The legs of the person start moving along with the bus, but the rest portion of his body tends to remain at rest. Because of this, a person falls backward if he is not alert.

(2) A person standing in a moving bus falls forward if the driver applies brakes suddenly.

Inertia

The unwillingness (or inability) of an object to change its state of rest or uniform motion along a straight line is called the inertia of the object.

It is the inherent property of all the objects. Newton’s first law of motion is also known as Galileo’s law of inertia.

The inertia of an object is measured by its mass. Inertia is directly proportional to the mass. It means that inertia increases with an increase in mass and decreases with a decrease in mass. A heavy object will have more inertia than a lighter one.

Types of Inertia

Inertia is divided into three types as given below:

1. Inertia of rest The tendency of a body to oppose any change in its state of rest is known as inertia of rest. For Example.

• When a bus suddenly starts moving forward, then the passengers in the bus fall backward.
• The carpet is beaten with a stick to remove the dust particles.
•  When a tree is vigorously shaken, then some of the leaves fall from the tree.

2. Inertia of motion The tendency of a body to oppose any change in its state of uniform motion is known as inertia of motion, For Example.

• The passengers fall forward when a fast-moving bus stops suddenly.
• A person falls forward while getting down from a moving bus or train.
•  Luggage is usually tied with a rope on the roof of a bus.

3. Inertia of direction The tendency of a body to oppose any change in its direction of motion is known as inertia of direction, For Example.

• When a fast-moving bus negotiates a curve on the road, then passengers fall away from the center of the curved road.
• The sparks produced during the sharpening of a knife against a grinding wheel leave tangentially to its rim.
• A stone tied to a string is whirling in a horizontal circle. If the string breaks, then the stone flies away tangentially.

Momentum

Momentum measures the quantity of motion possessed by a body. It is defined as the product of the mass and velocity of the body. Besides magnitude, momentum also has a direction.

At any instant, its direction is the same as the direction of the velocity. If a body of mass m moves with a velocity v, then momentum p is given by p=mv

The SI unit of momentum is kg-m/s.

Example 1. A car of mass 1000 kg is moving with a velocity of 72 km/h. Find its momentum.
Answer:

Given, mass, m = 1000 kg

Velocity, v=72 km/h =72 \(\times \frac{5}{18}\)=20 m /s and Momentum,  p= ?

Momentum = Mass x Velocity => p – mv p =1000 x 20 = 20000 kg-m/s Thus, the momentum of the car is 20000 kg-m/s.

Newton’s Second Law of Motion

The second law of motion states that the rate of change of momentum of an object is directly proportional to the applied external force and takes place in the direction in which external force acts.

Mathematical Formulation of the Second Law of Motion

If a body of mass m moving at initial velocity u accelerates uniformly with an acceleration a for time t, so that its final velocity changes to v, then

Initial momentum,  \(p_1\)= mu

Final momentum, \(p_2\)= mv

Change in momentum = \(p_2-p_1\) = mv – mu  = m(v-u)

According to the second law of motion,

Force, F \(\propto \frac{\text { change in momentum }}{\text { time }}\)

⇒ \(F \propto \frac{p_2-p_1}{t} \Rightarrow F \propto \frac{m(v-u)}{t}\)

⇒ \(F \propto m a [\frac{v-u}{t}=a]\)

F = k ma

The quantity k is a constant of proportionality.

One unit of force is defined as the amount that produces an acceleration of 1 \(\mathrm{~m} / \mathrm{s}^2\) on an object of 1 kg mass.

i.e. 1 unit of force =k \(\times 1 \mathrm{~kg} \times 1 \mathrm{~m} / \mathrm{s}^2 \Rightarrow k=1\)

Thus, the force can be written as F =ma

The SI unit of force is Newton, which is denoted by the symbol N and it is equivalent to kg-m/s2.

When the applied force F is zero, then the acceleration a is also zero and the body remains in its state of rest or of a uniform motion.

Applications of Newton’s Second Law of Motion

The following applications are based on Newton’s second law of motion:

1. A cricket player (or fielder) moves his hands backward while catching a fast cricket ball.
2. During athletics meets, athletes doing the high jump and long jump land on foam or a heap of sand to decrease the force on the body, and the landing is comfortable.

Newton’s First Law from Mathematical Expression of Second Law

The first law of motion can be mathematically stated from the mathematical expression of the second law of motion.

As we know, F = ma

F=\(\frac{m(v-u)}{t} [a=\frac{(v-u)}{t}]\)

Ft = mv – mu

From this equation, if F = 0, then v = u for any value of time. This means that, in the absence of an external force, the object will continue moving with uniform velocity u throughout the time t and if u is zero, then v will also be zero, i.e. the object will remain at rest.

Example 2. The force acts on an object of mass 4 kg and changes its velocity from 10 m/s to 20 m/s in 5 s. Find the magnitude of force.
Answer:

Given, the mass of an object, m = 4 kg

Initial velocity, u = 10 m/s; Final velocity, v = 20 m/s Time taken, t = 5 s

We know that, Newton’s first law of motion,

v = u + at or a=\(\frac{v-u}{t}\)

a=\(\frac{20-10}{5}=\frac{10}{5}=2 \mathrm{~m} / \mathrm{s}^2\)

Magnitude of force, F = Mass x Acceleration

F =ma = 4kgx2 m/s² = 8 N Hence, the magnitude of the force is 8 N.

Example 3. A force of 50 N acts on a stationary body of mass 10 kg for 2 s. Find the acceleration produced in the body and the velocity attained by it.
Answer:

Initial velocity, u- 0 [since the body is stationary]

Force, F – 50 N, mass, m —10 kg, time, t = 2 s

Final velocity, v = ?, acceleration, a = ? v -u

Acceleration, a=\(\frac{v-u}{t}\)

a=\(\frac{10-0}{2}=5 \mathrm{~m} / \mathrm{s}^2\)

Final velocity,v = \(\frac{F t}{m}+u {[F t=m(v-u)]}\)

v=\(\frac{50 \times 2}{10}+0=10 \mathrm{~m} / \mathrm{s}\)

Thus, the acceleration produced in the body is 5 \(\mathrm{~m} / \mathrm{s}^2\), and the velocity attained by it is 10 m/s.

Example 4 A bullet train is moving with a velocity of 180 km/h and it takes 5 s to stop after the brakes are applied. Find the force exerted by the brakes on the wheel of the train if its mass with the wagon is 2000 kg.
Answer:

Given, initial velocity u=180 km/h =180 \(\times \frac{5}{18}\)=50 m/s

Final velocity v=0,

time t=5 s , mass m = 2000 kg

From the first equation of motion,

v=u+a t

u=-a t

a=-\(\frac{u}{t}=\frac{-50}{5}=-10 \mathrm{~m} / \mathrm{s}\)

Now, the force exerted by the brakes on the wheel is given by Newton’s second law F – ma – 2000 x -10 = -20000 N Negative sign shows that the direction of force is opposite to the motion of the body.

Example 5. A force of 6N gives a mass mv an acceleration of 18 m/s² and a mass m², an acceleration of 24 m/s². What acceleration would it give if both the masses were tied together?
Answer:

Given, force F =6N

Acceleration of mass mx, ax =18 m/s² Acceleration of mass w², a² = 24 m/s²

Acceleration is produced when both masses are tied together a -?

From Newton’s second law of motion Reaction Ground ;

⇒ \(m_1=\frac{F}{a_1}=\frac{6}{18}=\frac{1}{3} \mathrm{~kg}\)

⇒ \(m_2=\frac{F}{a_2}=\frac{6}{24}=\frac{1}{4} \mathrm{~kg}\)

Total mass m=\(m_1+m_2=\frac{1}{3}+\frac{1}{4}=\frac{4+3}{12}=\frac{7}{12} \mathrm{~kg}\)

Acceleration produced in combined mass a=\(\frac{F}{m}\)

a=\(\frac{6}{7 / 12}=\frac{12 \times 6}{7}=10.28 \mathrm{~m} / \mathrm{s}^2\)

Example 6. The velocity-time graph of a ball of mass 30 g moving along a straight line on a long table is given figure. How much force does the table exert on the ball to bring it to rest?

Answer:

The initial velocity of the ball =30 cm/s

Final velocity v=0 and Time t=6 s

Acceleration a=\(\frac{v-u}{t}=\frac{(0-30)}{6}=-5 \mathrm{~cm} / \mathrm{s}^2=-0.05 \mathrm{~m} / \mathrm{s}^2\)

The force exerted on the ball, F = ma

F = ma =\(\left(\frac{30}{1000}\right) \times(-0.05)=-1.5 \times 10^{-3} \mathrm{~N}\)

The negative sign shows that the frictional force exerted by the table is opposite to the direction of motion of the ball.

Impulse

It is termed as the total impact of force. This is equal to the change in momentum of the body. In other words, impulse is defined as the product of force and a small time in which force acts.

According to Newton’s second law, F =ma

F=\(\frac{m(v-u)}{t}\)

F=\(\frac{m v-m u}{t}\)

Ft = mv – mu

Impulse, I=F t=\(p_2-p_1\)

or Impulse = Change in momentum

The SI unit of impulse is N-s or kg – m/s

Example 7. If a force of 1000 N is applied over a vehicle of 500 kg, then for how much time the velocity of the vehicle will increase from 2 m/s to 10 m/s? Also, find the impulse.
Answer:

Given, F = 1000 N, mass m = 500 kg

Final velocity =10 m/s

Initial velocity u = 2 m/s

F=\(\frac{m(v-u)}{t}\)

t=\(\frac{m(v-u)}{F}=\frac{500 \times(10-2)}{1000}=4 \mathrm{~s}\)

Impulse I = Ft =1000 x 4 = 4000 N-s

Thus, the time required by the vehicle is 4s, and its impulse is 4000 N-s.

Newton’s Third Law of Motion

The third law of motion states that, whenever one object exerts a force on another object, then the second object exerts an equal and opposite force on the first object.

Spring balance action and reaction forces are equal and opposite

Thus, action and reaction forces are equal in magnitude and opposite in direction. They still do not cancel each other’s effect because they act on different objects.

Applications of Newton’s Third Law of Motion

1. Collision of two persons If two persons walking or running in opposite directions collide with each other, then both feel hurt because they apply force to each other. Two opposing forces are in action and reaction pairs.
2.  Walking of a person A person is able to walk because of Newton’s third law of motion. During walking, a person pushes the ground in a backward direction, and in the reaction, the ground also pushes the person with an equal magnitude of force but in the opposite direction. This enables him to move in the forward direction against the push.
3. The recoil of a gun When a bullet is fired from a gun, then the bullet also pushes the gun in the opposite direction with an equal magnitude of force. Since the gun has a greater mass than a bullet, the acceleration of the gun is much less than the acceleration of the bullet.
4. Propulsion of a boat in a forward direction Sailor pushes water with oar in a backward direction resulting in water pushing the oar in the forward direction. Consequently, the boat is pushed in the forward direction. The force applied by oar and water are of equal magnitude but in opposite directions.
5. Rocket propulsion The propulsion of a rocket is based on the principle of action and reaction. The rapid burning of fuel produces hot gases which rush out from the nozzle at the rear end at a very high speed. The equal and opposite reaction force moves the rocket upward at a great speed.

Activity 1

Objective: To demonstrate the property of inertia of rest using carom board and coin.

Materials Required:

1. Carom board
2.  Carom coins

Procedure

1. Make a pile of similar carom coins on a carom board as shown.
2. Attempt a sharp horizontal hit at the bottom of the pile using another carom coin or the striker.
3. If the hit is strong enough, the bottom coin moves out quickly.
4. Once the lowest coin is removed, the inertia of the other coins makes them fall vertically on the table.

 

Discussion/Conclusion

1. If the hit is strong enough, the bottom coin moves quickly on the carom table without disturbing the upper carom coins. Once the lowest coin is removed, the upper carom coins fall vertically on the table due to the inertia of rest.
2. If the hit is weak, the disturbance may be communicated to upper carom coins, which may fall randomly with or without the actual movement of the coin.

Question 1. What does the Inertia of an object tend to cause?
Answer:

The inertia of an object tends to resist any change in its state of rest or motion.

Question 2. What happens if the hit is weak?
Answer:

If the hit is weak, the upper carom coins may fall randomly with or without actual movement of the coin.

Question 3. On what factor does the inertia of a body depend?
Answer:

The inertia of the body depends on the mass of the body.

Activity 2

Objective: To demonstrate the property of inertia of rest using glass and coins.

Materials Required:

1. Glass tumbler
2. Stiff card
3. Five-rupees coin
4. Table

Procedure

1. Set a five-rupee coin on a stiff card covering an empty glass tumbler, standing on a table.
2. Give a sharp horizontal flick to the card with a finger. If we do it fast, then the card shoots away allowing the coin to fall vertically into the glass tumbler due to its inertia.
3. The inertia of the coin tries to maintain its state of rest even when the card flows off.

Discussion/Conclusion

When the card is given a sharp horizontal flick with a finger, the card underneath the five-rupee coin shoots away. The coin falls vertically into the glass tumbler due to the inertia of the rest.

The basic thing is that force is applied to the card. This is why the coin tends to remain at rest.

Question 1. Which law is associated with this activity?
Answer:

Newton’s first law of motion is associated with it.

Question 2. Name the concept involved in this activity.
Answer:

The concept of inertia is involved in this activity.

Question 3. There are two bodies of mass m₂ >m₂) present on a rough surface, which of them has more inertia and why?
Answer:

Since, m₁ > m₂, so mass m₁ has more inertia than m₂.

Activity 3

Objective: To demonstrate the forces of action and reaction are equal and opposite by throwing the bag full of sand.

Materials Required

1. Two students
2. Two carts
3. A bag full of sand
4. White paint

Procedure

1. Request two children to stand on two separate carts
2. Give them a bag full of sand. Ask them to play a game of catching the bag.
3. Does each of them receive an instantaneous reaction as a result of throwing the sandbag (action)?
4. You can paint a white line on cartwheels to observe the motion of the two carts when the children throw the bag towards each other.

Discussion/Conclusion

Yes, in this case, each of them receives an instantaneous reaction as a result of throwing the sandbag. When the boy throws the bag towards the girl, the cart of the boy moves back. Similarly, when the girl throws the bag towards the boy, the cart of the girl moves back.

Thus, it explains Newton’s third law of motion, i.e. the force is exerted forward in throwing the bag, and the person who is throwing it gets pushed backward. Action and reaction are taking place simultaneously on two different bodies.

Question 1. What happens to the cart when the two children are playing the game of catching the bag?
Answer:

The cart of each child moves outwards when both the children are playing the game of catching the bag.

Question 2. Name the type of reaction involved in throwing the bag.
Answer:

An instantaneous type of reaction is involved in throwing the bag.

Question 3. Are the forces of action and reaction equal?
Answer:

Yes, the forces of action and reaction are equal in magnitude but opposite in direction.

Force And Laws Of Motion Question And Answers

Question 1. Which of the following has more inertia?

1. A rubber ball and a stone of the same size
2. A bicycle and a train
3. A five-rupees coin and a one-rupee coin
4. Give reasons for your answer.

Answer:

1. The inertia of an object is proportional to its mass.
2. A stone of the same size as a of rubber ball will have greater mass, so the stone will have more inertia.
3. A train has much greater mass than of a bicycle, so the train will have more inertia.
4. A five-rupee coin has more mass than a one-rupee coin, so a five-rupee coin will have greater inertia.

Question 2. In the following example, try to identify the number of times the velocity of the ball changes.

A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the ball and kicks it towards a player of his team. Also, identify the agent supplying the force in each case.

Answer:

There are several times at which the direction and magnitude of velocity of the ball changes, which are as given below:

Whenever a force is applied to the ball, then the velocity of the ball will change.

1. When the first player kicks the ball toward another player of his team, then the velocity of the ball will change because the first player applies some force on the ball.
2. When another player kicks the ball toward the goal, then the velocity of the ball will change, here again, the force is applied to the ball.
3. When the goalkeeper of the opposite team collects the ball, then the velocity of the ball will be changed, it becomes zero. Here, the goalkeeper applies some force on the ball to stop.
4. When the goalkeeper kicks the ball towards his own team, then the velocity of the ball changes because the goalkeeper applies some force on the ball.

Question 3. Explain, why some of the leaves may get detached from a tree, if we vigorously shake its branch.

Answer:

Leaves have the inertia of rest. When the branch is shaken, it tends to remain in the same state and gets detached when the position of the branch changes.

Question 4. Why do you fall in the forward direction when a moving bus breaks to a stop and fall backward when it accelerates from rest?

Answer:

When the moving bus brakes to a stop, then the passengers who had inertia of motion, oppose a change in their state. However, the lower portion of their body in contact with the bus comes to rest.

So, they fall forward. When the bus accelerates from rest, then the passengers who had inertia of rest, oppose a change in their state. However, the lower portion of their body starts moving with the bus. Hence, they fall backward.

Question 5. If action is always equal to the reaction, then explain how a horse can pull a cart.

Answer:

The horse pushes the ground in a backward direction. The ground exerts a reaction force on the horse and cart system to push them forward.

When the reaction force exceeds the force of friction between the wheels of the cart and the ground, then the cart is pushed forward.

Question 6. Explain, why is it difficult for a fireman to hold a hose pipe, which ejects a large amount of water with a high velocity.

Answer:

A fireman finds it difficult to hold a hose-pipe which is ejecting a large amount of water at high velocity. Because the stream of water rushing out of the pipe in the forward direction exerts a large force on the pipe.

Due to the reaction of the forward force, a force is applied to the pipe in the backward direction. Therefore, the fireman struggles to keep the hose pipe at rest.

Exercises

Question 1. An object experiences a net zero external unbalanced force. Is it possible for the object to be traveling with a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason.

Answer:

As per Newton’s first law of motion, no force is needed to move an object that is already moving with a constant (non-zero) velocity. So, when an object experiences a net zero external unbalanced force, then it can move with a non-zero velocity. When an external force is zero then the velocity of the object remains the same both in magnitude and direction.

Question 2. When a carpet is beaten with a stick, dust comes out of it. Explain.

Answer:

When a carpet is beaten with a stick, then the fibers of the carpet attain the state of motion while the dust particles remain at rest due to the inertia of rest, and hence dust particles get detached.

Question 3. Why is it advised to tie any luggage kept on the roof of a bus with a rope?

Answer:

• When the bus stops suddenly, the bus comes to a state of rest but the luggage remains in a state of motion. So, due to inertia of motion, the luggage moves forward and may fall down from the roof of the bus.
• If the bus starts suddenly, then the bus comes in a state of motion but the luggage remains in the state of rest. Due to the inertia of rest, the luggage does not move in the forward direction and may fall down.

Question 4. A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest. Why does the ball slow down to stop?

1. The batsman did not hit the ball hard enough
2. Velocity is proportional to the force exerted on the ball
3. There is a force on the ball opposing the motion
4. There is no unbalanced force on the ball, so the ball would want to come to rest.

Answer:

The ball slows down to stop because the force of friction between the ground and the ball acts as an external force that opposes the motion of the ball.

Question 5. A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 seconds. Find its acceleration. Find the force acting on it, if its mass is 7 tonne.

(Hint: 1 tonne = 1000 kg)

Answer:

The truck starts from rest, so initial velocity, u = 0,

Distance, s = 400 m , Time, t = 20 s Mass, m – 1 tonne = 7 x 1000 = 7000 kg

From Newton’s second law of motion, s=u t+\(\frac{1}{2} a t^2\)

400 =0 \(\times 20+\frac{1}{2} \times a \times(20)^2\)

400 =200 a

a =2 \(\mathrm{~m} / \mathrm{s}^2\)

From Newton’s second law of motion, the force acting on the truck,

F=m a=7000 \(\times 2 =14000 \mathrm{~N}\)

=1.4 \(\times 10^4 \mathrm{~N}\)

Thus, the acceleration of the truck is 2 m \(s^2\), and the force acting on it is 1.4 \times 10^4 N.

Question 6. A stone of size 1 kg is thrown with a velocity of 20 m/s across the frozen surface of a lake and comes to rest after traveling a distance of 50 m. What is the force of friction between the stone and the ice?

Answer:

Given, the mass of the stone, m-1 kg,

Initial velocity, u = 20 m/s

Final velocity, v = 0 [since stone comes to rest]

Distance covered, s = 50 m

From Newton’s third law of motion, v² = u² + 2as

(0)² =(20)² + la (50)

⇒ 100 a =-400

⇒ a = – 4 m/s²

Here, a negative sign shows that there is a retardation in the motion of the stone.

Force of friction between stone and ice = Force required to stop the stone

= ma – 1 x (-4) = – 4 N

Question 7. An 8000 kg engine pulls a train of 5 wagons, each of 2000 kg along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate

1. The net accelerating force.
2. The acceleration of the train and
3. The force of wagon 1 on wagon 2.

Answer:

(1) Net accelerating force = Force exerted by the engine -Friction force

[Here, frictional force is subtracted because it opposes the motion]

= 40000 -5000 =35000 = 3.5 x 104 N

(2) From Newton’s second law of motion,

Accelerating force = mass of the train x acceleration of train F

a =\(\frac{F}{m}\)

Mass of train =5 x Mass of one wagon

=5 x 2000=10000 kg

Acceleration =\(\frac{35000}{10000}=3.5 \mathrm{~m} / \mathrm{s}^2\)

(3) Force of wagon 1 on wagon 2

=(35000-3.5×2000) N = 28000 N

Question 8. An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road, if the vehicle is to be stopped with a negative acceleration of 1.7 m/s²?

Answer:

Given, mass, m~ 1500 kg, acceleration, = – 1.7 m/s²

From Newton’s second law of motion,

F = ma – 1500 x (-1.7) = -2550N

Question 9. What is the momentum of an object of mass m moving with a velocity v?

1. \(m^2 v^2\)
2. \(m v^2\)
3. \(\frac{1}{2} m v^2\)
4. mv

Answer: 4. mv

The momentum of an object of mass m moving with a velocity v is given by p = mv.

Question 10. Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?

Answer:

The cabinet will move across the floor with constant velocity if there is no net external force applied to it. Here, a horizontal force of 200 N is applied to the cabinet, so for the net force to be zero, an external force of 200 N should be applied to the cabinet in the opposite direction.

Thus, the frictional force = 200 N [frictional force always acts in the direction opposite to the direction of motion]

Question 11. According to the third law of motion when we push an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.

Answer:

1. The logic given by the student is not correct because two equal and opposite forces cancel each other only in the case that they act on the same body. Action and reaction force always act on two different bodies, so they cannot cancel each other.
2. When a massive truck is pushed, then the truck may not move because the force applied is not sufficient to overcome the force produced by friction acting opposite to the applied force to move the truck.

Question 12. A bullet of mass 10 g traveling horizontally with a velocity of 150 m/s strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also, calculate the magnitude of the force exerted by the wooden block on the bullet.

Answer:

Given, the mass of the bullet,

m =10 g

= \(\frac{10}{1000} \mathrm{~kg}=0.01 \mathrm{~kg}\)

Initial velocity, u= 150m/s

Final velocity, v = 0 [since bullet comes to rest]

Time, t = 0.03 s

From Newton’s first law of motion, v = u + at 0 =150+ a x 0.03

a=\(\frac{-150}{0.03}=-5000 \mathrm{~m} / \mathrm{s}^2\)

Distance covered by the bullet before coming to rest is given by \(v^2=u^2+2 a s\)

0 =\((150)^2+2(-5000) \mathrm{s}\)

s \(\frac{(150)^2}{10000}=2.25 \mathrm{~m}\)

The magnitude of the force applied by the bullet on the block,

F = ma= 0.01 x -5000 = -50 N

Question 13. An object of mass 100 kg is accelerated uniformly from a velocity of 5 m/s to 8 m/s in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.

Answer:

Given, Mass of the object, m = 100 kg Initial velocity, u = 5 m/s Final velocity, v = 8 m/s,

Time, t = 6 s

(1) Initial momentum = mu

= 100 x 5 = 500 kg – m / s and final momentum = mv

= 100 x 8 = 800 kg – m / s

(2) From Newton’s second law, the force exerted on the object = rate of change of momentum = Change in momentum

=\(\frac{\text { Change in momentum }}{\text { Time }}\)

=\(\frac{\text { Final momentum }- \text { Initial momentum }}{\text { Time }}\)

=\(\frac{800-500}{6}=\frac{300}{6}=50 \mathrm{~N}\)

Question 14. Akhtar, Kiran, and Rahul were riding on a motorcar that was moving with a high velocity on an expressway when an insect hit the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar.

Akhtar said that, since the motorcar was moving with a larger velocity, it exerted a larger force on the insect and as a result insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and change in their momentum. Comment on these suggestions.

Answer:

1. According to the law of conservation of momentum, the momentum of the car and insect system before collision = momentum of the car and insect system after collision
2. Hence, the change in momentum of the car and insect system is zero. The insect gets stuck on the windscreen. This means that the direction of the insect is reversed. As a result, the velocity of the insect changes by a great amount.
3. On the other hand, the car continues moving with a constant velocity. Kiran’s suggestion is wrong.
4. Akhtar made a correct conclusion because the mass of the car is very large compared to the mass of the insect. Rahul gave a correct explanation.

Question 15.  How much momentum will a dumbbell of mass 10 kg transfer to the floor, if it falls from a height of 80 cm? Take, its downward acceleration to be 10 m/s².

Answer:

Given the mass of the dumb-bell, m = 10 kg

Initial velocity, u = 0 [because it falls from rest]

Distance covered, s = 80 cm = 0.8 m

Acceleration, a = 10 m/s²

From Newton’s third law of motion,

v² =u² + 2 as

v²= 0 + 2 x 10 x  0.8 =16 ⇒ = -v 16 = 4 m/s

The momentum of the dumbbell just before it touches the floor is given by p = mv = 10 x 4 = 40 kg-m/s When the dumbbell touches the floor, then its velocity becomes zero and hence the momentum. Thus, the total momentum of the dumbbell is transferred to the floor.

So, the momentum transferred to the floor is 40 kg- m/s

Question 16. The following is the distance-time table of an object in motion:

1. What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing or zero?
2. What do you infer about the force acting on the object?

Answer:

(1) Here, initial velocity, u=0

Using Newton’s second law of motion,

s=u t+\(\frac{1}{2} a t^2=\frac{1}{2} a t^2 [ u=0]\)

We get, a=\(\frac{2 s}{t^2}\)

Thus, acceleration is increasing.

(2) Since, acceleration is increasing, so net unbalanced force is acting on the object.

Question 17.  Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2 m/s². With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort.)

Answer: 

Given, the mass of the motorcar, m =1200 kg

Acceleration produced, a = 0.2 m/s²

Force applied on the car by three persons,

F = ma = 1200 x 0.2 = 240 N

Force applied on the car by one person = \(\frac{240}{3}\) = 80 N

Each person pushes the motorcar with a force of 80 N.

Question 18. A hammer of mass 500 g moving at 50 m/s strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer?

Answer:

Given, Mass of the hammer = 300 g= 0.5 kg

The initial velocity of the hammer, u = 50 m/s

The final velocity of the hammer, v = 0 [because the hammer stops]

Time, t = 0.01 s

According to Newton’s second law of motion, the force of the nail on the hammer = rate of change of momentum of the hammer

= \(\frac{m v-m u}{t}=\frac{0.5 \times 0-0.5 \times 50}{0.01}\)

= \(-\frac{25}{0.01}=-2500 \mathrm{~N}\)

The force of the nail on the hammer is equal and opposite to that of the hammer on the nail.

Question 19. A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. its velocity is slowed down to 18 km/h in 4 s by an unbalanced: external force. Calculate the acceleration and change in momentum. Also, calculate the magnitude of the force required

Answer:

Given, mass, m =1200 kg

Initial velocity, u=90 \(\mathrm{~km} / \mathrm{h}=90 \times \frac{5}{18}=25 \mathrm{~m} / \mathrm{s}\)

Final velocity, v=18 \(\mathrm{~km} / \mathrm{h}=18 \times \frac{5}{18}=5 \mathrm{~m} / \mathrm{s}\)

Time, t=4 s

(1) Acceleration, a=\(\frac{v-u}{t}=\frac{5-25}{4}\)

=-\(\frac{20}{4}=-5 \mathrm{~m} / \mathrm{s}^2\) [here, the negative sign indicates that the velocity decreases]

(2) Change in momentum

= Final momentum – Initial momentum

= m v-m u=m(v-u)

= 1200(5-25)

= 1200 \(\times(-20)=-24000 \mathrm{~kg}-\mathrm{m} / \mathrm{s}\)

(3)  Magnitude of the force required

= Rate of change of momentum

=\(\frac{\text { Change in momentum }}{\text { Time }}\)

=\(\frac{-24000}{4}=-6000 \mathrm{~N}\)

Summary

Any action which causes pull, hit, or push on a body is called force. Types of forces

1. Balanced forces When the net effect produced by many forces acting on a body is zero, then the forces are said to be balanced forces.
2. Unbalanced forces When the net effect produced by many forces acting on a body is non-zero, then the forces are said to be unbalanced forces.
3. Newton’s first law of motion States that an object will continue to remain in its state of rest or in a uniform motion along a straight line or path unless an external force acts on it.
4. The unwillingness (or inability) of an object to change its state of rest or of uniform motion along a straight line is called the inertia of the object.

Types of inertia

1. The inertia of Rest The tendency of a body to oppose any change in its state of rest is known as the inertia of rest.
2. The inertia of Motion The tendency of a body to oppose any change in its state of uniform motion is known as the inertia of motion.
3.  The inertia of Direction The tendency of a body to oppose any change in its direction of motion is known as the inertia of direction

Momentum Measures the quantity of motion possessed by a body. It is defined as the product of the mass and velocity of the body.

Momentum, p = mv

The SI unit of momentum is kg-m/s.

Newton’s second law of motion States that the rate of change of momentum of an object is proportional to the applied external force and takes place in the direction in which external force acts.

Force applied on an object is equal to the product of the mass of the object and acceleration produced in it. Force, F = ma

The SI unit of force is Newton (N).

Impulse is defined as the product of force and the small time in which the force acts.

Impulse, I = Ft – p₂ – p₁ or Impulse = Change in momentum The SI unit of impulse is N-s or kg-m/s.

Newton’s third law of motion States that whenever one object exerts a force on another object, then the second object exerts an equal and opposite force on the first object.

Gravitation Multiple-Choice Questions

Question 1. Newton’s law of gravitation is universal because it

1. acts on all bodies and particles in the universe
2. it acts on all the masses at all distances and is not affected by the medium
3. is always attractive
4. None of the above

Answer: 2. it acts on all the masses at all distances and is not affected by the medium

According to Newton’s law of gravitation, the force between two masses in the universe, F=\(\frac{G m_1 m_2}{r^2}\)

Which does not depend on medium and G is the same in the whole universe.

Question 2. The weakest force in the following is

1. magnetic force
2. nuclear force
3. gravitational force
4. electric force

Answer: 3. gravitational force

Gravitational force is the weakest force among given forces.

Question 3. The atmosphere is held to the earth by

1. earth’s magnetic field
2. earth’s rotation
3. gravity
4. earth’s electric field

Answer: 3. gravity

The atmosphere is held to the earth by gravity.

Question 4. Between the two gravitational constants G and g, which is called the universal gravitational constant?

1. g
2. G
3. Both (1) and (2)
4. None of these

Answer: 2. G

G is called the universal gravitational constant because its value is the same in the whole universe.

G=6.67 \(\times 10^{-11} \mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2\)

Question 5. The value of gravitational acceleration (g) is

1. highest at poles
2. highest at equator
3. lowest at poles
4. lowest at equator

Answer: 1. highest at poles

Gravitational acceleration g is given by g = \(\frac{G m_e}{R_e^2}\)

Since the radius of the earth at the equator is greater than the poles, the value of g at the poles is greater than the equator

Question 6. The law of gravitation gives the gravitational force between

1. The Earth and other planets only
2. the earth and sun only
3. two charged bodies only
4. any two bodies having some mass

Answer: 4. any two bodies having some mass

The law of gravitation is applicable to all bodies having some mass.

Question 7. The gravitational force between two objects is 10 N. If the masses of both objects are doubled without changing the distance between them, then the gravitational force would become

1. 2.5 N
2. 20 N
3. 40 N
4. 10 N

Answer: 3. 40 N

Gravitational force, \(F_1=\frac{G m_1 m_2}{r^2}\)

When masses are doubled, then

⇒ \(F_2=\frac{G 2 m_1 2 m_2}{r^2}=4 \frac{G m_1 m_2}{r^2}\)

⇒ \(F_2=4 F_1=4 \times 10=40 \mathrm{~N}\)

Question 8. The magnitude of the weight of a body at the centre of the earth is

1. zero
2. equal to the mass of the body
3. greater than g
4. less than g

Answer: 1. zero

The value of gravitational acceleration (g) at the centre is zero, hence weight, w =mg = mx 0 = 0.

Question 9. When a ball is fully immersed in a liquid, its weight decreases. It happens due to

1. gravitational force
2. magnetic force
3. buoyant force
4. friction force

Answer: 3. buoyant force

When a ball is fully immersed in a liquid, its weight decreases due to buoyant force.

Question 10. Why the dam of the water reservoir is thick at the bottom?

1. The quantity of water increases with the depth
2. The density of water increases with the depth
3. The pressure of water increases with the depth
4. Temperature of water increases with depth

Answer: 3. Pressure of water increases with depth

The dam of water is made thick at the bottom, because due to maximum depth, the pressure of water is maximum at the bottom and to bear this maximum pressure dam should be thick at the bottom.

Question 11. Three spheres have radii of 1 cm, 2 cm and 3 cm, respectively. Which sphere exerts maximum pressure on earth?

1. First
2. Second
3. Third
4. All Equal

Answer: 1. First

We know that, pressure =\(\frac{\text { force }}{\text { area }}=\frac{F}{A}\)

or p=\(\frac{F}{A} \Rightarrow \) p \(\propto \frac{1}{A}\)

p \(\propto \frac{1}{r^2}\)

Therefore, the sphere which has a lesser area will exert maximum pressure. Since the first sphere has a small radius, its area will be small. Hence, it will exert maximum pressure.

Question 12. A petrol cane of mass 5 kg has a volume of 2880. What is the density of the cane?

1. 17.3 \(\mathrm{~g} / \mathrm{cm}^3\)
2. 16.5 \(\mathrm{~g} / \mathrm{cm}^3\)
3. 1.65 \(\mathrm{~g} / \mathrm{cm}^3\)
4. 1.73 \(\mathrm{~g} / \mathrm{cm}^3\)

Answer: 4. 1.73 \(\mathrm{~g} / \mathrm{cm}^3\)

Given, mass of cane, m=5 kg =5000 g, volume of cane, V=2880 \(\mathrm{~cm}^3\)

Density of cane, \(\rho=\frac{m}{V}\)

⇒ \(\rho=\frac{5000}{2880}=1.73 \mathrm{~g} / \mathrm{cm}^3\)

Question 13. The pressure exerted by the weight of a cubical box of side 2 cm on the surface of table is 4 Pa. What is the weight of the box?

1. 1.6 \(\times 10^{-4} \mathrm{~N}\)
2. 1.6 \(\times 10^{-5} \mathrm{~N}\)
3. 16 \(\times 10^{-4} \mathrm{~N}\)
4. 15 \(\times 10^{-4} \mathrm{~N}\)

Answer: 3. 16 \(\times 10^{-4} \mathrm{~N}\)

Given, pressure, p = 4 Pa and side of cube,

a = 2 cm = 0.02 m

Area of cubical box, A =\(a^2=(0.02)^2\)

=4 \(\times 10^{-4} \mathrm{~m}^2\)

Since, pressure, p =\(\frac{F}{A}\)

F =p \(\times A=4 \times 4 \times 10^{-4}\)

=16 \(\times 10^{-4} \mathrm{~N}\)

So, the weight of the block is \(16 \times 10^{-4} \mathrm{~N}\).

Question 14. A large ship can float, but a steel needle sinks, because of

1. mass
2. volume
3. density
4. None of these

Answer: 4. None of these

This concept is explained by Archimedes’ principle which states that when a body is fully or partly submerged in a fluid, a buoyant force from the surrounding fluid, acts on the body.

Question 15. A wooden object of mass 700 g is thrown into the water tub. What is the buoyant force acting on the object, if it displaces 700 g of water?

1. 5 N
2. 8 N
3. 7 N
4. 10 N

Answer: 3. 7 N

Given, the mass of the object (m) = 700 g = 0.7 kg

Acceleration due to gravity (g) =10 m/s²

According to Archimedes’ principle, the buoyant force is equal to the weight of displaced water,

i.e. weight of displaced water = mg = 0.7 x 10 = 7 N So, buoyant force is 7 N.

Question 16. A ball weighs 90 g in air, 60 g in water and 40 g in a liquid. If the density of water is 1 g cm-3, then what is the density of this liquid?

1. 3 \(\mathrm{~g} \mathrm{~cm}^{-3}\)
2. 1 \(\mathrm{~g} \mathrm{~cm}^{-3}\)
3. 1.66 \(\mathrm{~g} \mathrm{~cm}^{-3}\)
4. 2.66 \(\mathrm{~g} \mathrm{~cm}^{-3}\)

Answer: 3. 1.66 \(\mathrm{~g} \mathrm{~cm}^{-3}\)

1. As given, when immersed in water, the ball displaces 90 – 60 = 30 g of water.
2. So, it’s volume = volume of water displaced = \(\frac{30}{1 \mathrm{~g} \mathrm{~cm}^{-3}}=30 \mathrm{~cm}^3\)
3. Density of ball = \(\frac{90}{30}=3 \mathrm{~g} \mathrm{~cm}^{-3}\)
4. So, the ball displaces (90-40)=50 \(\mathrm{~g} \) of liquid with density =  \(\frac{50 \mathrm{~g}}{30 \mathrm{~cm}^3}=1.66 \mathrm{~g} \mathrm{~cm}^{-3} \)

Question 17. If two forces in the ratio 3 :8 act on two surface areas in ratio 2 :5. What is the ratio of pressure exerted?

1. 3: 8
2. 6: 40
3. 15: 16
4. 4: 8

Answer: 3. 15: 16

Ratio of areas =2: 5, i.e. \(\frac{A_1}{A_2}=\frac{2}{5}\)

Ratio of forces =3: 8, i.e. \(\frac{F_1}{F_2}=\frac{3}{8}\)

Ratio of pressure =\(\frac{p_1}{p_2}=\frac{F_1}{F_2} \times \frac{A_2}{A_1}\)

=\(\frac{3}{8} \times \frac{5}{2}=\frac{15}{16}\)

Class 9 Science Chapter 9 Very Short Answer Type Questions

Question 1. When do we use the term force of gravity rather than force of gravitation?
Answer:

We use the term force of gravity rather than force of gravitation for the force of attraction between two bodies in which one body has an infinitely large mass.

Question 2. Name the scientist who determined the value of the universal gravitational constant.
Answer:

Henry Cavendish determined the value of the universal gravitational constant.

Question 3. Which force brings tides into the ocean?
Answer:

The gravitational force of the moon brings tides into the ocean.

Question 4. Is it possible to shield a body from gravitational effects?
Answer:

No, as gravitational interaction does not depend on the nature of the intervening medium.

Question 5. Which force keeps the moon in a uniform circular motion around the Earth?
Answer:

The gravitational force between the moon and the Earth keeps the moon in a uniform circular motion around the Earth.

Read and Learn More Class 9 Science Solutions

Question 6. Suppose gravity of the earth suddenly becomes zero, then which direction will the moon begin to move, if no other celestial body affects it?
Answer:

The moon will begin to move in a straight line in the direction in which it was moving at that instant because the circular motion of the moon is due to the centripetal force provided by the gravitational force of the earth.

Question 7. A stone and the earth attract each other with an equal and opposite force. Why, then we see only the stone falling towards the earth but not the earth rising towards the stone?
Answer:

As the mass of the earth is very large, the acceleration produced in the earth is too small, hence we see only the stone falling towards the earth but not the earth rising towards the stone.

Question 8. Two objects kept in a room do not move towards each other as per the universal law of gravitation. Why?
Answer:

The size of the bodies is very small. Therefore the force of attraction between them is very small. So, both objects do not move towards each other.

Question 9. The earth is acted upon by the gravitation of the sun even, then it does not fall into the sun. Why?
Answer:

The earth does not fall into the sun because the earth remains in its circular orbit due to the gravitational force acting on it.

Question 10. Write the direction of acceleration due to gravity.
Answer:

The direction of acceleration due to gravity is downwards i.e. towards the centre of the earth.

Question 11. Two objects of masses m₁ and m₂ are dropped in a vacuum from a height above the surface of the earth (m₁ is greater than m₂). Which one will reach the ground first and why?
Answer:

Both will reach the ground at the same time because the acceleration due to gravity is independent of the masses of freely falling bodies.

Question 12. When a body is thrown upwards, its velocity becomes zero at the highest point. What will be its acceleration at this point?
Answer:

The acceleration at this point is equal to the value of the downward direction.

Question 13. Does the velocity of a body during free fall remain constant? Why/Why not?
Answer:

No, the velocity of a body increases at every point of its motion during free fall as acceleration due to gravity acts on it.

Question 14. At which place on the earth, the acceleration due to gravity is zero?
Answer:

At the centre of the earth, the acceleration due to gravity is zero.

Question 15. Name the place on the earth’s surface, where the weight of a body is maximum and minimum.
Answer:

Weight is maximum at the poles and minimum at the equator.

Question 16. Anu buys 300 g of gold at the poles. What will be the weight of gold at the equator?
Answer:

The value of acceleration due to gravity (g) is less at equators than at poles, so the weight of 300 g gold will be less at the equator

Question 17. The astronauts in space feel weightless. Why?
Answer:

They do not exert any force/weight on their spaceship in the absence of gravity in space.

Question 18. What is the final velocity when a body is dropped from a height?
Answer:

The final velocity is zero when a body is dropped from a height.

Question 19. Which of the two will double the pressure? Doubling the area or making the area half.
Answer:

Making the area half \(\left[ p=\frac{F}{A}\right]\)

Question 20. A girl stands on a box having 60 cm length, 40 cm breadth and 20 cm width in three ways i.e. when length and breadth form the base, when breadth and width form the base, and when width and length form the base. In which condition, the pressure exerted by the box will be maximum?
Answer:

Pressure =\(\frac{\text { Thrust }}{\text { Area }}\)

The pressure exerted by the box will be maximum when the area is small.

The area will be minimal when breadth and width form the base.

Question 21. When a floating body is pressed down a little, which force will increase at that time?
Answer:

Upthrust.

Question 22. A hydrometer is made heavy near the bottom.
Answer:

A hydrometer is made heavy near its bottom so that it can float with the stem in the vertical position.

Question 23. A bucket of water weighs less inside the well water. Why?
Answer :

Due to the upthrust exerted by the well water on the bottom of the bucket in an upward direction.

Question 24. If the density of a body is 800 kg m^-3. Will it sink or float when dipped in a bucket of water? (Take, density of water = 1000 kg^-3)
Answer:

Since its density is less than that of water, hence it will float.

Question 25. If 100 cc of water is heated to three different temperatures i.e. 4°C, 20°C and 45°C. At what temperature, its density will be maximum?
Answer:

4° C; because density increases with a decrease in temperature.

Question 26. The density of glass is 3.5 g cm^-3. What does it mean?
Answer:

3.5 g cm^-3 density of glass means that the volume of 3.5 glass is 1 cm³.

Question 27. An object is dropped one by one in three liquids having different densities. The object floats with \(\frac{1}{9}, \frac{2}{11}\) and \(\frac{3}{7}\) parts of their volumes outside the liquid surface, in liquids of densities \(d_1, d_2\) and \(d_3\), respectively. Arrange them in increasing order.
Answer:

In a liquid of higher density, more part of the object remains outside the liquid. Since the order of part of their volume outside the liquid is \(\frac{1}{9}<\frac{2}{11}<\frac{3}{7}\)

Thus, the order of densities is \(d_1<d_2<d_3\).

Question 28. Name the instrument which is used to determine the density of the liquid.
Answer:

The instrument which is used to determine the density of liquid is a “hydrometer.”

Question 29. Arrange the following in the increasing order of their relative densities.

Iron, air and water

Answer:

The increasing order of the relative densities is as follows: Air < water < iron

Question 30. A body projected horizontally moves with some horizontal velocity although it is under the action of the force of gravity, why?
Answer:

The force of gravity has no effect on the horizontal velocity because it acts in a vertically downward direction. So the body moves with uniform horizontal velocity.

Question 31. A stone dropped from a tree takes 2 seconds to reach the ground. Find its velocity on striking the ground.
Answer:

Given, u =0, t =2 s, a = g =9.8 m/s², v = ?

From the first equation of motion, v =u + at

= 0 + 9.8x 2 = 19.6 m/s

Question 32. Find pressure, when a thrust of 20 N is applied on a surface area of 10 cm².
Answer:

Area, A=10 \(\mathrm{~cm}^2=10 \times 10^{-4} \mathrm{~m}^2[1 \mathrm{~m}=100 \mathrm{~cm}]\)

Pressure(p) =\(\frac{\text { Force }(F)}{\text { Area }(A)}=\frac{20}{10 \times 10^{-4}}\)

=2 \(\times 10^4 \mathrm{~Pa}\)

Pressure is 2 \(\times 10^4 \mathrm{~Pa}\).

Question 33. A metal cuboid of mass M kg rests on a table. A surface area of 40 cm² is in contact with the table. The pressure exerted by the cuboid on the table surface is 10000 Pa. Find the value of M is (given that, g=10 \(\mathrm{~ms}^{-2}\)  )
Answer:

Given, A=40 \(\mathrm{~cm}^2=0.004 \mathrm{~m}^2, p=10000 \mathrm{~Pa}\)

p=\(\frac{F}{A}=\frac{M g}{A}\) [e F=M g]

M=\(\frac{p \times A}{g}=\frac{10000 \times 0.004}{10}=4 \mathrm{~kg}\)

Question 34. Find the mass of 5 \(\mathrm{~m}^3\) of cement of density 3000 kg \(\mathrm{~m}^{-3}\).
Answer:

Given, volume, V=5 \(\mathrm{~m}^3\)

Density, \(\rho=3000 \mathrm{~kg} \mathrm{~m}^{-3}\), mass, m= ?

As, \(\rho=\frac{m}{V}\)  or m=\(\rho\) V

= 3000×5 =15000 kg

Question 35. An object weighs 10 N in air. When immersed fully in water, it weighs only 8 N. What will be the weight of the liquid displaced by the object?
Answer:

Weight of liquid displaced, F =10 – 8 = 2 N

Class 9 Science Chapter 9 Gravitation Short Answer Type Questions

Question 1. “All the objects in the universe attract each other.”

1. What is the force of attraction called?
2. Name any two factors on which this force of attraction depends.

Answer:

1. This force of attraction is called as force of gravitation.
2.  The force of gravitation depends on two factors:
   1.  Directly proportional to the product of their masses.
   2. Inversely proportional to the square of the distance between them.

Question 2. What is the source of centripetal force that a planet requires to revolve around the sun? On what factors does that force depend?
Answer:

The motion of the planet around the sun is due to the centripetal force. This centripetal force is provided by the gravitational force between the planet and the sun.

This force depends on the mass of the sun and mass of the planet and the distance between the sun and the planet.

Question 3. State the source of centripetal force that a planet requires to revolve around the sun. On what factors does the force depend? Suppose this force suddenly becomes zero, then in which direction will the planet begin to move, if no other celestial body affects it?

Answer:

The source of centripetal force is the gravitation force. It depends upon the following factors:

1. Mass of the planet and the sun.
2.  Distance between the planet and the sun.

If this force suddenly becomes zero, then the planet will begin to move in a straight line in the direction in which it was moving at that instant.

Question 4. (1) Seema buys a few grains of gold at the poles as per the instruction of one of her friends. She hands over the same when she meets her at the equator. Will the friend agree with the weight of gold bought? If not, why?

(2) If the moon attracts the Earth, then why does the Earth not move towards the moon?

Answer:

1.  No, her friend will not agree with the weight of the gold bought because the weight at the poles is greater than the weight at the equator.
2. We know that the gravitational force is always attractive, still, the moon does not fall on the Earth because the gravitational force between the Earth and the moon works as the necessary centripetal force for the moon to make it revolve around the Earth.

Question 5. Give two reasons for the variation of g at the equator and at the poles.

Answer:

The variation of g at the equator and at the poles are

1. due to the difference in the radius and
2. due to the rotation of the earth.

Question 6. Two solid objects of masses 1 kg and 2 kg are dropped from a helicopter at the same time. Which one will reach the ground earlier? Justify your answer with a suitable reason.

Answer:

Both will reach the ground at the same time as we know that an object experiences acceleration during free fall. This acceleration experienced by an object is independent of its mass because g=\(\frac{G M}{R^2}\)

As they are dropped at the same time, they will reach the ground at the same time.

Question 7. Give reasons.

1.  The moon does not have an atmosphere.
2.  If you jump on the moon, you will rise much higher than, if you jump on the earth.

Answer:

Moon does not have strong gravity to hold atmospheric gases.

Acceleration due to gravity g is much less on the moon’s surface than that of the earth’s surface. v² -u²

Hence, h=\(\frac{v^2-u^2}{2 g}\) is larger.

Question 8. Give one example of each where the same force acting on

1. a smaller area exerts a larger pressure.
2. a larger area exerts a smaller pressure.

Answer:

1.  Needles have sharp tips having smaller areas to exert a larger pressure.
2. School bags have broad bases and wide straps having larger areas to exert less pressure.

Question 9. Explain, why a camel walks easily on a sandy surface than a man in spite of the fact that a camel is much heavier than a man.

Answer:

The feet of the camel are larger and so cover a larger area, which results in low pressure on the sand bed as compared to human beings and thus, enables them to walk easily on sand without sinking.

Question 10. Give a brief description of why the bottom of dams is broad.

Answer:

As we know, pressure is the force acting per unit area. Dams have large water storage. If the bottom of the dams is not made broad, the large hydraulic pressure may cause dams to sink into the Earth’s basin. So, an increase in the base area decreases the pressure exerted by large water storage.

Question 11. Why pressure exerted by the solid and fluid are different?

Answer:

1. Solids exert pressure because of their weight, i.e. pull of mass by the centre of the Earth with an acceleration of g = 9.8 ms^-2.
2. Similarly, fluids, i.e. gases and liquids both exert equal pressure in all directions over the inner walls of the container in which it is kept.

Question 12. Name two forces which act on a body immersed in a liquid. Give the directions in which they act.

Answer:

The two forces are

1. Weight of the body acting downwards.
2. The buoyant force acts upwards.

Question 13. State the condition under which an object floats on the surface of a liquid. What is the volume of the liquid displaced by the object?

Answer:

1. When the upthrust on the body in a liquid is greater than the weight of the body, then the body floats on the surface of the liquid.
2. The volume of the body = volume of liquid displaced.

Question 14. (1) Name the SI units of thrust and pressure.

(2) In which situation, do we exert more pressure on the ground when we stand on one foot or on both feet? Justify your answer.

Answer:

1. The SI unit of thrust is Newton (N). The SI unit of pressure is Nm^-2or pascal (Pa).
2.  We exert more pressure on the ground when we stand on one foot than both feet, as the area of one foot is half that of two feet as p \(\propto \frac{1}{A}\).

Question 15. (1) Explain, why a completely immersed bottle in water bounces back on the surface.

(2) Why does a bucket of water weigh less inside the well water?

Answer:

1. Since it is known that a body can sink in water only when its weight is greater than the upthrust acted on it by the water. But in this case, the upthrust act on the bottle is greater than its weight, that’s why, it bounces back on the water surface.
2. A bucket of water weighs less inside the well water, it is because when the bucket immersed in the water fully, upthrust “acts on it by water which reduces its actual weight.

Question 16. When a fresh egg is put into a beaker filled with water, it sinks in water. But when a lot of salt is dissolved in the water, the egg begins to rise and then floats. Why?

Answer:

On dissolving a lot of salt in water, the density of salt solution becomes higher than that of pure water. Due to its much higher density, the salt solution exerts much more upward buoyant force on the egg, making the egg rise and then float.

Question 17.  Two different bodies are completely immersed in water and undergo the same loss in weight. Is it necessary that their weights in the air should also be the same? Justify your answer.

Answer:

No, their weights in air don’t need to also be the same. This is because the two bodies have undergone the same loss in weight on completely immersing in water due to their equal volumes, not due to their equal weights. So, they may have different weights in the air.

Question 18. (1) Why does a bucket of water feel heavier when taken out of water?

(2) Lead has greater density than iron and both are denser than water. Is the buoyant force on a lead object greater than, less than or equal to the buoyant force on an iron object of the same volume?

Answer:

1.  A bucket of water feels heavier when taken out of water because when immersed in water, an upward force, i.e. buoyant force acts on it which is equal to the weight of water displaced by the bucket.
2. The buoyant force on a lead object is lesser than the buoyant force on an iron object because lead has greater density, so it displaces a lesser amount of water consequently lesser amount of buoyant force acts on it.

Question 19. Verify Archimedes’ principle of buoyancy with an activity. For the activity, you are provided with a piece of stone, a rubber string and a container filled with water.

Answer:

1. First of all, tie up the stone with the rubber string and hold it against a scale fixed on a wall. Put a mark on the elongated rubber string when the stone is tied.
2. Repeat this experiment but this time suspend the stone in a beaker, filled with water. Now, compare the markings.
3. Explanation: As the buoyant force is acting on the stone in an upward direction, this, gravitational pull decreases and in turn, the stretch of the rubber is lost.

Question 20. State Archimedes’ principle. Write its two applications.

Answer:

Archimedes’ principle states that “when a body is immersed fully or partially in a liquid, it experiences an upward force that is equal to the weight of the liquid displaced by it.”

Applications

1. It is used in designing ships and submarines.
2. It is used in making lactometers, which are used to determine the purity of milk.

Question 21. When a metallic block is immersed below the surface of a liquid, state and define the upward force acting on it.

Answer:

When a metallic block is immersed below the surface of the liquid, it experiences an upward force known as buoyant force.

Buoyant force acting on the block =Weight of liquid displaced by object = mg=(V \(\rho\)) g

where, m – mass of the metallic block

⇒ \(\rho\) = density of metallic block V = volume

g = acceleration due to gravity.

Question 22. If a body is compressed to half its previous volume, what will be the effect on its density and why?

Answer:

1. Since, density \((\rho)=\frac{\text { mass }(m)}{\text { volume }(V)}\)
2. Therefore, if the volume of a body is compressed to half of its previous volume, then the density will be doubled.

Question 23. A block of ice is floating in a bucket of water full up to the brim. Some portion of the ice is visible above the water level. As it melts completely do you expect water to spill out? Give a reason for your answer.

Answer:

We know that the volume of ice is greater than that of water. So in this case when the ice melts, it occupies a volume less than that occupied by the ice dipped in it, so the water level will not spill out.

Question 24. Identical packets are dropped from two aeroplanes, one above the equator and the other above the North Pole, both at height h. Assuming all conditions are identical, will those packets take the same time to reach the surface of the earth? Justify your answer.

Answer:

1. No, those packets do not take the same time to reach the surface of the earth.
2. As the value of g is less at the equator than poles, so packets dropped at the poles reach the surface of the earth first.

Question 25. How will the weight of a body of mass 100 g change, if it is taken from the equator to the poles? Give reasons.

Answer:

Since, the acceleration due to gravity increases from the equator to the poles, so its weight increases because the radius of the earth is less at the poles than at the equator.

Question 26. When you immerse an empty plastic bottle in a bucket of water, it comes above the surface of the water. Why does this happen? How can it remain immersed in water and why?

Answer:

1. When we immerse an empty plastic bottle in a bucket of water, the upward force (upthrust or buoyant force) exerted by water on the bottle is greater than its own weight, therefore it comes above the surface of the water.
2. To keep the bottle completely immersed, an external force which is equal to the difference between the upward force and the weight of the bottle, must be applied to the bottle in a downward direction. This is because the upthrust on the bottle due to water must be balanced.

Question 27. What happens to the magnitude of the force of gravitation between two objects, if (1) distance between the objects is tripled? (2) mass of both objects is doubled?

Answer:

As we know,F=\(\frac{G m_1 m_2}{r^2}\) [symbols have their usual meanings]

(1) \(r^{\prime}=3 r \Rightarrow F^{\prime}=\frac{G m_1 m_2}{9 r^2}=\frac{F}{9}\) [force decreases by 9 times]

(2) \(m_1^{\prime}=2 m_1 and m_2^{\prime}=2 m_2\)

⇒ \(F^{\prime}=\frac{4 G m_1 m_2}{r^2}=4 F\) [force increases by 4 times]

To maintain the same force one of the mass is to be increased by 36 times.

Question  29. A body weighs 25 kg on the surface of the earth. If the mass of the earth is \(6 \times 10^{24}\) kg then the radius of the earth is 6.4 \(\times 10^{6}\) and the gravitational constant 6.67 \(\times 10^{-11} \mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2\).

Calculate

1. the mutual force of attraction between the body and the earth.
2. the acceleration produced in the body.

Answer:

Given, \(M_e=6 \times 10^{24} \mathrm{~kg}\),m=25 kg

⇒ \(R_e=6.4 \times 10^6 \mathrm{~m}\) and G=6.67 \(\times 10^{-11} \mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2\)

(1) Mutual force, F=G \(\frac{M_e}{R_e^2} m \)

= \(\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 25}{\left(6.4 \times 10^6\right)^2}=244 \mathrm{~N}\)

(2) Acceleration produced in the body,

a=\(\frac{F}{m}=\frac{244}{25}=9.8 \mathrm{~m} / \mathrm{s}^2\)

Question 30. Two bodies of masses 3 kg and 12 kg are placed at a distance of 12 m. A third body of mass 0.5 kg is to be placed at such a point that the force acting on this body is zero. Find the position of that point, for Given, mx =3 kg and m² =12 kg

Answer:

Given, \(m_1\)=3 kg and m_2=12 kg

Let the mass, \(m_3\)=0.5 kg be placed at a distance of x from m_1 as shown

Then, the force acting on \(m_3\) due to \(m_1\) is equal and opposite to the force acting on \(m_3\) due to \(m_2\).

⇒ \(F_{31}=F_{32}\)

⇒ \(\frac{G m_1 m_3}{x^2}=\frac{G m_3 m_2}{(12-x)^2} \Rightarrow \frac{3}{x^2}=\frac{12}{(12-x)^2}\)

⇒ \(\left(\frac{12-x}{x}\right)^2=\frac{12}{3}\) =4

⇒ \(\frac{12-x}{x}\) =2

12-x=2 x  \(\Rightarrow\) 12=3 x \(\Rightarrow x=4 \mathrm{~m}\)

The required point is at a distance of 4 m from the mass of 3 kg.

Question 31. Calculate the acceleration due to gravity on the surface of satellite having mass 7.4 x \(10^22\) kg and radius 1.74 x \(10^6\) cm. (Take, G = 6.7 x \(10^{-11} \mathrm{~N}-\mathrm{m} / \mathrm{kg}^2\))

Answer:

As we know, acceleration due to gravity, g=\(\frac{G M}{R^2}\).

For the satellite, R =1.7 .4 \(\times 10^6 \mathrm{~cm}=\frac{1.74 \times 10^6}{100}=1.74 \times 10^4 \mathrm{~m}\)

M =7.4 \(\times 10^{22} \mathrm{~kg}\)

g =\(\frac{6.67 \times 10^{-11} \times 7.4 \times 10^{22}}{1.74 \times 10^4 \times 1.74 \times 10^4}\)

=\(\frac{6.67 \times 7.4}{1.74 \times 1.74} \times 10^3 \mathrm{~g}=16.30 \times 10^3 \mathrm{~m} / \mathrm{s}^2\)

Question 32. What height above the surface of the earth, the value of g becomes 64% of its value at the surface of the earth? Take, the radius of the earth = 6400 km. r? Let g – acceleration due to gravity at the earth’s surface.

Answer:

g, = acceleration due to gravity at height

⇒ \(g_h\) = acceleration due to gravity at height

h =\(\frac{64}{100} \times g=0.64 g\)

g =\(\frac{G M}{R_e^2}\)

Similarly, \(g_h=\frac{G M}{\left(R_e+h\right)^2}\)

From Eqs. (1) and (2), we get

⇒ \(g_h=\frac{g R_e{ }^2}{\left(R_e+h\right)^2}\)

0.64 g=\(\frac{g R_e{ }^2}{\left(R_e+b\right)^2}\)

0.64\(\left(R_e+b\right)^2=R_e^2\)

0.8\(\left(R_e+h\right)=R_e\)

0.8 h=\(R_e-0.8\)

⇒ \(R_e=0.2 R_e\)

h=\(\frac{2 \times 6400}{8}=1600 \mathrm{~km} \left[ R_e=6400 \mathrm{~km}\right]\)

Thus, the height above the surface of the earth is 1600 km.

Question 33. Estimate the gravitational force between two protons (1.6x \(10^{-27}\) kg) separated by a distance of I A.

Answer:

Since, gravitational force, F=\(\frac{G M_p M_p}{r^2}\)

Given, mass of proton, \(M_p=1.6 \times 10^{-27} \mathrm{~kg}\)

Distance, r=1 \(Å=10^{-10} \mathrm{~m}\)

F =\(\frac{6.67 \times 10^{-11} \times 1.6 \times 10^{-27} \times 1.6 \times 10^{-27}}{\left(10^{-10}\right)^2}\)

=17.1 \(\times 10^{-45} \mathrm{~N}\)

Question 34. A ball is dropped from a height half of the earth’s radius. Find the value of g at this point.

Solution:

According to the question, the distance of the ball from the earth’s centre,

⇒ \((R+h)=R+\frac{R}{2}=\frac{3 R}{2} [height, h=R / 2]\)

From the formula of acceleration due to gravity,

g =\(\frac{G M}{(R+h)^2}=\frac{G M}{(3 R / 2)^2}=\frac{4 G M}{9 R^2}\)

= \(\frac{4}{9} \times 10=4.44 \mathrm{~m} / \mathrm{s}^2 \left[ \frac{G M}{R^2}=10 \mathrm{~m} / \mathrm{s}^2\right]\)

Thus, the value of g is 4.44 \(\mathrm{~m} / \mathrm{s}^2\).

Question 35. A body is dropped from a height of 100 m. What is its height above the ground after 2 seconds of its fall? (Take, g=10m/s² )

Solution:

Given, initial velocity, u = 0 Time taken, t – 2 s

Acceleration due to gravity, a = g

From the second equation of motion,

⇒ \(s=u t+\frac{1}{2} g t^2=0+\frac{1}{2} \times 10 \times(2)^2=20 \mathrm{~m}\)

The height of the body above the ground after 2 s of its fall, h =100 – 20 =80 m

Question 36.  A ball is thrown up with a velocity of 19.6 m/s.

1. How long will it take to reach the maximum height?
2. How high will it go?

Answer:

Given, initial velocity, u = 19.6 m/s

(1) Final velocity, v = 0 at maximum height.

So, t=\(\frac{u}{g}=\frac{19.6}{9.8}=2 \mathrm{~s}\) [ v=u-g t and v=0]

From the third equation of motion,

⇒ \(v^2=u^2-2 g h\) [negative sign is taken due to upward motion]

⇒ \(u^2=2 g h\) {[ v=0]}

h=\(\frac{u^2}{2 g}=\frac{(19.6)^2}{2 \times 9.8}=19.6 \mathrm{~m}\)

Question 37. Your mass on the earth is 50 kg. Planet M has two times the force of gravity of that of the Earth. What will be your mass and weight on planet M? (Take, g = 9.8 m/s²)

Answer:

As the mass remains constant, so on the planet M your mass will be 50 kg.

Given, that acceleration due to gravity on planet M is two times that of the Earth.

So, weight on planet M will be 50 x 2 x g = 50x2x9.8 =980

Question 38. Suppose that the radius of the earth becomes twice its original radius without any change in its mass. Then, what will happen to your weight?

Answer:

The weight of a body is the force with which a body is attracted towards the earth, w=\(\frac{G M m}{R^2}\)

If the radius of the earth becomes twice its original radius, then w=G \(\frac{M m}{(2 R)^2}=\frac{G M m}{4 R^2}=\frac{w}{4}\)

Thus, the weight will be reduced to one-fourth of the original

Question 39. 39 Find the weight of an 80 kg man on the surface of the moon. What should be his mass on the earth and on the moon?

(Take, \(g_e=9.8 \mathrm{~m} / \mathrm{s}^2, g_m=1.63 \mathrm{~m} / \mathrm{s}^2\) )

Answer:

Given, mass on the earth = mass on the moon =80 kg

Man’s weight on the earth,

⇒ \(w_e=9.8 \times\) 80=784 N

Man’s weight on the moon,

⇒ \(w_m=1.63 \times\) 80=130.4 N

Thus, the weight of the moon is 130.4 N.

Question 40. 40 The weight of any person on the moon is about l/6th of that on the earth. He can lift a mass of 15 kg on the earth. What will be the maximum mass which can be lifted by the same force applied by the person on the moon?

Answer:

Maximum weight which can be lifted = mg

=15 \(\mathrm{~kg} \times 9.8 \mathrm{~m} / \mathrm{s}^2=147 \mathrm{~N}\)

Mass which can be lifted on the moon

=147 \(\mathrm{~N} \times \frac{1}{g_{\text {earth }} / 6}\)

[at moon, acceleration due to gravity is 1/6 th to that of the earth]

= \(\frac{147 \times 6}{9.8}=\frac{147}{1.63}=90 \mathrm{~kg}\)

Thus, the maximum mass which can be lifted is 90 kg.

Question 41. Calculate the average density of the earth in terms of g, G and R.
Answer:

According to the formula, density =\(\frac{\text { mass }}{\text { volume }}\).

If the radius of the earth is R, then its volume =\(\frac{4}{3} \pi R^3\). [The shape of the earth is spherical]

Mass of the earth, M=\(\frac{g R^2}{G} [ g=\frac{G M}{R^2}]\)

Density =\(\frac{g R^2}{G} / \frac{4}{3} \pi R^3=\frac{3 g}{4 \pi R G}\)

Thus, the average density of the earth is \(\frac{3 g}{4 \pi R G}\)

Question 42. Which will exert more pressure, 100 kg mass on 10 m² or 50 kg mass on 4 m²? Give reason. (Take, g = 10 ms^-2)

Answer:

For 100 kg of mass,

Pressure, \(p_1=\frac{\text { Force }}{\text { Area }}=\frac{m_1 g}{A_1}=\frac{100 \times 10}{10}=100 \mathrm{~Pa}\)

For 50 kg of mass,

Pressure, \(p_2=\frac{m_2 g}{A_2}=\frac{50 \times 10}{4}=125 \mathrm{~Pa}\)

⇒ \(p_2>p_1\)

Hence, a 50 kg mass will exert more pressure.

Question 43. A block of wood of mass 5 kg and dimensions 40 cm x 20 cm x 10 cm is placed on a tabletop. Find the pressure exerted, if the block lies on the tabletop with sides of dimension

1. 40 cm x 20 cm
2. 40 cm x 10 cm (Take, g = 10 \(\mathrm{~ms}^{-2}\))

Answer:

Given, mass, m =5 kg, acceleration due to gravity, g =10 \(\mathrm{~ms}^{-2}\)

g=10 \(\mathrm{~ms}^{-2}\)

Weight =m g=5 \(\times 10=50 \mathrm{~N}\)

(1) Pressure =\(\frac{\text { Weight }}{\text { Area }}=\frac{50}{40 \times 20 \times 10^{-4}} \mathrm{~Pa}\)

=6.25 \(\times 10^2 \mathrm{~Pa}\)

(2) Pressure =\(\frac{\text { Weight }}{\text { Area }}=\frac{50}{40 \times 10 \times 10^{-4}} \mathrm{~Pa}\)

=1.25 \(\times 10^3 \mathrm{~Pa}\)

Question 44. If two forces in the ratio 5:9 act on two areas in the ratio 10:3, find the ratio of pressure exerted.

Answer:

Given, ratio of areas =10: 3$, i.e. \(A_1 / A_2\)=10 / 3

Ratio of forces =5: 9, i.e. \(F_1 / F_2\)=5 / 9

Ratio of pressures =\(\frac{p_1}{p_2}=\frac{F_1}{A_1} \times \frac{A_2}{F_2}\)

= \(\frac{F_1}{F_2} \times \frac{A_2}{A_1}=\frac{5}{9} \times \frac{3}{10}=\frac{1}{6}\)

Therefore, the ratio of the pressures is 1:6.

Question 45. A ball filled with air has a volume of 500 cm3. Calculate the minimum force applied by a child to put it completely inside the water. (Take, g = 10 \(\mathrm{~ms}^{-2}\))

Answer:

Given that,

Volume, V=500 cm³ =500 x 10^-6 m³,

g = 10 ms^-2, F =?

The force required to put the ball inside the water = Buoyant force

= Weight of water displaced – mg … (1)

Now, we know that,

Mass of water = Density of water x Volume m =pV

On substituting this value in Eq. (1), we get Force = p Vg

= (1000 kg m^-3) X (500 X 10^-6 m³) X (10 ms^-2)

= 1000 x 500 x 10~6 x 10 N =5 N

The minimum force applied by a child to put the ball completely inside the water is 5 N.

Question 46. A ball weighs 80 g in air, 60 g in water and 50 g in liquid. If the density of water is 1 g cm^-3, find the density of this liquid.

Answer:

As given, when immersed in water, the ball displaces 80 -60 = 20 g of water.

So, its volume = volume of water displaced

=\(\frac{20 \mathrm{~g}}{1 \mathrm{~g} \mathrm{~cm}^{-3}}=20 \mathrm{~cm}^3\)

Density of ball =\(\frac{80 \mathrm{~g}}{20 \mathrm{~cm}^3}=4 \mathrm{~g} \mathrm{~cm}^{-3}\)

So, the ball displaces (80-50) \(\mathrm{g}=30 \mathrm{~g}\) of liquid with density =\(\frac{30 \mathrm{~g}}{20 \mathrm{~cm}^3}=1.5 \mathrm{~g} \mathrm{~cm}^{-3}\).

Question 47. Prove that, if a body is thrown vertically upwards, then the time of ascent is equal to the time of descent.

Answer:

For the upward motion,

v=u-g \(t_1\),  0=u-g \(t_1\),  \(t_1=\frac{u}{g}\)

and the downward motion,

v=u+g \(t_2\), v=0+g \(t_2\)

The body falls back to the earth at the same speed as it was thrown vertically upwards.

v=u,  u=0+g \(t_2 \Rightarrow t_2=\frac{u}{g}\)

From Eqs. (1) and (2), we get \(t_1=t_2 \Rightarrow\) Time of ascent = Time of descent

Question 48. A ball is dropped from the edge of a roof. It takes 0.1s to cross a window of height 2.0 m. Find the height of the roof above the top of the window.

Answer:

Let AB be the window and suppose the roof T is at a height y above A. Also, suppose it takes a time TV for the ball to reach A. The velocity of the ball at A is

⇒ \(v_1=0+g t_1=9.8 t_1\) (1)

Now, consider the motion of the ball from A to B.

Here, the initial velocity is v1 the distance covered is 2 m and the time taken is 0.1 s.

From second equation of motion, s=u t+\(\frac{1}{2} g t^2\)

2.0=\(v_1(0.1)+\frac{1}{2} \times 9.8 \times(0.1)^2 =9.8 t_1(0.1)+0.049\)

⇒ \(t_1 =1.99 \approx 2 \mathrm{~s}\)

The height y is y=\(\frac{1}{2} g t_1^2\)

= \(\frac{1}{2} \times 9.8 \times(2)^2=19.6 \mathrm{~m}\)

The roof is at a height of 19.6 m above the top of the window.

Question 49. On the earth, a stone is thrown from a height in a direction parallel to the earth’s surface while another stone is simultaneously dropped from the same height. Which stone would reach the ground first and why?

Answer:

For both the stones, initial velocity, u- 0

Acceleration in a downward direction = g

Now, from the second equation of motion,

h=u t+\(\frac{1}{2} g t^2 \Rightarrow  h=0+\frac{1}{2} g t^2\)

h=\(\frac{1}{2} g t^2 \quad \Rightarrow t=\sqrt{\frac{2 h}{g}}\)

Both stones will take the same time to reach the ground because the two stones fall from the same height.

Question 50. A ball is thrown with some speed u m/s. Show that under the free fall, it will fall on the ground at the same speed.

Answer:

When the ball is thrown upwards, then it will reach a certain height h and start falling. At maximum height h, the final velocity will be v = 0.

The maximum height reached by the ball,

⇒ \(v^2-u^2=2 g h\)

⇒ \(0-u^2=-2 g h\) [acceleration =-g]

h=\(\frac{u^2}{2 g}\)  [using equation]

In the second case, when the ball starts to fall, then the initial velocity u=0. It will accelerate due to gravity, i.e. a=g and reach ground with speed (say \(v_2\) ).

Using equation, \(v_2^2-u^2\) =2 g h

⇒ \(v_2^2-0 \)=2 g h

⇒ \(v_2^2 =2 g\left(\frac{u^2}{2 g}\right)=u^2\) [{ from Eq. (1) })

⇒ \(v_2\) =u

Thus, the ball reaches the ground at the same speed.

Class 9 Science Chapter 9 Gravitation Long Question And Answers

Question 1. (1) Write the formula to find the magnitude of the gravitational force between the earth and an object on the earth’s surface.

(2) Derive how does the value of gravitational force F between two objects change when

1.  distance between them is reduced to half and
2. mass of an object is increased four times.

Answer:

(1) Formula to find the magnitude of gravitational Force, F=\(\frac{G M m}{R^2}\)

M=mass of the earth

m = mass of the object

R = radius of the earth

and universal gravitational constant, G=\(6.67 \times 10^{-11} \mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2\)

(2) (1) Let the gravitational force be F when the distance between them is R,

F=\(\frac{G M m}{R^2}\)  → Equation 1

Now, when the distance is reduced to half,

⇒ \(F^{\prime}=\frac{G M m}{\left(\frac{R}{2}\right)^2}=\frac{4 G M m}{R^2}\)  → Equation 2

On dividing Eq. (1) by Eq. (2), we get

⇒ \(\frac{F}{F^{\prime}}=\frac{G M m}{R^2} \times \frac{R^2}{4 G M m}\)

⇒ \(F^{\prime}\) =4 F

(2) When the mass becomes 4 times,

∴ \(\frac{F}{F^{\prime}}=\frac{G M m}{R^2} \times \frac{R^2}{4 G M m} \Rightarrow F^{\prime}=4 F\)

Question 2.  (1) Prove that, if the earth attracts two bodies placed at the same distance from the centre of the earth with equal force, then their masses will be the same.

(2) Mathematically express the acceleration due to gravity in terms of mass of the earth and radius of the earth.

(3) Why is G called a universal constant?

Answer:

Let the two bodies have masses m₁ and m₂ and they are placed at the same distance R from the centre of the earth. According to the question, if the same force acts on both of them, then

⇒ \(F_1=\frac{G M m_1}{R^2}\)

and\(F_2=\frac{G M m_2}{R^2}\)

As, \(F_1=F_2\)

Hence, \(\frac{G M m_1}{R^2}=\frac{G M m_2}{R^2}\)

So, \(m_1=m_2\), their masses will be the same.

(2) Mathematically, g=\(\frac{G M}{R^2}\).

where, g = acceleration due to gravity

G = universal gravitational constant M mass of the earth and R = radius of the earth

(3) G is known as the universal gravitational constant because its value remains the same all the time everywhere in the universe, applicable to all bodies whether celestial or terrestrial.

Question 3. (1) A person weighs 110.84 N on the moon, whose acceleration due to gravity is 1/6 of that of the earth. If the value of g on the earth is 9.8 m/s², then calculate

1.  g on the moon
2.  mass of people on the moon
3. weight of the person on the earth

(2) How does the value of g on the earth is related to the mass of the earth and its radius? Derive it.

Answer:

(1) g on the moon is given by

\(g^{\prime}=\frac{g}{6}=\frac{9.8}{6}=1.63 \mathrm{~m} / \mathrm{s}^2\)

(2) Mass of the person on the moon

=\(\frac{110.84}{1.63}=68 \mathrm{~kg}\)

(3) Weight of person on the earth = mg

=68 \(\times\) 9.8

=666.4 \(\mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2\)

Weight of a person on the earth will be

w=\(\frac{G M m}{R^2}\)

where, M=mass of the earth R = radius of the earth m = mass of person and G=6.67 \(\times 10^{-11} \mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2\)

Question 4. Two objects of masses mx and having the same size are dropped simultaneously from heights ha and h₂, respectively. Find out the ratio of time they would take to reach the ground. Will this ratio remain the same, if

1.  one of the objects is hollow and the other one is solid and
2. both of them are hollow, size remains the same in each case. Give reason.

Answer:

Height of object A, \(h_1=\frac{1}{2} g t_1^2\)

Height of object B, \(h_2=\frac{1}{2} g t_2^2\)

⇒ \(h_1: h_2=t_1^2: t_2^2\)

or \(t_1: t_2=\sqrt{h_1}: \sqrt{h_2}\)

(1) Acceleration due to gravity is independent of the mass of the falling body. So, the ratio remains the same.

(2) If bodies are hollow, then also ratio remains the same, \(t_1: t_2=\sqrt{h_1}: \sqrt{h_2}\)

Question 5. A stone is dropped from the edge of a roof.

1. How long does it take to fall 4: 9 am
2.  How fast does it move at the end of that fall?
3. How fast does it move at the end of 7.9 m?
4.  What is its acceleration after Is and after 2s?

Answer:

Given, initial velocity u=0

Acceleration g=9.8 \(\mathrm{~m} / \mathrm{s}^2\)

(1) We have, s =u t+\(\frac{1}{2} g t^2\)

4.9 =0 \(\times t+\frac{1}{2} \times 9.8 \times t^2\)

⇒ \(t^2 =\frac{9.8}{9.8}=1 \Rightarrow t=1 s\)

The stone takes 1 s to fall 4.9 m

(2) We have, \(v^2-u^2=2\) as

⇒ \(v^2-0^2 =2 \times 9.8 \times\) 4.9

v =\(\sqrt{96.04}=9.8 \mathrm{~m} / \mathrm{s}\)

At the end of 4.9 m, the stone will be moving at a speed of 9.8 m/s.

We have, \(v^2-u^2\)=2 as

⇒ \(v^2-0^2 =2 \times 9.8 \times 7.9\)

v =12.44 \(\mathrm{~m} / \mathrm{s}\)

=12.44 m/s

The stone will be moving with a speed of 12.44 m/s at the end of 7.9 m

(4) During the free fall the acceleration produced in a body remains constant.

So, acceleration after 1 s = 9.8 m/s² Acceleration after 2 s = 9.8 m/s²

Question 6. (1) A steel needle sinks in water but a steel ship floats. Explain, how.

(2) Why do you prefer a broad and thick handle for your suitcase?

Answer:

(1) The ship displaces more water than the needle as the volume of the ship is more than that of the needle. Since upthrust depends on the volume of the object (U = Vdg), so more the volume of the object, the more upthrust acts on it and the object floats.

(2) Since, pressure act on the body is inversely proportional to the surface area of contact, i.e. p \(\propto \frac{1}{A}\)

It means that the more the area of contact, the less pressure will act on the body. As the broad and thick handle of our suitcase has a large area, due to which less pressure acts on our hand and it is very easy to take from one place to another

Question 7. The radius of the earth at the poles is 6357 km and the radius at the equator is 6378 km. Calculate the percentage change in the weight of a body when it is taken from the equator to the poles.

Answer:

Let acceleration due to gravity at the equator,

⇒ \(g_e=\frac{G M_e}{R_e^2}\)

and acceleration due to gravity at poles,

⇒ \(g_p=\frac{G M_e}{R_p^2}\)

The variation of acceleration due to gravity,

⇒ \(\Delta g=g_p-g_e=G M_e\left(\frac{1}{R_p^2}-\frac{1}{R_e^2}\right)\)

e variation in g=\(\frac{G M_e\left(\frac{1}{R_p^2}-\frac{1}{R_e^2}\right)}{\frac{G M_e}{R_e^2}} \times 100\)

Percentage variation in g=\(\frac{G M_e\left(\frac{1}{R_p^2}-\frac{1}{R_e^2}\right)}{\frac{G M_e}{R_e^2}} \times 100\)

= \(\frac{R_e^2-R_p^2}{R_e^2 R_p^2} \times 100 \times R_e^2=\frac{R_e^2-R_p^2}{R_p^2} \times 100\)

= \(\frac{(6378)^2-(6357)^2}{(6357)^2} \times 100 \approx 0.7 \%\)

% variation in the weight of a body = % Change in  g=0.7 \%

Class 9 Science Notes For Chapter 9 Gravitation

It has been observed that an object dropped from a height falls towards the earth. Newton generalised this idea and said that not only the earth but every object in the universe attracts every other object.

This force of attraction between two objects is called the force of gravitation or gravitational force. In this chapter, we shall learn about gravitation and the universal law of gravitation.

Gravitation

Gravitation is defined as the force of attraction between any two bodies in the universe. The earth attracts (or pulls) all objects lying on or near its surface towards its centre. The force with which the earth pulls the objects towards its centre is called the gravitational force of the earth or gravity of the earth.

Universal Law of Gravitation

The universal law of gravitation was given by Isaac Newton. According to this law, the attractive force between any two objects in the universe is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

The direction of the force is along the line joining the centres of two objects. Consider two bodies A and B having masses m₁ and m_2, whose centres are at a distance d from each other.

Read and Learn  More Class 9 Science Notes

The gravitational force between two objects is directed along the line joining their centres Then, the force between two bodies is directly proportional to the product of their masses,

⇒ \(F \propto m_1 m_2\)  →  Equation 1

and the force between two bodies is inversely proportional to the square of the distance between them, i.e.

F \(\propto \frac{1}{d^2}\)

F \(\propto \frac{m_1 m_2}{d^2}\) or F=G\( \frac{m_1 m_2}{d^2}\)

where, G=6.67 \(\times 10^{-11} \mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2\) is called the universal

Combining Eqs. (1) and (2), we get

where G= 6.67 x 10_11N-m₂/kg² is called the universal gravitational constant.

Its value does not depend on the medium between the two bodies and the masses of the bodies or the distance between them. Suppose the masses of two bodies are 1 kg each and the distance d between them is 1 m, then

F = G [m₁ = m₂ = 1 kg and d = m]

Hence, the universal gravitational constant is defined as the gravitational force between two objects of unit masses separated by a unit distance from each other placed anywhere in space. The SI unit of G is N-m₂ kg. The value of G was found out by Henry Cavendish (173T1810) by using a sensitive balance.

Importance of Universal Law of Gravitation

1. The universal law of gravitation successfully explained several phenomena given below:
2. The force that binds us to the earth.
3. The motion of the moon around the earth.
4. The motion of planets around the sun.
5. The occurrence of tides is due to the gravitational force of attraction of the moon.
6. The flow of water in rivers is also due to the gravitational force of the earth on water.

Motion of Moon around Earth and Centripetal Force

The force that keeps a body moving along the circular path acting towards the centre is called centripetal (centre-seeking) force. The motion of the moon around the Earth is due to the centripetal force. The centripetal force is provided by the gravitational force of attraction of the earth. If there were no such force, then the moon would pursue a uniform straight-line motion.

Example 1. Find the gravitational force between the earth and an object of 2 kg mass placed on its surface. (Givenr mass of the earth = 6 x \(10^24\) kg and radius of the earth = 6.4 x \(10^6\) m)

Answer:

Given, mass of the earth, \(m_e=6 \times 10^{24} \mathrm{~kg}\)

Mass of an object, \(m_o=2 \mathrm{~kg}\)

Radius of the earth, R=6.4 \(\times 10^6 \mathrm{~m}\)

Gravitational force, F=?

F=G \(\frac{m_e m_o}{d^2}\) [by universal law of gravitation]

F=\(\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 2}{\left(6.4 \times 10^6\right)^2}=19.5 \mathrm{~N}\)

Example 2. The mass of Mars is 639 x 10²³ kg and that of Jupiter is 189 x 10²⁷ kg. If the distance between Mars and Jupiter is 749 x 1²³ m. Calculate the force exerted by the Jupiter on the mars. (G =6.7 x 10~n Nm² kg²)

Answer:

Given, the mass of the mars \(M_m=639 \times 10^{23} \mathrm{~kg}\)

The mass of the Jupiter \(M_j=1.89 \times 10^{27} \mathrm{~kg}\)

The distance between Mars and Jupiter

d=7.49 \(\times 10^5 \mathrm{~m}\)

G=6.7 \(\times 10^{-11} \mathrm{Nm}^2 / \mathrm{kg}^2\)

The force exerted by Jupiter on Mars

F =G \(\frac{M_m \times M_j}{d^2}\)

= \(\frac{6.7 \times 10^{-11} \times 639 \times 10^{23} \times 1.89 \times 10^{27}}{\left(7.49 \times 10^5\right)^2}\)

=1.44 \(\times 10^{29} \mathrm{~N}\)

Thus, the force exerted by the Jupiter on the mars is 1.44 \(\times 10^{29} N\).

Free Fall

When objects fall towards the earth under the influence of the earth’s gravitational force alone, then these are called freely falling objects and such a motion is called free fall.

Acceleration due to Gravity [g]

Whenever an object falls towards the earth, acceleration is involved. This acceleration is due to the earth’s gravitational pull and is called acceleration due to gravity. It is denoted by g.

The SI unit of is the same as that of acceleration, i.e. m/s². Let the mass of the earth be M and an object falling freely towards it be m. The distance between the centres of the earth and the object is R.

From Newton’s law of gravitation,

Also, from the second law of motion, the force exerted on an object, F =ma

Since, a = g (i.e. acceleration due to gravity)

F = mg Equation  (2)

Equating RHS of Equation. (1) and (2), we get GMm

m g=\(\frac{G M m}{R^2}\) or g=\(\frac{G M}{R^2}\)

From the formula, it is clear that acceleration due to gravity does not depend on the mass of a falling object. It depends only on the mass of the earth or celestial bodies.

Equations of Motion for Free Fall

The three equations of motion which we have derived earlier are for bodies under uniform acceleration. In the case of motion of bodies under free fall, there is a uniform acceleration, i.e. acceleration due to gravity (g) acting downward.

So, the previous three equations of motion can be applied to the motion of bodies under free fall as follows:

General equations Equations for body under free fall of motion

where h is the height from which the object falls, t is the time of fall, u is the initial velocity and v is the final velocity when the body accelerates at g.

In solving numerical problems, we should remember the following points:

1. If an object falls vertically downwards, then acceleration due to gravity is taken as positive, since its velocity increases while falling.
2.  If an object is thrown vertically upwards, then acceleration due to gravity is taken as negative, since its velocity decreases as it moves upward.

Example 3. A car falls off a ledge and drops to the ground in 0.6 seconds. The value of g is 10 m/s2 (for simplifying the calculation).

1.  What is its speed on striking the ground?
2. What is its average speed during the 0.6 s?
3. How high is the ledge from the ground?

Answer:

Initial velocity u = 0

Acceleration due to gravity g=10 \(\mathrm{~m} / \mathrm{s}^2\)

(1) Speed v=u+g t

v=0+10 \(\times\) 0.6

v=6 \(\mathrm{~m} / \mathrm{s}\)

(2) Average speed =\(\frac{u+v}{2}=\left(\frac{0+6}{2}\right)=3.0 \mathrm{~m} / \mathrm{s}\)

(3) Distance travelled h=u t+\(\frac{1}{2} g t^2\)

h=0+\(\frac{1}{2} g t^2\)

h=\(\frac{1}{2} \times 10 \times(0.6)^2=5 \times 0.36=1.80 \mathrm{~m}\)

Example 4. An object is thrown vertically upwards and rises to a height of 13.07 m. Calculate

1. The velocity with, which the object was thrown upwards
2. The Time taken by the object to reach the highest point.

Answer:

Distance travelled, h =13.07 m

Final velocity v = 0

Acceleration due to gravity g=-9.8 \(\mathrm{~m} / \mathrm{s}^2\) (upward motion)

(1) \(v^2=u^2+2 g h\)

0=\(u^2+2 \times(-9.8) \times 13.07\)

⇒ \(u^2=256 \Rightarrow u=16 \mathrm{~m} / \mathrm{s}\)

(2) v=u+a t

0=16-9.8 \(\times t \Rightarrow t=1.63 \mathrm{~s}\)

Example 5. A ball is thrown vertically upwards with a velocity of 25 m/s. If g is 10 m/s², then calculate

1. the height it reaches.
2. time taken to return back.

Answer:

Given, initial velocity, u = 25 m/s, final velocity, v =0

If a body is thrown upwards, then its velocity becomes zero at the highest point, where it reaches, acceleration due to gravity,

g=-10 \(\mathrm{~m} / \mathrm{s}^2\)

(1) Height, h =\(\frac{v^2-u^2}{2 g}\)

=\(\frac{0-(25)^2}{2(-10)}=\frac{-625}{-20}=31.25 \mathrm{~m}\)

Height, h=\(\frac{v^2-u^2}{2 g}\)

(2) Time, t=\(\frac{v-u}{g}=\frac{0-25}{-10}=2.5 \mathrm{~s}\)

Time taken to return back,

T = time of ascent + time of descent =21 Time taken to return back, T = 2×2.5 = 5s

Weight

The weight of an object is the force with which it is attracted towards the earth.

Weight of an object, w = mg

where, m = mass, g = acceleration due to gravity or w=\(\frac{G M m}{R^2}\)

Here, M = mass of the earth and R = radius of the earth

Important points regarding weight are as follows:

1. Weight is a vector quantity, it acts in a vertically downward direction, and its SI unit is newton (N). Weight of 1 kg mass is 9.8 N. (i.e. 1 kg-wt =9.8 N)
2. The weight of an object is not constant, it changes from place to place.
3.  In the space, where g = 0, the weight of an object is zero.
4.  At the centre of the earth, weight becomes zero. This is due to the fact that on going down to the earth, the value of g decreases and at the centre of the earth, g = 0.

Weight of an Object on the Moon

Let the mass of an object be m and its weight on the moon be wm. Suppose the mass of the moon is Mm and its radius is Rm. According to the universal law of gravitation, the weight of an object on the moon will be

⇒ \(w_m=G \frac{M_m \times m}{R_m^2}\)

Let the weight of the same object on the earth be we. Let the mass of the earth be Me and the radius of the earth be Re.

⇒ \(\frac{w_m}{w_e} =\frac{\frac{G M_m m}{R_m^2}}{\frac{G M_e m}{R_e^2}}=\frac{M_m}{M_e} \times \frac{R_e^2}{R_m^2}\)

Now, \(M_m =5.98 \times 10^{24} \mathrm{~kg} ; M_e=736 \times 10^{22} \mathrm{~kg}\)

⇒ \(R_m =1.74 \times 10^6 ; R_e=6.37 \times 10^6 \)

⇒ \(\frac{w_m}{w_e} =\frac{5.98 \times 10^{24}}{736 \times 10^{22}} \times \frac{\left(637 \times 10^6\right)^2}{\left(1.74 \times 10^6\right)^2} \approx \frac{1}{6}\)

Thus, the weight of an object on the moon is one-sixth of its weight on the Earth.

Example 6. The mass of an object is 12 kg. Calculate

1. Its weight on the earth.
2. Its weight on the moon.

Answer:

Given, m= 12 kg

(1) Acceleration due to gravity on earth, ge =9.8 m/s² weight earth we = mg =12 x 9.8 = 117.6 N

(2) Acceleration due to gravity on the moon \(g_m=\frac{g_e}{6}\)

⇒ \(g_m=\frac{9.8}{6} \mathrm{~m} / \mathrm{s}^2\)

Weight on moon \(w_m=m g_m=12 \times \frac{9.8}{6}=9.8 \times 2=19.6 \mathrm{~N}\)

Example 7. A man weighs 600 N on the earth. What is his mass, if g is 10 m/s²? On the moon, his weight would be 100 N. What is the acceleration due to gravity on the moon?

Answer:

Given, the weight of a man on the earth, \(w_e=600 \mathrm{~N}\)

Acceleration due to gravity on the earth, \(g_e=10 \mathrm{~m} / \mathrm{s}^2\)

Weight of man on the moon, \(w_m=100 \mathrm{~N}\)

Acceleration due to gravity on the moon, \(g_m\)=?

As we know, mass of the man, m=\(\frac{w_e}{g_e}=\frac{600}{10}=60 \mathrm{~kg}\)

⇒ \(g_e=\frac{w_e}{m}\)

Similarly, for the moon \(g_m=\frac{w_m}{m}\)

⇒ \(g_m=\frac{100}{60}=1.66 \mathrm{~m} / \mathrm{s}^2\)

Thus, acceleration due to gravity on the moon is 1.66 \(\mathrm{~m} / \mathrm{s}^2\), i.e. \(g_m=\frac{g_e}{6}\).

Example 8. A particle weighs 120 N on the surface of the earth. At what height above the earth’s surface will its weight be 30 N? Radius of the earth = 6400 km.

Answer:

The weight of a particle on the surface of the earth is

w=m g=\(\frac{m M G}{R^2} [g=\frac{G M}{R^2}]\)

Let \(w_1\) be the weight of a particle at height h above the earth’s surface.

So, \(\frac{w}{w_1} =\frac{G \frac{M}{R^2}}{G \frac{M}{(R+h)^2}}=\frac{(R+h)^2}{R^2}\)

⇒ \(\frac{120}{30} =\left(\frac{R+h}{R}\right)^2\)

4=\(\left(\frac{R+h}{R}\right)^2 \Rightarrow \quad 2=\frac{R+h}{R}\)

2 R=R+h \(\Rightarrow\) R=h

Height of the particle, h= Radius of the earth,

b=6400 km.

Thrust and Pressure

1. Thrust is the force acting on an object perpendicular to its surface. The effect of thrust depends on the area on which it acts.
2. The unit of thrust is the same as that of force, i.e. the SI unit of thrust is Newton (N). It is a vector quantity.
3. Pressure is the force acting perpendicularly on a unit area of an object.
4. Pressure(p)=\(\frac{\text { Force }(F)}{\text { Area }(A)}=\frac{\text { Thrust }}{\text { Area }}\)
5. The SI unit of pressure is Nm -2, which is also called Pascal (Pa) named after the scientist Blaise Pascal. It is a scalar quantity.
6. \(1 \mathrm{~Pa}=1 \mathrm{Nm}^{-2}\)
7. From the formula of pressure, it is clear that the same force can produce different pressures depending on the area over which it acts. A force acting on a smaller area exerts a large pressure while the same force acting on a larger area exerts small pressure.

Example 9. A Force of 200 N is applied to an object of an area of 4 m₂. Find the pressure.

Answer:

Given, force, F=200 \(\mathrm{~N}\), area, A=4 \(\mathrm{~m}^2\)

Now, Pressure, p=\(\frac{\text { Force }(F)}{\text { Area }(A)}=\frac{200}{4}=50 \mathrm{Nm}^{-2}\)

Example 10. A woman is wearing sharp-heeled sandals (stilettos). If the mass of a woman is 60 kg and the area of one heel is 1 cm², find out the pressure exerted on the ground, when the woman stands on just one heel. (Given, g = 10ms^-2)

Answer:

In the given case, the force will be the weight of the woman which is given by m x g [where m is the mass of the woman and g is the acceleration due to gravity].

Force, F = mx g [weight of woman]

= 60 x 10 N [given, m = 60 kg, A = 1 cm²]

= 600 N

And area, A=1 \(\mathrm{~cm}^2=\frac{1}{10000} \mathrm{~m}^2\)

Pressure, p=\(\frac{\text { Force }(F)}{\text { Area }(A)}=\frac{600 \times 10000}{1}\)

=6000000 \(\mathrm{Nm}^{-2}\) (or } 6000000Pa

Thus, the pressure exerted by the woman (of 60 kg) standing on only one heel of area 1 cm² is 6000000 Nm-2.

Example 11 A block of wood is kept on a tabletop. The mass of the wooden block is 6 kg and its dimensions are 50 cm x 30 cm x 20 cm.

Find the pressure exerted by the wooden block on the tabletop if it is made to lie on the tabletop with its sides of dimensions

1. 30 cm x 20 cm
2. 50 cm x 30 cm

Answer:

The mass of the wooden block = 6 kg

The dimensions = 50 cm x 30 cm x 20 cm Thrust F = mg = 6 x 9.8 = 58.8 N

(1) Area of a side = Length x Breadth

=30 \(\times 20=600 \mathrm{~cm}^2=0.06 \mathrm{~m}^2\)

Pressure \(p_1 =\frac{F}{A}=\frac{58.8}{0.06}\)

=980 \(\mathrm{~N} / \mathrm{m}^2\)

(2) When the block lies on its side of dimensions 50 \(\mathrm{~cm} \times 30 \mathrm{~cm}\), it exerts the same thrust

Area = Length x Breadth

=50 \(\times\) 30=1500 \(\mathrm{~cm}^2=0.15 \mathrm{~m}^2\)

Pressure \(p_2=\frac{F}{A}=\frac{58.8}{0.15}=392 \mathrm{~N} / \mathrm{m}^2\)

Some Daily Life Applications of Pressure

1. The handles of bags, suitcases, etc., are made broad so that less pressure is exerted on the hand.
2. Buildings are provided with broad foundations so that the pressure exerted on the ground becomes less.
3. Railway tracks are laid on cement or iron sleepers so that the pressure exerted by the train could spread over a larger area and thus pressure decreases.
4. Pins, needles and nails are provided with sharp pointed ends to reduce the area and hence to increase the pressure.
5. Cutting tools have sharp edges to reduce the area so that with lesser force, more pressure can be exerted.
6. Pressure on the ground is more when a man is walking than when he is standing because, in the case of walking, the effective area is less.
7. Depression is much more common when a man stands on the cushion than when he lies down on it because in the standing case, the area is lesser than in the case of lying.
8. The tractors have broad tyres, to create less pressure on the ground so that tyres do not sink into comparatively soft ground in the field.

Pressure in Fluids

All liquids and gases are together called fluids. Water and air are the two most common fluids. Solids exert pressure on a surface due to their weight.

Fluids also have weight, therefore fluids also exert pressure on the base and walls of the container in which they are enclosed. Fluids exert pressure in all directions.

Buoyancy

1. The tendency of a liquid to exert an upward force on an object immersed in it is called buoyancy. Gases also exhibit this property of buoyancy.
2. Buoyant Force is an upward force which acts on an object when it is immersed in a liquid. It is also called upthrust.
3. It is the buoyant force due to which a heavy object seems to be lighter in water. As we lower an object into a liquid, the liquid underneath it provides an upward force.
4. For Example, A piece of cork is held below the surface of water. When we apply a force by our thumb, the cork immediately rises to the surface. This is due to the fact that every liquid exerts an upward force on the objects immersed in it.

Factors Affecting Buoyant Force

The magnitude of buoyant force depends on the following factors:

(1) Density of the Fluid

The liquid having higher density exerts more upward buoyant force on an object than another liquid of lower density. This is the reason why it is easier to swim in seawater in comparison to normal water.

The Sea water has higher density and hence exerts a greater buoyant force on the swimmer than the freshwater having lower density.

(2) Volume of Object Immersed in the Liquid

As the volume of a solid object immersed inside the liquid increases, the upward buoyant force also increases. The magnitude of buoyant force acting on a solid object does not. depend on the nature of the solid object. It depends only on its volume.

For Example, When two balls made of different metals having different weights but equal volume are fully immersed in a liquid, they will experience an equal upward buoyant force as both the balls displace an equal amount of the liquid due to their equal volumes.

Floating or Sinking of Objects in Liquid

When an object is immersed in a liquid, the following two forces act on it:

Weight of the object which acts in a downward direction, i.e. it tends to pull down the object.

Buoyant force (upthrust) which acts in an upward direction, i.e. it tends to push up the object.

Whether an object will float or sink in a liquid, depends on the relative magnitudes of these two forces which act on the object in opposite directions. There are three conditions of floating and sinking of objects. These are:

1. If the buoyant force or upthrust exerted by the liquid is less than the weight of the object, the object will sink into the liquid.
2.  If the buoyant force is equal to the weight of the object, the object will float in the liquid.
3. If the buoyant force is more than the weight of the object, the object will rise in the liquid and then float.

Density

1. The density of a substance is defined as mass per unit volume.
2. Density =\(\frac{\text { Mass of the substance }}{\text { Volume of the substance }}\) or \(\rho=\frac{m}{V}\)
3. The SI unit of density is kilogram per metre cube (kg/m³). It is a scalar quantity. The density of a substance under specified conditions always remains the same.
4. Hence, the density of a substance is one of its characteristic properties. It can help us to determine its purity. It is different for different substances. The lightness and the heaviness of different substances can be described by using the word density.
5. Objects having a density less than that of a liquid, float on the liquid. Objects having greater density than that of liquid, sink in the liquid. It decreases with an increase in temperature.

Example 12. A sealed can of mass 700 g has a volume of 500 cm³. Will this can sink in water? (The density of water is 1 g cm^-3)

Answer:

Given, the mass of the can, m=700 g

Volume of can, V=500 \(\mathrm{~cm}^3\)

Density of can, \(\rho=\frac{m}{V}\)

=\(\frac{700}{500}=1.4 \mathrm{~g} \mathrm{~cm}^{-3}\)

Since the density of the can is greater than the density of the water, the can will sink into the water.

Archimedes’ Principle

The statement ‘When an object is fully or partially immersed in a liquid, it experiences a buoyant force or upthrust, which is equal to the weight of the liquid displaced by the object’, i.e.

Buoyant force or upthrust acting on an object = Weight of liquid displaced by the object

Even gases like air, exert an upward force or buoyant force on the objects placed in them. It is buoyant force or upthrust due to displaced air which makes a balloon rise in air.

Applications of Archimedes’ Principle

1. Archimedes’ principle is used in:
2. Designing ships and submarines.
3. lactometer (a device used to determine the purity of milk). hydrometer (a device used for determining the density of liquid).

Example 13 If an iron object is immersed in water, it displaces 8 kg of water. How much is the buoyant force acting on the iron object in Newton? (Given, g = 10 ms^-2)

Answer:

According to Archimedes’ principle, “the buoyant force acting on this iron object will be equal to the weight of water displaced by this iron object.”

We know that, weight, w = mx g. Here, the mass of water, m= 8kg

Acceleration due to gravity, g =10 ms^-2

On putting values, w = 8 x 10 = 80 N

Since the weight of water displaced by the iron object is 80 N, therefore the buoyant force acting on the iron object (due to water) will also be 80 N.

Activity Zone

Activity 1

Objective: To describe a circular path in the motion of a stone.

Materials Required:

Thread and stone.

Procedure

1. A stone describing a circular path with a velocity of constant magnitude
2. Take a piece of thread.
3. Tie a small stone at one end and hold the other end of the thread. Whirl it as shown in the figure.
4.  Observe the motion of the stone.
5. Release the thread.
6. Again note the direction of motion of the stone.

Observation

1. The stone moves in a circular path with a certain speed and changes direction at every point.
2. The change in direction involves a change in velocity or acceleration.
3. The force that causes this acceleration and keeps the body moving along the circular path is acting towards the centre.
4. This force is called centripetal force. When the thread is released, the stone makes a tangent to the circle and falls down.

Question 1. Name the motion of the stone.
Answer:

The motion of the stone is along a circular path.

Question 2. What happens to the stone if centripetal force vanishes?
Answer:

Due to the absence of centripetal force, the stone flies off along a tangent of a circle.

Question 3. Name the straight line that meets the circle at one point.
Answer:

The tangent is a straight line that meets the circle at one point.

Activity 2

Objective: To study free-falling or rising bodies.

Material Required: Stone

Procedure

1. Take a stone.

2. Throw it upwards.

3. It reaches a certain height and then it starts falling down.

Observations

1. The stone is attracted by the earth hence it falls down due to the earth’s gravitational force.
2. The stone gets accelerated, when it goes upwards, g is negative when it falls towards the earth, g is positive.

Question 1. Which force accelerates a body in free fall?
Answer:

The force of gravity on earth accelerates a body in free fall.

Question 2. What is the direction of motion of an object if g is taken negative?
Answer:

The direction of motion of an object is upwards if g is taken negative.

Question 3. What is the direction of acceleration due to gravity?
Answer:

The direction of acceleration due to gravity is always downward.

Activity 3

Objective: To study any object that may be hollow or solid, big or small should fall at the same rate.

Materials Required

1. Glass jar
2. One sheet of paper

Procedure

1. Take a sheet of paper and a stone.
2. Drop them simultaneously from the first floor of a building.
3. Observe whether both of them reach the ground simultaneously.
4. Also, repeat in a glass jar and then observe what happens.

Observation

1. The paper reaches the ground a little later than the stone. This is due to air resistance. The air offers resistance due to friction to the motion of the falling objects.
2. The resistance offered by air to the paper is more than the resistance offered to the stone.
3. In a glass jar, the air is sucked out, the paper and the stone falls at the same rate. When the earth attracts the paper sheet and crumbled paper, they experience the same acceleration during free fall.
4. This acceleration experienced by an object is independent of its mass. Thus, any object whether it be hollow or solid, big or small should fall at the same rate.

Question 1. Which one reaches the ground first?
Answer:

The stone reaches the ground first.

Question 2. The paper reaches the ground later, why?
Answer:

The paper reaches the ground later due to air resistance.

Question 3. What happens when air is sucked in along a glass jar?
Answer:

When air is sucked into a long glass jar both stone and paper reach at the bottom at the same time.

Activity 4

Objective: To understand buoyant force.

Materials Required: Plastic bottle, stopper, bucket and water.

Procedure

1. Take an empty plastic bottle. Close the mouth of the bottle with an airtight stopper. Put it in a bucket filled with water. You will see that the bottle will float.
2. Push the bottle into the water. You will feel an upward push. Try to push it further down.
3. You will find it difficult to push deeper and deeper. This indicates that water exerts a force on the bottle in the upward direction. The upward force exerted by the water goes on increasing as the bride is pushed deeper till it is completely immersed.
4. Now, release the bottle. It bounces back to the surface.

Observation

1. The force due to the gravitational attraction of the Earth acts on this bottle in a downward direction. But the water exerts an upward force on the bottle. This upthrust is known as buoyant force.
2. When the bottle is immersed, the upward force exerted by the water on the bottle is greater than its weight. Therefore, it rises up when released.
3. To keep the bottle completely immersed, the upward force on the bottle due to water must be balanced. This can be achieved by an externally applied force acting downwards.
4. This force must at least be equal to the difference between the upward force and the weight of the bottle.

Conclusion

It is the buoyant force which keeps the object float.

Question 1. When a body is fully immersed in a liquid, what happens to its weight?
Answer:

When a body is fully immersed in a liquid, its weight decreases due to buoyant force.

Question 2. What are the factors affecting buoyant force?
Answer:

The buoyant force depends on

1.  the density of the liquid
2. the volume of the solid immersed in liquid.

Question 3. What happens, if the density of an object is more than the density of the liquid in which it is immersed?
Answer:

The object will sink into the liquid.

Question 4. Who exerts buoyant force?
Answer:

The buoyant force is exerted by the liquid molecules on the body when it is dipped in that liquid.

Question 5. By which force does a body start floating?
Answer:

Buoyant force helps the body to float in a liquid in which it is immersed.

Activity 5

Objective: To know why objects float or sink when placed on the surface of water.

Materials Required: Two beakers filled with water, an iron nail and a piece of cork.

Procedure

1. Take a beaker filled with water.
2. Take an iron nail.
3. Place it on the surface of the water.
4. Take another beaker filled with water.
5. Take a piece of cork and an iron nail of equal mass.
6. Place them on the surface of the water.

Observation

1. The nail sinks.
2. The cork floats (as shown alongside).

Reasons

1. The force due to the gravitational attraction of the Earth on the iron nail, pulls it downwards. There is an upthrust of water on the nail, which pushes it upwards.
2. However, the downward force acting on the nail is greater than the upthrust of water on the nail. So, it sinks. This happens because of the difference in their densities.
3. The density of cork is less than the density of water. This means that the upthrust of water on the cork is greater than the weight of the cork. So, it floats.
4. The density of an iron nail is more than the density of water. This means that the upthrust of water on the iron nail is less than the weight of the nail. So, it sinks.

Conclusion

The object of density less than that of a liquid floats on the surface of the liquid. Whereas, the object of density greater than that of a liquid sinks in the liquid.

Question 1. What happens to buoyant force, when we increase the density of the fluid?
Answer:

The buoyant force increases with an increase in the density of the fluid.

Question 2. What is the condition for floating of a body?

Answer:

If the density of a body is less than that of the liquid in which it is immersed, then the body floats in liquid.

Question 3. Give the condition for the sinking of a body.

Answer:

If the density of the body is greater than that of the liquid, it sinks into the liquid.

Question 4. If an object floats with \(\frac{1}{9}, \frac{2}{11}\) and \(\frac{3}{7}\) parts of its volume
Answer:

1. Outside the surface of the liquid of densities \(d_1, d_2\) and \(d_3\). Then, arrange the densities in increasing order.
2. Upthrust due to liquid on an object is directly proportional to the density of the liquid. Therefore, densities in increasing order are of \(d_1<d_2<d_3\)

Question 5. A nail sinks while a cork floats. Why?
Answer:

The density of the nail is greater than that of liquid, so it sinks. But the density of cork is less than that of water, so it floats.

Gravitation Question And Answer

Question 1. State the universal law of gravitation.
Answer:

The universal law of gravitation states that the force of attraction between any two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. This law is applicable to any two objects anywhere in the universe.

Question 2. Write the formula for the magnitude of the gravitational force between the earth and an object on the surface of the earth.
Answer:

The formula for the magnitude of the gravitational force between the earth and an object on the surface of the earth is given by  F=G \(\frac{M m}{R^2}\)

where, M = mass of the earth

m = mass of an object

G = gravitational constant

= \(6.67 \times 10^{-11} \mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2\)

and R = distance between the centres of the earth and an object.

Question 3. What do you mean by free fall?
Answer:

The falling of a body from a height towards the earth under the gravitational force of the earth is called free fall. Hence, the motion of a particle falling down or going up under the action of gravity means the body is in free fall.

Question 4. What do you mean by acceleration due to gravity?
Answer:

The acceleration of a body during free fall towards a celestial body is called acceleration due to gravity. Its value is 9.8 m/s².

For objects on or near the surface of the earth,

m g=G \(\frac{M \times m}{R^2}\)

where, g = acceleration due to gravity

M = mass of the earth

m = mass of an object

and R = radius of the earth

Hence, g=\(\frac{G M}{R^2}\)

Question 5. What are the differences between the mass of an object and its weight?
Answer:

Differences between the mass of an object and its weight are as follows:

Question 6. Why is the weight of an object on the moon 1/6 its weight on the earth?
Answer:

Weight of an object, w = mg [where, m = mass of an object and g — acceleration due to gravity]

The mass of an object m remains constant at all places. Acceleration due to gravity changes from place to place. So, we can say that the weight of an object depends on the acceleration due to gravity.

On the moon, the acceleration due to gravity is l/6th that of the earth, this is the reason why the weight of an object on the moon is 1 /6th its weight on the earth.

Question 7. Why is it difficult to hold a school bag having a strap made of thin and strong string?
Answer:

It is difficult to hold a school bag having a strap made of a thin and strong string because the area under the strap is small. Hence, a large pressure is exerted by the strap on the fingers or shoulders. Due to the large pressure, the strap tends to cut the skin and hence pain is caused.

Question 8. What do you mean by buoyancy?
Answer:

The upward force exerted by a liquid on any object immersed in it is called buoyancy or upthrust.

Question 9. Why does an object float or sink when placed on the surface of water?
Answer:

When an object is placed on the surface of water, two forces act on the object.

1. The weight of the object acts vertically downwards.
2. The upthrust of the water acts vertically upwards.

The object will float on the surface of the water if the upthrust is greater than the weight of the object. The object will sink if the weight of the object is more than the upthrust of the water.

Question 10. You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?
Answer:

Mass is more than 42 kg. As the buoyant force due to air is acting on us, the reading of the weighing machine will be less than the actual mass of a person.

Question 11. You have a bag of cotton and an iron bar, each indicating a mass of 100 kg when measured on a weighing machine. In reality, one is heavier than the other. Can you say which one is heavier and why?
Answer:

The cotton bag is heavier as the buoyant force acting on the cotton bag is more as its surface area is more than that of the iron piece.

Exercises

Question 1. How does the force of gravitation between two objects change when the distance between them is reduced to half?
Answer:

The force of gravitation between two objects is given by

F=G \(\frac{m_1 m_2}{r^2}\)

If the distance is reduced to half, i.e. \(r^{\prime}=r / 2\). Then, a new force of gravitation,

⇒ \(F^{\prime}=\frac{G m_1 m_2}{r^{\prime 2}}=\frac{G m_1 m_2}{(r / 2)^2}=4 \times \frac{G m_1 m_2}{r^2}=4 F\)

i.e. The force of gravitation becomes 4 times the original value.

Question 2. Gravitational force acts on all objects in proportion- to their masses. Why, then a heavy object does not fall faster than a light object?
Answer:

Acceleration due to gravity (g) is independent of the mass of the falling object and is equal for all objects at a point. So, a heavy object falls with the same acceleration as a light object.

Question 3. What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Take, the mass of the earth is 6 \(\times 10^{24}\) kg and the radius of the earth is 6.4 \(\times 10^6 \mathrm{~m}\).)
Answer:

The gravitational force between the earth and an object is given by F=\(\frac{G M m}{R^2}\).

where, G = gravitational constant

=6.67 \(\times 10^{-11} \mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2\)

M = mass of the earth =6 \(\times 10^{24} \mathrm{~kg}\)

R = radius of the earth =\(6.4 \times 10^6 \mathrm{~m}\)

and m = mass of an object =1 kg

F =\(\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 1}{\left(6.4 \times 10^6\right)^2}\)

=\(9.77 \approx 9.8 \mathrm{~N}\)

Thus, the magnitude of the gravitational force between the earth and a 1 kg object is 9.8 N.

Question 4. The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater than smaller than or equal to the force with which the moon attracts the earth? Why?
Answer:

When two objects attract each other, then the gravitational force of attraction applied by the first object on the second object is the same as the force applied by the second object on the first object. So, both the earth and moon attract each other by the same gravitational force of attraction.

Question 5. If the moon attracts the earth, then why does the earth not move towards the moon?
Answer:

The earth does not move towards the moon in spite of the attraction by the moon because the mass of the earth is much greater than the mass of the moon and for a given force, acceleration is inversely proportional to the mass of the object.

Question 6. What happens to the force between two objects, if

1. the mass of one object is doubled?
2. the distance between the objects is doubled and tripled?
3. the masses of both objects are doubled?

Answer:

The force of attraction between two objects is given by

F=\(\frac{G m_1 m_2}{r^2}\)

where, \(m_1\) and \(m_2\) = masses of the objects

r= distance between the objects

and

G= gravitational constant

(1) If the mass of one object is doubled, then the new force,

\(F^{\prime} =\frac{G\left(2 m_1\right) m_2}{r^2}\)

=2 \(\times \frac{G m_1 m_2}{r^2}\)=2 F  i.e. Force becomes double.

(2) If the distance between the objects is doubled, then the new force,

⇒ \(F^{\prime}=\frac{G m_1 m_2}{(2 r)^2}=\frac{G m_1 m_2}{4 r^2}=\frac{1}{4} \cdot \frac{G m_1 m_2}{r^2}\)

=\(\frac{F}{4}\)

i.e. Force becomes one-fourth. If the distance between the objects is tripled, then new force,

⇒ \(F^{\prime} =\frac{G m_1 m_2}{(3 r)^2}=\frac{G m_1 m_2}{9 r^2}\)

= \(\frac{1}{9}\left(\frac{G m_1 m_2}{r^2}\right)=\frac{F}{9}\)

i.e. Force becomes one-ninth.

(3) If the masses of both objects are doubled, then the new force,

⇒ \(F^{\prime} =\frac{G\left(2 m_1\right)\left(2 m_2\right)}{r^2}\)

=4 \(\times \frac{G m_1 m_2}{r^2}=4 F\)

i.e. Force becomes four times.

Question 7. Calculate the force of gravitation between the earth and the sun given that the mass of the earth = 6 \(\times 10^{24}\) kg and of the sun = 2 \(\times 10^{30}\) kg. The average distance between the two is 1.5 \(\times 10^{11}m\).
Answer:

The force of attraction between the Earth and the sun is given by

F=\(\frac{G M_s M_e}{r^2}\)

where, G=6.67 \(\times 10^{-11} \mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2\)

Mass of the sun, \(M_s=2 \times 10^{30} \mathrm{~kg}\)

Mass of the earth, \(M_e=6 \times 10^{24} \mathrm{~kg}\)

Average distance between the Earth and the sun

F =\(\frac{6.67 \times 10^{-11} \times 2 \times 10^{30} \times 6 \times 10^{24}}{\left(1.5 \times 10^{11}\right)^2}=1.5 \times 10^{11} \mathrm{~m}\)

=3.6 \(\times 10^{22} \mathrm{~N}\)

Thus, the force between the earth and the sun is 3.6 \(\times 10^{22} \mathrm{~N}\).

Question 8. What is the importance of the universal law of gravitation?
Answer:

1. The universal law of gravitation successfully explained several phenomena given below:
2.  The force that binds us to the earth.
3. The motion of the moon around the earth.
4. The motion of planets around the sun.
5. The tides are due to the moon and the sun.
6. The flow of water in rivers is also due to the gravitational force of the earth on water.

Question 9. What is the acceleration of free fall?
Answer:

The acceleration of free fall is the acceleration produced in the motion of an object when it falls freely towards the Earth. It is also called acceleration due to gravity. Its value on the earth’s surface is 9.8 m/s².

Question 10. What do we call gravitational force between the earth and an object?
Answer:

The gravitational force between the earth and an object is called the force of gravity or gravity.

Question 11. Amit buys a few grams of gold at the poles as per the instruction of one of his friends. He hands over the same when he meets him at the equator. Will the friend agree with the weight of gold bought? If not, why?

[Hint The value of g is greater at the poles than at the equator.]

Answer:

No, his friend will not agree with the weight of gold. w = mg

⇒ \(g \propto \frac{1}{R^2}\)

The value of g is greater at the poles than at the equator. Therefore, gold at the equator weighs less than that at the poles. Thus, Amit’s friend will not agree with the weight of the gold bought.

Question 12. Why does a sheet of paper fall slower than one that is crumpled into a ball?
Answer:

The sheet of paper falls slower than one that is crumpled into a ball because in the first case, the area of the sheet is more, so it experiences a large opposing force due to air. In contrast, the sheet crumpled into a ball experienced less opposing force due to the small area. This opposing force arises due to air resistance or air friction.

Question 13. The gravitational force on the surface of the moon is only 1/6 as strong as gravitational force on the earth. What is the weight in Newton of a 10 kg object on the moon and on the earth?
Answer:

Given, the mass of the object, m=10 kg

Weight on the earth, w=m g=10 \(\times 9.8=98 \mathrm{~N}\)

Weight on the moon =\(\frac{1}{6}\) of the weight on the earth

=\(\frac{1}{6} \times\) 98=16.33 N

Question 14. A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate

1. the maximum height to which it rises.
2. the total time it takes to return to the surface of the earth.

Answer:

Given, initial velocity, u = 49 m/s

(1) At the maximum height velocity becomes zero.

Final velocity, v = 0

⇒ \(v^2 =u^2-2 g h\)

0 =\((49)^2-2 \times 9.8 \times h\)

h =\(\frac{(49)^2}{2 \times 9.8}=122.5\) m

Maximum height attained =122.5 m

(2) Time taken by the ball to reach the maximum height.

From the first equation of motion, v=u-g t or

0=49-9.8 \(\times\) t

t=\(\frac{49}{9.8}\)=5 s

For the motion against gravity, the time of descent is the same as the time of ascent. So, the time taken by the ball to fall from maximum height is 5 s

Total time taken by the ball to return to the surface of the earth =5+5=10 s.

Question 16. A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g=10 m/s2, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?
Answer:

Given, initial velocity, u = 40 m/s

Final velocity becomes zero, i.e. v = 0 [at maximum height]

From the third equation of upward motion,

⇒ \(v^2 =u^2-2 g h\)

⇒ \((0)^2 =(40)^2-2 \times 10 \times h\)

0 =1600-20 h

h =\(\frac{1600}{20}=80 \mathrm{~m}\)

The maximum height reached by the stone = 80 m.

After reaching the maximum height, the stone will fall towards the earth and will reach the earth’s surface covering the same distance.

So, the distance covered by the stone = 80 + 80 = 160 m. Displacement of the stone = 0.

Because the stone starts from the earth’s surface and finally reaches the earth’s surface again, i.e. the initial and final positions of the stone are the same.

Question 17. A stone is allowed to fall from the top of a tower 100 m high and at the same time, another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.
Answer:

Let after time t both stones meet and s be the distance travelled by the stone dropped from the top of the tower at which the stones will meet.

Distance travelled by the stone dropped = s

Distance travelled by the stone projected upwards = (100-5) m

For the stone dropped from the tower,

s=u t+\(\frac{1}{2} g t^2=0+\frac{1}{2}(10) t^2\)

[ u=0 because the stone is dropped, i.e. it starts from rest] s=5 \(t^2\)

For the stone projected upwards, \(s^{\prime}=u t-\frac{1}{2} g t^2\)

[due to upward motion, negative sign is taken] (100-s) =25 \(t-\frac{1}{2} \times 10 t^2\)

100-s = \(25 t-5 t^2\)

From Eq. (1), we get

⇒ \(100-5 t^2=25 t-5 t^2\)

25 t=100

t=4 s

So, the stones will meet after 4 s.

s=5 \(t^2=5 \times(4)^2=80 \mathrm{~m}\)

So, the stones will be at a distance of 80 m from the top of the tower or 20 m (100 m – 80 m) from the base of the tower.

Question 18. A ball thrown up vertically returns to the thrower after 6 seconds, find

1. the velocity with which it was thrown up,
2. the maximum height it reaches and
3. its position after 4 s.

Answer:

Total time taken =6 s

Time taken to reach the maximum height = \(\frac{6}{2}\) = 3 s [ time of ascent = time of descent]

(1) From the first equation of motion, v = u = gt [negative sign is taken due to upward motion]

0 =U -9.8 x 3 [v at maximum height, v = 0]

⇒ u- 29.4 m/s

(2) From the third equation of motion,

⇒ \(v^2=u^2-2\) g h [negative sign is taken due to upward motion]

0 =\((29.4)^2-2 \times 9.8 \times h\)

h =\(\frac{(29.4)^2}{2 \times 9.8}=44.1 \mathrm{~m}\)

The maximum height attained by the ball is 44.1 m.

(3) In the initial 3 s, the ball will rise, and then in the next 3 s it falls toward the earth.

The position after 4 s

= Distance covered in Is in the downward motion

From the second equation of motion,

h=u t+\(\frac{1}{2} g t^2=0+\frac{1}{2} \times 9.8 \times(1)^2=4.9 \mathrm{~m}\)

i.e. The ball will be at 4.9 m below the top of the tower or the height of the ball from the ground will be at (44.1-4.9) =39.2

Question 19. In what direction, does the buoyant force on an object immersed in a liquid act?
Answer:

The buoyant force on an object immersed in a liquid always acts in the vertically upward direction.

Question 20. Why does a block of plastic released under water, come up to the surface of water?
Answer:

The upthrust or buoyant force acting on the block of plastic by the water is greater than the weight of the plastic block. So, the plastic block comes up to the surface of the water.

Question 21. The volume of 50 g of a substance is 20 \(\mathrm{~cm}^3\). If the density of water is 1 g \(\mathrm{cm}^{-3}\), will the substance float or sink?
Answer:

Given, the mass of the substance, m = 50 g

Volume of substance, V=20 \(\mathrm{~cm}^3\)

Density of substance,

⇒ \(\rho=\frac{\text { Mass }}{\text { Volume }}=\frac{50}{20}=2.5 \mathrm{~g} \mathrm{~cm}^{-3}\)

i.e. The density of the substance is greater than the density of water, so it will sink in water.

Question 22. The volume of a 500 g sealed packet is 350 cm³. Will the packet float or sink in water, if the density of water is 1 g \(\mathrm{~cm}^{-3}\)? What will be the mass of the water displaced by this packet?
Answer:

Given, the mass of the packet =500 g

Volume of packet =350 \(\mathrm{~cm}^3\)

Density of packet, \(\rho=\frac{\text { Mass }}{\text { Volume }}\)

=\(\frac{500 \mathrm{~g}}{350 \mathrm{~cm}^3}\)

=1.43 \(\mathrm{~g} \mathrm{~cm}^{-3}\)

i.e. The density of the packet is greater than the density of the water, so it will sink in the water.

The mass of water displaced by the packet

= Volume of packet x Density of water =350 x 1 = 350 g

Summary

• Gravitation is defined as the non-contact force of attraction between any two bodies in the universe.
• According to the universal law of gravitation, the attractive force between any two objects in the universe is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
• Mathematically, F = \(\frac{G M m}{d^2}\) where G is called the universal gravitational constant and its value is 6.67 \(\times 10^{-11} \mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2\)
• Free fall Whenever objects falls towards the earth under the earth’s gravitational force alone, then such a motion is called free fall.
• The acceleration with which an object falls towards the earth due to the earth’s gravitational pull is called acceleration due to gravity. It is denoted by g.
• At the surface of earth, g = \(\frac{G M}{R^2}\)
• The SI unit of g is \(\mathrm{ms}^{-2}\) and its value is 9.8 \(\mathrm{~m} / \mathrm{s}^2\).

Equations of motion for freely falling bodies

v =u+g t

h =\(u t+\frac{1}{2} g t^2\)

⇒ \(v^2 =u^2+2 g h\)

• where h is the height from which the object falls, t is the time of fall, u is the initial velocity and v is the final velocity when the body accelerates at g.
• The total amount of matter contained in an object is called its mass.
• The SI unit of mass is kilogram (kg). Mass is a scalar quantity.
• The weight of an object is the force with which it is attracted towards the earth i.e. weight of an object, w = mg The SI unit of weight is Newton (N) Weight is a vector quantity.
• The weight of an object on the moon is 1 / 6 th of its weight on the earth.
• Thrust is the force, acting on an object perpendicular to its surface.
• The SI unit of thrust is Newton (N). Thrust is a vector quantity.
• Pressure is the force acting perpendicularly on a unit area of an object.
• Pressure can be calculated as, (p)=\(\frac{\text { Force }(F)}{\text { Area }(A)}=\frac{\text { Thrust }}{\text { Area }}\)
• The SI unit of pressure is \(\mathrm{Nm}^{-2}\) or pascal(Pa).
• Pressure is a scalar quantity.
• All the liquids and gases are called fluids.
• Buoyant force is an upward force, which acts on an object when it is immersed in a liquid. It is also called upthrust. Factors Affecting the Buoyant Force Density of the Fluid.
• The volume of the object immersed in the liquid.
• If the buoyant force exerted by the liquid is less than the weight of the object, the object will sink in the liquid.
• If the buoyant force is equal to the weight of the object, the object will float in the liquid.
• If the buoyant force is more than the weight of the object, the object will rise in the liquid and then float.
• The density of a substance is defined as mass per unit volume. Density = \(\frac{\text { Mass of the substance }}{\text { Volume of the substance }}\)
• The SI unit of density is kilogram per metre cube (kg/m3) and it is a scalar quantity.
• According to Archimedes’ principle, when an object is fully or partially immersed in a liquid, it experiences a buoyant force or upthrust which is equal to the weight of liquid displaced by the object.

Work, Energy And Power Multiple Choice Question And Answers

Question 1. When a horse-pulls a cart, who does the work ?

1. Cart
2. Wheels
3. Road
4. Horse

Answer: 4. Horse

When a horse-pulls a cart, work is being done by horse.

Question 2. If 10 N of force is applied to an object, but the object does not move, then how much work being done by the force ?

1. Zero
2. 10 J
3. 10N
4. 20 J

Answer: 1. Zero

If there is no displacement due to application of force, then net work done will be zero.

Question 3. A 4 N of force displaces a body by 2m, the work done will be

1. 5 J
2. 8 J
3. 8 N
4. 5 N

Answer: 2. 8 J

Work done by given force =force x displacement = 4 x 2 =8J

Question 4. A car is moving uniformly in a circular racing track of radius 1 km. If the car makes two complete round. How much work is done by the car, if car has a force of 20 kN?

1. Zero
2. 20 kJ
3. 200 kJ
4. 2000 kJ

Answer: 1. Zero

During a circular motion, force is perpendicular to the displacement, so net work done is zero.

Question 5. What kind of energy is possessed by a running horse?

1. Kinetic energy
2. Potential energy
3. No energy
4. Heat energy

Answer: 1. Kinetic energy

A running horse has kinetic energy.

Question 6. If two stones A and B are dropped from a tower, then which one has maximum kinetic energy ?

1. Lighter stone
2. Heavier stone
3. Both have equal
4. None of the above

Answer: 2. Heavier stone

If two stones are dropped from some height, then the heavier stone has greater kinetic energy w.r.t. lighter one.

Question 7. A monkey weighing 50 kg climbs up a vertical tree of height 300 m. How much potential energy does it gain? [Take, g=9.8 m / \(s^2\) ]

1. 147 \(\times 10^4 \mathrm{~J}\)
2. 14.7 \(\times 10^3 \mathrm{~J}\)
3. 14.7 \(\times 10^2 \mathrm{~J}\)
4. 14.7 \(\times 10^4 \mathrm{~J}\)

Answer: 4. 14.7 \(\times 10^4 \mathrm{~J}\)

Given, m=50 kg, g=9.8 \(ms^{-2}\), b=300 m

Work done by the body = mgh

=50 \(\times 9.8300=14.7 \times 10^4 \mathrm{~J}\)

Gain in PE = Work done =14.7 \(\times 10^4\) J

Question 8. A scooter of mass 150 kg is travelling at 10 m/s. If its speed increases to 15 m/s, then by how much amount does its kinetic energy increases ?

1. 9500 J
2. 9425 J
3. 9375 J
4. 9753 J

Answer: 3. 9375 J

Given, m=150 kg, \(v_1\)=10 m / s and \(v_2\)=15 m/ s

The change in kinetic energy

=\(\mathrm{KE}_2-\mathrm{KE}_1=\frac{1}{2} m v_2^2-\frac{1}{2} m v_1^2\)

=\(\frac{1}{2} \times 150 \times 15^2-\frac{1}{2} \times 150 \times 10^2\)

= 16875-7500= 9375 J

Question 9. A ball is allowed to fall freely from a tower. Which energy is possessed at the middle point during the fall ?

1. Kinetic only
2. Potential only
3. Both (1) and (2)
4. Heat only

Answer: 3. Both (1) and (2)

At middle point of fall, the ball has both kinetic and potential energy.

Question 10. In head phone, the electrical energy is converted to what ?

1. Light energy
2. Sound energy
3. Chemical energy
4. Heat energy

Answer: 2. Sound energy

Electrical energy changes into sound energy.

Question 11. Which mathematical relation of energy of a stone of mass m fading freely from height h remains conserved at every point in its downward motion?

1. E = mgh
2. E=\(\frac{1}{2} m v^2\)
3. E=\(m g h+\frac{1}{2} m v^2\)
4. None of these

Answer: 3. E=\(m g h+\frac{1}{2} m v^2\)

Total energy ot the free hilling stone is conserved.

So, total energy = kinetic energy + potential energy.

Or E=\(m g+\frac{1}{2} m v^2[latex]

Question 12. If a stone of mass 5 kg drops from height 20 m, then what will be the velocity at the surface of earth ? (hike, g = 10 m/s²)

1. 20 m/s
2. 30 m/s
3. 35 m/s
4. 40 m/s

Answer: 1. 20 m/s

Given, mass of stone, m = 5 kg

Height, h = 20 m and g =10 m/s²

So, potendal energy at highest point = mgh

= 5 x 10 x 20 = 1000 J

Now, according to conservation of energy,

kinetic energy at surface of the earth = potential energy at height h

So, 1/2 m [latex]v^2\) =\(m g h^h\)

⇒ \(v^2 =\frac{2 m g^b}{m}\)

v =\(\sqrt{2 g^b}=\sqrt{2 \times 10 \times 20}=20 \mathrm{~m} / \mathrm{s}\)

Question 13. Mukesh drops a ball of 200 g from a tower of height 20 m. What will be its kinetic energy at the height of 5 m ?

1. 30 J
2. 50 J
3. 60 J
4. 65 J

Answer: 1. 30 J

Given, m = 200 g = 0.2 kg, height, h = 20 m

When ball reaches at height of 5 m, its velocity is given by

⇒ \(v^2=v^2+2 g\)

⇒ \(v^2=0^2+2 \times 10 \times 15\)

v=\(\sqrt{300}=\sqrt{100 \times 3}=10 \sqrt{3}\)

So, kinetic energy \(\left.=\frac{1}{2} m v^2=\frac{1}{2} \times 0.2 \times 300=30\right]\)

Question 14. An object of mass 50 kg is raised to a height of 7 m above the ground. What is its potential energy? If object is allowed to fall, then its kinetic energy when it is half way down.

1. 3500 J, 1650 J
2. 3800 J, 1750 J
3. 3800 J, 1650 J
4. 3500 J, 1750 J

Answer: 4. 3500 J, 1750 J

Given, mass, m = 50 kg, height, h = 7 m

As, potential energy is given by PR = mgh =50x 10x 7 = 3500J

When it is allowed to fall its potential energy gets converted into kinetic energy.

So, when it reaches the half-way, half of its potential energy gets converted to kinetic energy.

Question 15. A horse does 5000 J of work in 100 s. What is its power ?

1. 50 W
2. 50 J
3. 10W
4. 10J

Answer: 1. 50 W

Given, work done (W) = 5000 J

Time taken (r) =100 s

We know that,

Power,P=\(\frac{W}{t}=\frac{5000}{100}=50 \mathrm{~J} / \mathrm{s}\)

Thus, power, P=50 W

Question 16. The power uses of household water pump is 200 W. What is the mass of water, the pump can lift per minute from a 7 m deep tank? [Take, g =10 m/s²]

1. 172 kg
2. 272 kg
3. 150 kg
4. 250 kg

Answer: 1. 172 kg

Given,power,.P = 200 W^ = 60 s,height,h = 7m

Energy supplied to the pump = Power x Time = 200 Wx 60s =12000J

So, energy =mgh

12000 = m x 10 x 7

m = 172 kg

So, mass of the water is 172 kg.

Question 17. A certain household has consumed 200 units of energy during a month. How much energy is then in joules?

1. 72 \(\times 10^5 \mathrm{~J}\)
2. 72 \(\times 10^7 \mathrm{~J}\)
3. 7.2 \(\times 10^8 \mathrm{~J}\)
4. 72 \(\times 10^{\mathrm{B}} \mathrm{J}\)

Answer: 3. 7.2 \(\times 10^8 \mathrm{~J}\)

Given, energy consumed = 200 units

= 200 kWh = 200x 1000×3600

= 7.2 x 108 J

Question 18. Ankit does 500 J of work in 15 min and Bablu does 700 J of work in 30 min. Who expends more power, Ankit or Bablu ?

1. Ankit
2. Bablu
3. Both same
4. Insufficient information

Answer: 1. Ankit

Power expended by Ankit =\(\frac{W}{t}=\frac{500}{15 \times 60}\)=0.55J

and power expended by Bablu =\(\frac{700}{30 \times 60}\)=0.38 J

Ankit expends more power.

Class 9 Science Chapter 10 Work, Energy And Power Very Short Answer Type Questions

Question 1. What is the work done when you apply an ION force on a wall?
Answer:

Zero work because there is no displacement of the wall.

Question 2. Why do we say work done against gravity is negative?
Answer:

It is because force and displacement are in opposite directions to each other.

Question 3. A man is holding a suitcase in his hand at rest. What is the work done by him?
Answer:

Zero, as displacement is zero.

Question 4. A boy is running along a circular path at a uniform speed. What is the work done by the boy?
Answer:

Work done by the boy is zero because the angle between the force (centripetal) and displacement in case of circular motion is 90°.

Question 5. Think of two situations for each of the following given conditions.

1. Large mass but low kinetic energy.
2. Small mass but high kinetic energy.

Answer:

1. A shot put thrown by an athlete.
2. A hockey ball which has been hit hard by a player and is travelling fast.

Read and Learn More Class 9 Science Solutions

Question 6. What will cause greater change in kinetic energy of a body? Changing its mass or changing its velocity.
Answer:

Change in velocity will cause greater change in kinetic energy because KE = 1/2 mv2

Question 7. Name the type of energy stored in spring of a watch.
Answer:

Elastic potential energy is stored in spring of a watch.

Question 8. In which situation, the potential energy of a spring be minimum?
Answer:

When a spring is at its natural length (i.e. neither stretched nor compressed), the potential energy of a spring will be minimum.

Question 9. If a body is thrown vertically upwards, its velocity goes on decreasing. What happens to its kinetic energy when it stops at the top and its velocity becomes zero?
Answer:

The whole of its kinetic energy gets converted into potential energy (mgh), where m is the mass of body and h is the height and g is the acceleration due to gravity.

Question 10. Can any object have mechanical energy even, if its momentum is zero?
Answer:

1. Since, mechanical energy is the sum of kinetic energy and potential energy. And as given that, momentum of the body is zero, it means velocity of the body is zero, so it has kinetic energy equals to zero.
2. But it may have potential energy. So, even if the momentum of the body is zero, it may have mechanical energy.

Question 11. The head of a nail becomes hot when it is hammered into a plank of wood. What energy transformation takes place in this case?
Answer:

Kinetic energy of falling hammer is converted into heat energy when it strikes the head of a nail. Therefore, the head of a nail becomes hot.

Question 12. A car and a truck have the same speed of 60 ms”1. If their masses are in the ratio 1:4. Find the ratio of their KE.
Answer:

As, \(\frac{\mathrm{KE}_1}{\mathrm{KE}_2}=\frac{1 / 2 m_2 v^2}{1 / 2 m_2 v^2}=\frac{m_1}{m_2}=\frac{1}{4}\)=1: 4

Question 13. If a speed of a particle is doubled, what will be the ratio of its kinetic energy to its momentum?
Answer:

Kinetic energy, \(\mathrm{KE}^{\prime} =\frac{1}{2} m\left(v^{\prime}\right)^2 [v^{\prime}=2 v]\)

= \(\frac{1}{2} m(2 v)^2=4 \mathrm{KE}\)

and momentum, \(p^{\prime =m v^{\prime}}\)

= \(m(2 v)=2 p [ v^{\prime}=2 v]\)

⇒ \(\frac{\mathrm{KE}^{\prime}}{p^{\prime}} =\frac{4 \mathrm{KE}}{2 p}=2 \times\left(\frac{\mathrm{KE}}{p}\right)\)

Hence, ratio gets doubled.

Question 14. If a 5 kg mass is raised to a height of 2 m, calculate the work done against the force of gravity. (Given, g = 9.8 ms^-2)
Answer:

Given, mass, m= 5 kg,

height, h = 2m, W =?

We know that work, W = mgh = 5 x 9.8 x 2 = 98 J

Question 15. A ball is dropped from a height of 10 m. If the energy of the ball reduces by 40% after striking the ground, how much high can the ball bounce back? (Take, g=10m^-2)
Answer:

If the energy of the ball reduces by 40% after striking the ground, then remaining energy of the ball will be 60% of initial energy.

Let initial energy of the body of mass (m) at height (h) is (mgh).

According to the question,

mgh’ = 60% of mgh [given, h =10 m and g =10 ms^-2]

∴ \(h^{\prime}=60 \% \times h=\frac{60}{100} \times 10=6 \mathrm{~m}\)

Question 16. An object of mass 2 kg is dropped from a height of lm. What will be its kinetic energy as it reaches the ground? (Take, g =9.8 ms^-2)
Answer:

Given, mass, m = 2 kg and height, h =1 m.

On reaching the ground, the kinetic energy of object is converted into its potential energy such that KE= YE-mgh = 2 x 9.8 x 1 = 19.6 J

Question 17. A girl weighing 50 kg runs up a hill raising herself vertically 10 m in 20 s. What is the power expended by girl?
Answer:

Power expended by girl =\(\frac{\text { Work doac by girl }}{\text { Time raken }}\)

= \(\frac{F \times s}{t}=\frac{m g \times s}{t}\)

= \(\frac{50 \times 9.8 \times 10}{20}\)

= 245 W[ F=m g]

Class 9 Science Chapter 10 Work, Energy And Power Short Answer Type Questions

Question 1. Is it possible that an object is in the state of accelerated motion due to external force acting on it, but no work is being done by the force? Explain it with an example.
Answer:

Yes, when a force acts in a direction perpendicular to the direction of displacement e.g. Earth revolves around the Sun under the gravitational force of Sun on Earth, but no work is done by the Sun, though Earth has a centripetal acceleration

Question 2. At the bottom of the waterfall, the water is warmer than at the top. Give reason.

Answer: 

When water falls on the ground, its mechanical energy (KE + PE) is converted into heat energy, due to which the temperature of water at the bottom of the waterfall increases.

Question 3. What is meant by the transformation of energy? Explain with the help of two suitable examples.
Answer:

One form of energy can be converted into other forms of energy, this phenomenon is called transformation of energy.

1.  When we throw a ball, muscular energy which is stored in our body, gets converted into the kinetic energy of the ball.
2. The wound spring in the toy car possesses potential energy. As the spring is released, its potential energy changes into kinetic energy due to which, the toy car moves.

Question 4. A light and a heavy object have the same momentum. Find out the ratio of their kinetic energies. Which one has a larger kinetic energy?
Answer:

Suppose m₁ and m₂ are masses of light and heavy objects, respectively. As we know,

kinetic energy, K=\(\frac{1}{2} m v^2\) [where, y= velocity of objects.] …(1)

and momentum, p = mv

On multiplying and dividing with m in Eq. (1), we get

So, K=\(\frac{1}{2} \frac{m v^2 \times m}{m}\)

K=\(\frac{1}{2} \frac{(m v)^2}{m}\) as from Eq. (2) [ p=m v]

So, K=\(\frac{p^2}{2 m}\)

We have, kinetic energy, K=\(\frac{p^2}{2 m}\)

Momentum is the same for light and heavy bodies.

So, kinetic enetgy, K \(\propto \frac{1}{m}\)

Thus, kinetic energy is inversely proportional to the mass.

So, a lighter body has larger kinetic energy.

Question 5. When a force retards the motion of a body, what is the nature of work done by force? State reason. List two examples of such a situation.
Answer:

The nature of work done in the case of retarding motion is negative.

Suppose a force F brings a body moving with velocity v to rest (retards), then work done

= change in KE

=\(\mathrm{KE}_f-\mathrm{KE}_i\)

W =\(0-\frac{1}{2} m x^2=-\frac{1}{2} m s^2\)

For Example

When we apply the brakes of a car, the work done is negative.

Work done by frictional force is negative.

Question 6. A car is moving on a levelled road and gets its velocity doubled. In this process,

1. how would the potential energy of the car change?
2. how would the kinetic energy of the car change?

Answer:

1. The potential energy of the car remains the same, since PE (= mgh) is independent of velocity.
2. The kinetic energy of the car becomes four times since KE \(\left(=\frac{1}{2} m \nu^2\right)\) is proportional to the square of velocity.

Question 7. A girl sits and stands repeatedly for 6 min. Draw a graph to show the variation of the potential energy of her body with time.

Answer:

1. From the graph shown above, we can take the sitting position of the girl as the position of zero potential energy.
2. Let m be the mass of the girl and h be the position of the centre of gravity while standing above the sitting position.
3. The PE while standing is + mgh and while sitting is zero. We can assume that there is no acceleration or deceleration while standing and sitting, this is repeated after every minute.

Question 8. (1) The potential energy of a freely falling object decreases progressively. What happens to its

1.  kinetic energy,
2. total mechanical energy?

State the law on which your answer is based.

(2) A household consumes 1 kWh of energy per day How much energy is this in joules?

Answer:

(1) (1) As potential energy decreases with decreasing height, the speed of the object will increase and hence its kinetic energy will increase.

(2) Total mechanical energy will remain constant. It is based on the law of conservation of energy which states that energy can neither be created nor be destroyed. It can only be transformed from one form to another.

(2) 1 kWh = 3.6xl0 6J

Question 9. If a body falls from a height bounces from the ground and again goes upwards with the loss of a part of its energy.

1. How will its potential energy change?
2. What are various energy conversions taking place?

Answer:

1. When it strikes the ground, its PE is zero and after bouncing, its potential energy increases gradually.
2. At the time it strikes the ground, it has maximum KE and after it bounces, its KE starts changing into potential energy.

Question 10. Why do we say that a lift delivers more power in taking a man up than that delivered by a man climbing stairs through the same height?

Answer:

It is true to say that a lift delivers more power in taking a man up than that delivered by him climbing the stairs at the same height. It is because, a lift takes less time in taking the man up, so it delivers more power.

Question 11. When an arrow is shot, from where does the arrow acquire its kinetic energy?
Answer:

A stretched bow has potential energy due to the change in its shape. To shoot the arrow, the bow has to be released.

Therefore, its potential energy is converted into the kinetic energy of the arrow.

Question 12. Seema tried to push a heavy rock of 100 kg for 200 seconds but could not move it. Find the work done by Seema at the end of 200 s.

Answer:

The work done by Seema in pushing a rock for 200 s is zero. Because there is no displacement in this case, i.e. W = Fx s =0 [v s =0]

Question 13. A coolie lifts a box of 15 kg from the ground to a height of 2 m. Calculate the work done by coolie on the box. (Given, g = 9.8 ms^-2)

Answer:

Given, mass, m = 15 kg, distance, s =2m, g =9.8 ms-2, W=>

Now, force applied,

F=mg= 15 x 9.8 = 147 N

Work done, W = F x s = 147 x 2 = 294 N

Question 14. How is work done by a force measured? A porter lifts luggage of 20 kg from the ground and puts it on his head 1.7 m above the ground. Find the work done by the porter on the luggage. (Take, g = 10 ms^-2 )

Answer:

Work done by a force is measured by the product of force and displacement when they are in the same direction or opposite direction.

Given, m = 20 kg

s =1.7 m [where, m = mass of luggage, g = acceleration due to gravity.]

W = Fs = mgs [ Fand s arc in the same direction.]

= 20x 10x 1.7= 340 J

Question 15. A boy is moving on a straight road against a frictional force of 5 N.

After travelling a distance of 1.5 km, he forgot the correct path at a roundabout of a radius of 100 m as shown in the figure.

However, he moves on the circular path for one and a half cycles and then he moves forward up to 2 km. Calculate the work done by him.

Answer:

Given, the force applied by the boy against friction = 5 N

Displacement on the circular path

= One cycle + Half cycle

= 0 + Half cycle

= 0 + Diameter of circular path [ Displacement depends on initial and final point]

= 0 + 2 r =0 + 2 x 100 [given, r =100 m]

= 0 + 200 =200 m

Total displacement = 1.5 km+ 200 m+ 2 km

= 1.5 x 1000 +200+ 2x 1000 m [1km = 1000 m]

= 3700 m

Work done by boy = Fs

= 5 x 3700 =18500J

Question 16. A bus and a car have the same KE. Which of the two is moving fast? Explain.

Answer:

The KE is given by KE=\(\frac{1}{2} m v^2 \Rightarrow v=\sqrt{\frac{2 \mathrm{KE}}{m}}\) As KE of both bus and car is same, so v \(\propto \frac{1}{\sqrt{m}}\)

Since the mass of the car is less than that of the bus, the car moves faster than the bus.

Question 17. Compute the speed of a 2 kg ball having a kinetic energy of 4 J.

Answer:

Given, the mass of the ball, m = 2 kg

The kinetic energy of the ball, KE = 4 J

As, KE =\(\frac{1}{2} m v^2\)

⇒ \(v^2=\frac{2 \mathrm{KE}}{m}=\frac{2 \times 4}{2}=4\)

v=2 \(\mathrm{~ms}^{-1}\)

Question 18. A car and a truck have the same speed of 30 ms^-1. If their masses are in the ratio of 1:3, find the ratio of their kinetic energies.

Answer:

Given, \(v_1=v_2=30 \mathrm{~ms}^{-1}\)

According to question, \(\frac{m_1}{m_2}=\frac{1}{3}\)

⇒ \(\frac{\mathrm{KE} \text { of car }}{\mathrm{KE} \text { of truck }}=\frac{\frac{1}{2} m_1 v_1^2}{\frac{1}{2} m_2 v_2^2}=\left(\frac{m_1}{m_2}\right)=\frac{1}{3}\)

Question 19. If two bodies have masses in the ratio 1: 8 and have their speed in the ratio 4:5, find the ratio of their KE.

Answer:

Given, \(m_1 / m_2=1: 8, and v_1 / v_2=4: 5\)

⇒ \(\frac{\mathrm{KE}_1}{\mathrm{KE}_2} =\frac{1 / 2 m_1 v_2^2}{1 / 2 m_2v_2^2}=\frac{m_2}{m_2}\left(\frac{v_1}{v_2}\right)^2\)

= \(\frac{1}{8} \times\left(\frac{4}{5}\right)^2\)

= \(\frac{1}{8} \times \frac{16}{25}=\frac{2}{25}=2: 25\)

Question 20. A rocket is moving up with a velocity v. If the velocity of this rocket is suddenly tripled, what will be the ratio of the two kinetic energies?

Answer:

Given, \(v_1=v\) and \(v_2=3 v\)

Kinetic energy of rocket, K=\(\frac{1}{2} m v^2\)

The ratio of two kinetic energies,

⇒ \(\frac{K_1}{K_2}=\frac{\frac{1}{2} m v_1^2}{\frac{1}{2} m v_2^2}\)

⇒ \(\frac{K_1}{K_2}=\frac{v_1^2}{v_2^2} \quad\left[\text { put } v_2=3 v \text { and } v_1=v\right ]\)

We get =\(\frac{v^2}{(3 v)^2}=\frac{v^2}{9 v^2}=\frac{1}{9} \Rightarrow \frac{K_1}{K_2}=\frac{1}{9}\)
Thus, the ratio of two kinetic energies \(K_1: K_2=1: 9\)

Question 21. A child drops a stone of 1 kg from the top of a tower. Find its kinetic energy, 5 s after it starts falling. (Take, g = 10 ms^-2)

Answer:

Given, the mass of the stone, m = 1 kg

Initial velocity, u =0, time, t = 5 s Acceleration due to gravity, g = 10 ms^-2

As from the equation of motion.

r=\(N+s^t\) [for downward motion]

r=0+10 x 5

r=50 \(\mathrm{~ms}^{-1}\) [for downward motion]

The kinetic energy of the stone is given by

⇒ \(\frac{1}{2} m v^2=\frac{1}{2} \times 1 \times(50)^2=1250 \mathrm{~J}\)

Question 22. A mass of 20 kg is dropped from a height of 0.5 m. Find its (1) velocity and

(2) KE as it just reaches the ground.

Answer:

Given, height, h=0.5 m, mass, m=20 kg

(1) \(p^2=2 g h=2 \times 10 \times\) 0.5=10

⇒ \({\left[ g=10 \mathrm{~ms}^{-2}\right]}\)

∴ \(\nu=\sqrt{10}=3.16 \mathrm{~ms}^{-1}\)

(2) We know that, KE =\(\frac{1}{2} m \nu^2\)

= \(\frac{1}{2} \times 20 \times 10=100 \mathrm{~J}\)

Question 23. What is the amount of work done in the following cases? Justify your answer by giving the appropriate reason.

1. By an electron revolving in a circular orbit of radius r around a nucleus.
2.  By the force of gravity, when a stone of mass m is dropped from the top of a multi-storeyed building of height h.

Answer:

(1) Work done is zero as shown in the figure below.

When an electron revolves around the nucleus, a centripetal force F acts along the radius towards the centre O.

The displacement (s) acts tangentially, therefore the angle between the force and the displacement is 90°. Therefore, W= 0

(2) We know that, W = mgh

As the stone is dropped, its PE starts to convert into KE.

Let its speed be v, then from \(v^2\)=2 g h (when stone reaches the ground)

KE =\(\frac{1}{2} m v^2\)

= \(\frac{1}{2} \times m \times 2 g h=m g h \Rightarrow W\) = mgh

Question 24. A boy lifts a suitcase of 20 kg from the ground to a height of 1.2 m. Calculate the work done by him on the suitcase. (Given, g=10ms^-2)

Answer:

Given, mass, m = 20 kg, height, h = 1.2 m, g =10 ms^-2

Work done, W = mgh = 20 x 10 x 1.2 = 240 J

Question 25. A shotput player throws a shotput of mass 3 kg. If it crosses the top of the wall 2 m high at a speed of 4 ms^-1. Compute the total mechanical energy gained by the shotput when it crosses the wall.(Given, g =9.8 \(\mathrm{~ms}^{-2}\))

Answer:

Given, m=3 \(\mathrm{~kg}, h=2 \mathrm{~m}, y=4 \mathrm{~ms}^{-1}, g=9.8 \mathrm{~ms}^{-2}\)

Total mechanical energy =\(\mathrm{KE}+\mathrm{PE}=\frac{1}{2} m v^2+m g h\)

= \(\frac{1}{2} \times 3 \times 16+3 \times 9.8 \times 2\)

= 24+58.8=82.8 J

Question 26. Avinash can run with a speed of 8 ms^-1 against the frictional force of 10 N and Kapil can move with a speed of 3 ms^-1 against the frictional force of 25 N. Who is more powerful and why?

Answer:

Given, the force applied by Avinash = 10 N

Speed of Avinash = 8 m s^-1

Power of Avinash =F. v =10 x 8 =80 W

Now, force applied by Kapil = 25 N

Speed of Kapil = 3 m s^-1

Power of Kapil = F . v

= 25 x 3 = 75 W

Since, Avinash has more power (80 – 75), i.e. 5 W than Kapil. So, Avinash is more powerful.

Question 27. The human heart does 1.5 J of work in every beat. How many times per minute does it beat, if its power is 2 W?

Answer:

Given, power, P=2 W, time, t=1 min =60 s

Total work, W=\(P_t \quad\left[ P=\frac{W}{t}\right]\)

=2 x 60 s=120J

1.5 J work is done in 1 beat.

120 J work will be done in \(\frac{1 \times 120}{15}\)=80 beats

Therefore, number of beats per min = 80

Question 28. For an experiment to measure his power, a student records the time taken by him to run up a flight of steps on a staircase. Use the following data to calculate the power of the student.

Number of steps =28

Height of each step =20 cm

Time taken =5.4 s

Mass of student =55 kg

Acceleration due to gravity =9.8 \(\mathrm{~ms}^{-2}\)

Answer:

Given. n=28, b =20 cm =0.2 m, t=5.4 s

m =55 kg, g=9.8 \(\mathrm{~ms}^{-2}\)

We know that the power of student is given by

P =n \(\times \frac{W}{r}=n \times \frac{m g b}{t}\)

= \(\frac{28 \times 55 \times 9.8 \times 0.2}{5.4}\) [ W = mgh]

=559 W

Question 29. A boy X can run with a speed of 8 ms^-1 against the frictional force of 10 N and another can move with a speed of 3 ms^-1 against the frictional force of 20 N. Find the ratio of powers of X and Y.

Answer:

Given, the distance travelled by the boy X in 1 s = 8 m Distance travelled by the boy Fin ls=3m

As we know, work done by the boy X to run against the frictional force of 10 N = 10 Nx 8 m= 80 J

So, power of 80 J of work done by X=\(\frac{W}{t}=\frac{80 \mathrm{~J}}{1 \mathrm{~s}}\)

=80 W

Similarly, work done by the boy Y to run against the frictional force of 20 \(\mathrm{~N}=20 \mathrm{~N} \times 3 \mathrm{~m}=60 \mathrm{~J}\)

Power of Y=\(\frac{60 \mathrm{~J}}{1 \mathrm{~s}}=60 \mathrm{~W}\)

So, the ratio of two values of powers is given by

∴ \(\frac{\text { Power of } X}{\text { Power of } Y}=\frac{80}{60}=\frac{4}{3}=4: 3\)

Question 30. The power of a motor pump is 60 W. Find the mass of water, the pump can lift per minute from a 10 m deep well. (Take, g =10 ms^-2)

Answer:

Given, power, P = 60 W, time, t = 60 s Height, h =10m

The energy supplied to the pump = Power x Time

= 60 W x 1 min [vlmin=60s]

= 60 Wx 60s

= 3600 Ws= 3600J

So, energy, E = mgb

3600 = mx 10×10

m = 36 kg

So, the mass of the water is 36 kg.

Question 31.

1.  What is meant by the potential energy of a body?
2. A body of mass m is raised to a vertical height of h through two different paths A and B.

What will be the potential energy of the body in the two cases? Give a reason for your answer.

Answer:

1. The energy possessed due to the position of a body is called potential energy.
2. The work done against gravity in both cases is mgh. It is independent of the path along which the body is moved and it depends only on the initial and final positions of the body.

Question 32. At a height of 20 m above the ground, an object of mass 4 kg is released from rest. It is travelling at a speed of 20 ms^-1 when it hits the ground. The object does not rebound and the gravitational field strength is 10Nkg^-1.

How much energy is converted into heat and sound on impact?

Answer:

Given, height above the ground, b = 20 m Mass of the ball, m = 4 kg

Speed of the ball while striking the ground, v =20 ms^-1

Acceleration due to gravity,

g =10 Nkg^-1

According to the law of conservation of energy, all the energy of the ball will be converted into sound and heat energy because the ball does not rebound.

Energy of the ball =\(\frac{1}{2} m v^2\)

=\(\frac{1}{2} \times 4 \times(20)^2=800 \mathrm{~J}\)

Hence, 800 J of energy will be converted into heat and sound.

Question 33. A labourer whose own mass is 50 kg carries a load of an additional 60 kg on his head to the top of a building 15 m high. Find the total work done by him. Also, find the work done by him, if he carries another additional block of mass 10 kg to the same height. (Take, g =10 ms^-1)

Answer:

Given, total mass, m =50 + 60 =110 kg

Displacement, s =15 m

Work done by him is given by W =0 [since force and displacement are perpendicular to each other.]

If an additional block of mass 10 kg is carried by him to the same height, then also work done by him remains zero because force and displacement are perpendicular to each other.

Question 34. A girl with a mass of 35 kg sits on a trolley of mass 5 kg. The trolley is given an initial velocity of 4 ms^-1 by applying a force.

The trolley comes to rest after traversing a distance of 16 m.

1.  How much work is done on the trolley?
2. How much work is done by the girl?

To find the value of acceleration (a) by using the third equation of motion, i.e. v² = u² + 2as and then putting this value in W =mass to get the required work.

Answer:

Given, w = 4 ms^-1,v =0 and 5 =16 m From the third equation of motion,[for retardation, the acceleration is negative, i.e. a=-a]

⇒ \(v^2=u^2-2 a \Rightarrow(0)^2=(4)^2-2 a \times 16\)

0=16-32 a \(\Rightarrow a=\frac{16}{32}=0.5 \mathrm{~ms}^{-2}\)

where, u = initial velocity, v = final velocity, a = acceleration and s = displacement.

(1) Total mass = 35 + 5 = 40 kg

Work is done on the trolley,

W = F-d =ma s [v F =ma]

= 40×0.5×16 =320 J

(2) Given, the mass of the girl, m =35 kg

Work done by the girl,

W =F-d =ma r = 35×0.5×16 = 280]

Question 35. Calculate the kinetic energy of a car of mass 750 kg moving with a velocity of 54 km. Find the new kinetic energy of the car, if a passenger of mass 50 kg sits in the car.

Answer:

Given, mass, m=750 kg, velocity, v=54 \(\mathrm{kmh}^{-1}\)

=54 \(\times \frac{5}{18} \mathrm{~ms}^{-1}=15 \mathrm{~ms}^{-1}\)

Kinetic energy, \(\mathrm{KE}=\frac{1}{2} m v^2\)

KE =\(\frac{1}{2} \times 750 \times(15)^2=84375 \mathrm{~J}\)

If a passenger of mass 50 kg sits in the car, then total mass becomes (750+50) kg, i.e. 800 kg

New KE =\(\frac{1}{2} \times 800 \times(15)^2=90000 J\)

Question 36. On a level road, a scooterist applies brakes to slow down from a speed of 54 kmh^-1 to 36 kmh-1. What amount of work is done by the brakes? (Assuming the mass of the empty scooter is 86 kg and that of the scooterist and petrol is 64 kg.)

Answer:

Here, total mass, w = 86 + 64 = 150 kg

Initial velocity, u=54 \(\mathrm{kmh}^{-1}\)

=\(\frac{54 \times 1000}{3600}=15 \mathrm{~ms}^{-1}\)

Final velocity, v=36 \(\mathrm{kmh}^{-1}\)

=36 \(\times \frac{1000}{3600}=10 \mathrm{~ms}^{-1}\)

Work done by brakes = KE lost by the scooter

= Final KE – Initial KE

=\(\frac{1}{2} m v^2-\frac{1}{2} m u^2=\frac{1}{2} m\left(v^2-u^2\right)\)

=\(\frac{1}{2} \times 150\left[(10)^2-(15)^2\right]=\frac{1}{2} \times 150(100-225)\)

= -75×125 = -9375 J

Question 37. (1) Define potential energy.

(2) Give an example where potential energy is acquired by a body due to a 75 m change in its shape

(3) A skier of mass 50 kg stands at A, at the top of a ski jump. He takes off from A for his jump to B. Calculate the change in his gravitational potential energy between A and B.

Answer:

(1) Potential Energy It is the energy possessed by a body by virtue of its position or shape.

(2) In a toy car, the wound spring possesses potential energy. When spring is released, its potential energy changes into kinetic energy due to which the toy car moves.

(3) Given, m=50 \(\mathrm{~kg} . h=75 \mathrm{~m}_1 h_2=60 \mathrm{~m}\)

At point A, \(\mathrm{PE}_1=m g h_h=50 \times 10 \times 75\)

=37500 J

At point B, \(\mathrm{PE}_2=m g h_2=50 \times 10 \times 60\)

=30000 J

Change in PE =\(\mathrm{PE}_1-\mathrm{PE}_2\)=37500-30000 =7500 J

Question 38. A body of mass 5 kg is thrown vertically upwards with an initial velocity of 50 ms^-1. What will be its potential energy at the end of 5 seconds?

Answer:

Given, m=5 \(\mathrm{~kg}, w=50 \mathrm{~ms}^{-1}, t=5 \mathrm{~h} g=-10 \mathrm{~ms}^{-2}\)

Height covered by the body in 5 s is h=w t+\(\frac{1}{2} g t^2\)

50 \(\times 5-\frac{1}{2} \times 10 \times(5)^2\)

=250-125=125 m

Therefore, PE of the body after 5 s

= mgh =5 x 10 x 125 = 6250 J

Question 39. A body of mass 1.5 kg is thrown vertically upwards with an initial velocity of 15 ms^-1. What will be its potential energy at the end of 2 s? (Take, g =10 ms^-2)

Answer:

Given, mass, m = 1.5 kg

Initial velocity, u =15 ms^-1

Time, t = 2s

From Newton’s second equation of motion,

t=w t+\(\frac{1}{2} g t^2\)

s=15 \(\times 2+\frac{1}{2} \times 10 \times 4=50 \mathrm{~m}\)

Potential energy, PE = mgh

=1.5 \(\times 10 \times 50\) [ b=t]

=750 J

Question 40. Four men lift a 250 kg box to a height of 1 m and hold it without raising or lowering it.

1. How much work is done by the man in lifting the box?
2. How much work do they do in just holding it?
3.  Why do they get tired while holding it? (Given, g = 10 ms^-2)

Answer:

1. Given, mass, m = 250 kg, height, A = lm and acceleration due to gravity, g =10 m s^-2
2. Work done by the man in lifting the box W = Potential energy of box W = mgh = 250 x 10 x 1 = 2500 J (it) Work done is zero in holding a box because displacement is zero.
3. In holding the box, the energy of each man is lost. Due to loss of energy, they felt tired.

Question 41. The weight of a person on planet A is about half that on the Earth. He can jump up to 0.4 m height on the surface of the Earth. How high can he jump on planet A?

Answer:

It is given that the weight of a person on the Earth = w (i .e.w =mg)

And as he can jump up to height (hy) = 0.4 m So, potential energy at this point

= mgh = mgh x 0.4  → Equation .(1)

And it is given that

weight of the person on the other planet = \(\frac{w}{2}\)  → Equation 2

And if he could jump to height \(\left(b_2\right)\) its potential energy would be \(\frac{w}{2} h_2=\frac{m g}{2} b_2\)

Since he applied the same amount of effort in both cases to lift his body, its potential energy will be the same.

From Eqs. (1) and (2), we get

m g \(\times 0.4=\frac{m g}{2} b_2 \Rightarrow b_2=0.4 \times 2=0.8 \mathrm{~m}\)

Question 42. 300 J of work is to be done in lifting a bag of mass 5 kg in weight up to a height of 4 m from the ground. What will be the acceleration with which the bag was revised? (Take, g = 10 ms^-2)

Answer:

Given, mass, m=5 kg, work, W=300 J

h=4 \(\mathrm{~m}, g=10 \mathrm{~ms}{ }^{-2}\)

Work done, W=m g h+m a b=m(g+a) b

300=5 \(\times(10+a) \times 4\)

300=20(10+a)

10+a=\(\frac{300}{20}\)=15

a=15-10=5 \(\mathrm{~ms}^{-2}\)

Question 43. Shyam drops a ball of 100 g from a building of a height of 10 m. What will be its kinetic energy at the height of 4 m? What will happen to its total mechanical energy? Give reasons to justify your answer.

Answer:

Given, mass, m = 100 g = 0.1 kg, height, A = 10m

Potential energy, PE = mgh = 0.1x10x10 = 10 J [v g =10 ms^-2]

When the body reaches a height of 6 m, its velocity is given by

⇒ \(r^2=s^2+2 s^h\)

⇒ \(r^2=0^2+2 \times 10 \times 6\) [h=10-4 or h=6]

v=\(\sqrt{2 \times 10 \times 6}=2 \sqrt{30}\)

Now, its kinetic energy at this point is given by

=\(\frac{1}{2} m r^2=\frac{1}{2} \times 0.1 \times 4 \times 30=6 \mathrm{~J}\)

As, mechanical energy = KE + PE [at highest point] Mechanical energy = (0 + 10)J=10J Mechanical energy at 4 m height

= KE+PE = 6+0.1 x 10x 4= 10]

Hence, it is seen that total mechanical energy always remains constant during the motion.

Question 44. A mass of 10 kg is dropped from a height of 50 cm. Find its

1. potential energy just before dropping.
2. kinetic energy just touching the ground.
3. the velocity with which it hits the ground (Take, g = 10 ms^-2)

Answer:

Given, mass of the object, m= 10 kg Height, h =50 cm= 0.3 m

(1) As potential energy is given by PE = mgh =10 x 10 x 0.5 = 50]

(2) From the law of conservation of energy,

the total energy of the ball just before dropping =total energy of the ball just on touching the ground KE + PE of the ball just before dropping

= KE of the ball just on touching the ground => KE = 50J

(3) As we know, KE = 50 J

So, \(\frac{1}{2} m v^2=50 \Rightarrow v^2=\frac{50 \times 2}{10}=10\)

So, the velocity with which it hits the ground,

∴ \(\mathrm{v}=\sqrt{10}=3.16 \mathrm{~ms}^{-1}\)

Question 45. A small child tends to mimic his father by lifting a mass of 10 kg on his head. As soon as he succeeds in lifting it, he loses the object and falls back to the ground. If the child has a height of 90 cm, find the kinetic energy

1. At half the height of the child,
2. With which the object strikes the ground.

Answer:

Given, the mass of the object, m =10 kg Height of the child, h- 90 cm = 0.9 m Total energy of the object at the head of the child is given by

= PE+ KE = mgh + 0 [v KE=0, asrr =0] = mgh= lOx 9.8x 0.9= 88.2 J

(1) At half the height of the child, i.e. at 0.45 m

Total energy, TK = PK + KE = mgh + KE

= 10 x 9.8 x 0.45+ KE

Now, from the law of conservation of energy,

KE+ 10x 9.8x 0.45= 88.2 or KE+ 44.1 = 88.2

KE= 88.2-44.1 =44.1 J

(2) When the object strikes the ground,

Total energy = PE + KE = 0 + KE

= KE [PE = 0, as h =0]

From the law of conservation of energy,

KE= 88.2 J [v TE =88.2 J]

Question 46. (1) State and define the SI unit of power.

(2) A person carrying 10 bricks each of mass 2.5 kg on his head moves to a height 20 m in 50 s. Calculate the power spent in carrying the bricks of the person. (Given, g =10 ms^-2)

Answer:

(1) The SI unit of power is watt.

1 watt is the power of a body which does work at the rate of 1 joule per second.

wart =\(\frac{1 \text { Joule }}{1 \text { Second }}\)

(2) Given, the mass of one brick = 2.5 kg

Mass of 10 bricks = 2.5 x 10 = 25 kg

Height, h = 20 m, time, t = 50 s, power, P = ?

Power P =\(\frac{m g h}{t}\)

= \(\frac{25 \times 10 \times 20}{50}=100 \mathrm{Js}^{-1}\)

Question 47. Mohan lifts his cell phone to a height of 1 m. He takes Is to do this. After lowering the phone, he then lifts it 2 m in 2 s. Has he generated more power in doing the second task? Give a reason to justify your answer.

Answer:

Let, the mass of cellphone be m kg

Work done (W) by Mohan = mgh

where, h = height at which the cellphone is raised and

g = acceleration due to gravity

W = mg x 1 = mg

Power delivered (R)=\(\frac{\text { Work done }(W)}{\text { Time }(T)}\)

=\(\frac{m g}{1}=m g\)

Work done to lift cell phone by 2 m

= mg x 2 = 2 mg

Power delivered \(\left(R_2\right)=\frac{\text { Work done (IV) }}{\text { Tine (T) }}\)

=\(\frac{2 m g}{2}\)=m g

So in both cases, the power delivered by Mohan is the same.

Question 48. A force applied on a body of mass 4 kg for 5 s changes its velocity from 10 \(\mathrm{~ms}^{-1}\) to 20 \(\mathrm{ms}^{-1}\). Find the power required.

Answer:

Given, m=4 kg, t=5 s, w=10 \(ms^{-1}\),

v=\(20 \mathrm{~ms}^{-1}\), P=?

Power, \(P=\frac{W}{t}=\frac{\text { Change in } \mathrm{KE}}{\text { Time taken }}\)

P =\(\frac{1}{2} \times \frac{m\left(v^2-u^2\right)}{t}=\frac{1}{2} \times \frac{4\left\{(20)^2-(10)^2\right]}{5}\)

=\(\frac{1}{2} \times \frac{4 \times 300}{5}=120 \mathrm{Js}{ }^{-1}\)

Question 49. A boy of mass 50 kg runs up a staircase of 45 steps in 9 seconds. If the height of each step of the staircase is 15 cm, find the power of the boy. (Given, g=10 \(\mathrm{~ms}^{-2}\) )

Answer:

Given, the mass of the body, m=50 kg

b=45 \(\times 15=675 \mathrm{~cm}=6.75 \mathrm{~m}\)

t=9 \(\mathrm{~s}, g=10 \mathrm{~ms}^{-2}\)

PE =m g b=50 \(\times 10 \times 6.75=3375 \mathrm{~J}\)

So, the energy of the boy is 3375 J.

Power of the boy,

y, P =\(\frac{\mathrm{PE}}{\text { Time }}=\frac{\text { Energy }}{\text { Time }}\)

=\(\frac{3375}{9}=375 \mathrm{~W}\)

Question 50. A machine gun takes 10 s to fire 30 bullets with a velocity of 500 \(\mathrm{~ms}^{-1}\). Find the power developed by the gun when each bullet has a mass of 100g.

Answer:

Given, mass of 30 bullets =30 x 100=3 kg

Velocity, v=500 \(\mathrm{~ms}^{-1}\), time, t=10s

Power developed by the gun,\(P=\frac{W}{t}\)

(Here, work done by a gun will be equal to the kinetic energy of all the bullets.]

P =\(\frac{K E}{t}=\frac{\frac{1}{2} m v^2}{t}=\frac{m v^2}{2 t}\)

= \(\frac{3 \times(500)^2}{2 \times 10}=\frac{3 \times 500 \times 500}{20}=37500 W\)

Class 9 Science Chapter 10 Work, Energy And Power Long Answer Type Questions

Question 1. The velocity of a body moving in a straight line is increased by applying a constant force F for some distance in the direction of the motion. Prove that the increase in the kinetic energy of the body is equal to the work done by the force on the body.

Answer:

Consider an object of mass m moving with a uniform velocity. Let, it now be displaced through a distance s, when a constant force F acts on it in the direction of its displacement.

From the third equation of motion, \(v^2=u^2+2 a t\)

⇒ \(v^2-\mu^2=2 a t \Rightarrow r=\frac{p^2-u^2}{2 a}\)

We know that the work done by F is

W=F r \(\cos \theta\) (since, force and displacement are in some direction, so \(\theta=0^{\circ} )\)

W =m a \(\times s\) [ F=m a]

= ma \(x\left(\frac{v^2-u^2}{2 a}\right) {[\text { from Eq. (1) }]}\)

W =\(\frac{1}{2} m\left(v^2-u^2\right)\)

If the object is starting from its stationary position, i.e. w=0, from Eq. (2), we get

W=\(\frac{1}{2} m v^2\)

It is clear that the work done is equal to the increase in the kinetic energy of an object.

Question 2. (1) Name two forms of mechanical energy. Define the SI unit of energy.

(2) A man of mass 50 kg jumps from a height of 0.5 m. If gr = 10 ms^-2, what will be his energy at the highest point?

(3) Calculate the energy of a body of mass 20 kg moving with a velocity of 0.1 ms^-1.

Answer:

(1) Two forms of mechanical energy are

•  Kinetic energy and
• Potential energy

The SI unit of energy is joule (J). 1 J is the amount of work done on an object when a force of IN displaces it by lm along the line of action of force.

(2) Given, w = 50kg, h = 0.5 m, g=10 ms^-2

At the highest point, kinetic energy is converted into potential energy.

PE =\(m g^b=50 \times 10 \times 0.5\)=250

(3) Given, m=20 kg, v=0.1 \(\mathrm{~mm}^{-1}\)

As we know, a moving body has kinetic energy.

KE =\(\frac{1}{2} m v^2=\frac{1}{2} \times 20 \times(0.1)^2\)

= \(\frac{1}{2} \times 20 \times 0.1 \times 0.1=0.1 \mathrm{~J}\)

Question 3. (1) Define the kinetic energy of an object. Can the kinetic energy of an object be negative? Give reason.

(2) A car weighing 1200 kg is uniformly accelerated from rest and covers a distance of 40 m in 5 s. Calculate the work, the car engine had to do during this time.

Answer:

(1) The energy possessed by a body by virtue of its motion is called its kinetic energy.

No, the kinetic energy of an object cannot be negative because both m and v2 are always positive and KE =\(\frac{1}{2} m v^2\)

(2) Given, m =1200 kg, r = 40 m, t= 5 s, u= 0, W = ?.

We know that, W=F s = mas [ F=m a]

or \(s=u t+\frac{1}{2} a t^2\)

40=0 \(\times t+\frac{1}{2} \times d \times(5)^2\)

⇒ \(a=\frac{40 \times 2}{25}=3.2 \mathrm{~ms}^{-2}\)

⇒ \(W=F_5= matr\)

=1200 \(\times 3.2 \times 40=153600 \mathrm{~J}\)

Question 4. A vehicle of 1 tonne travelling with a speed of 60 ms^-1 notices a cow on the road 9 m ahead and applies brakes. It stops just in front of the cow.

1.  Find out the KE of the vehicle before applying brakes.
2. Calculate the retarding force provided by the brakes.
3.  How much time did it take to stop after the brakes were applied?
4.  What is the work done by the braking force?

Answer:

Given, the mass of the vehicle, m= tonne = 1000 kg

Initial speed, u = 60 ms^-1

Distance between vehicle and the cow, r=9 m

Final velocity, v=0

(1) KE of the vehicle before applying brakes is given by

=\(\frac{1}{2} m u^2=\frac{1}{2} \times 1000 \times 60 \times 60=1800000 \mathrm{~J} \text { (1W) }\)

(2) From the third equation of motion,

⇒ \(v^2-N^2=2 a\)

⇒ \((0)^2-(60)^2=2 \times a \times 9 \Rightarrow a=-200 \mathrm{~ms}^{-2}\)

So, retarding force provided by the brakes

∴ \(=m a d=1000 \mathrm{~kg} \times(-200) \mathrm{ms}^{-2}=-200000 \mathrm{~N}\)

(3) Now; again from the second equation of motion,

\(t=u t+\frac{1}{2} a t^2 \Rightarrow 9=60 t+\frac{1}{2} \times(-200) t^2\)

or 9=60 t-100\( t^2\)

or \(100 t^2-60 t+9=0 \Rightarrow(10 t-3)^2=0\)

or \(10 t-3=0 \Rightarrow t=\frac{3}{10}=0.3 \mathrm{~s}\)

(5) So, work done by the braking force is given by

=\(F_s=-200000 \mathrm{~N} \times 9 \mathrm{~m}=-1800000 \mathrm{~J}\)

Question 5. (1) A battery lights a bulb. Describe the energy changes involved in the process,

(2) Calculate the amount of work needed to stop a car of 500 kg moving at a speed of 36 kmh^-1.

Answer:

(1) When a battery lights a bulb, its chemical energy changes into light and heat energy.

(2) Here, m = 500 kg, v = 0 and

⇒ \(\mu=36 \mathrm{kmh}^{-1}=\frac{36 \times 1000}{3600}=10 \mathrm{~ms}^{-1}\)

Work done = Change in kinetic energy

=\(\frac{1}{2} m\left(v^2-v^2\right)=\frac{1}{2} \times 500\left(0-10^2\right)\)

=-\(\frac{1}{2} \times 500 \times 100=-25000 \mathrm{Js}^{-1}\)

So, it is negative because work is done to stop the car.

Question 6. A body of mass 20 kg is raised to the top of a building 15 m high and then dropped freely under gravity.

1. Find out the work done in raising the body to the top of the building.
2. What will be the value of gravitational potential energy at the top of the building?
3. By what factor will the gravitational potential energy of the same body increases, if it is raised to the top of a multi-storey building 45 m high?
4.  When will the kinetic energy of the body be maximum?

Answer:

Given, m=20 kg, h=15 m, g=10 \(ms^{-2}\)

(1) Work done =m g h=20 x 10 x 15=30001

(2) Gravitational PE = Work done=3000 J

(3) Gravitational PE at a height of 45m

= \(mgh^{\prime}=20 \times 10 \times 45=9000 \mathrm{~J}\)

Now,\(\frac{\text { Potential energy at height } 45 \mathrm{~m}}{\text { Potential energy at height } 15 \mathrm{~m}}=\frac{9000}{3000}\)=3

Therefore, PE increases by 3 times.

(4) KE will be maximum just before the body strikes the ground.

Question 7. An automobile engine propels a 1000 kg car A along a lavelled road at a speed of 36 kmh^-1. Find the power, if the opposing frictional force is 100 N.

Now, suppose after travelling a distance of 200 m, this car collides with another stationary car B of the same mass and comes to rest. Let its engine also stop at the same time.

Now, car B starts moving on the same level road without getting its engine started. Find the speed of the car B just after the collision.

Answer:

Given, the mass of the car = 1000kg

Mass of car B =1000 kg

The force applied by car A = 100 N

Speed of car \(A,\left(v_A\right)=36 \mathrm{~km} \mathrm{~h}^{-1}\)

=36 \(\times \frac{5}{18}=10 \mathrm{~m} \mathrm{~s}^{-1}\)

∴ [\(1 \mathrm{kmh}^{-1}=\frac{5}{18} \mathrm{~ms}^{-1}]\)

Power of car A, \(P_A=F \cdot v_A=100 \times 10=1000 \mathrm{~W}\) [F=force exerted by the car A against friction]

Again, for car A Newton’s second law, F = ma

100 =1000 x a

a =\(\frac{100}{1000}\)

a =\(\frac{1}{10} \mathrm{~m} \mathrm{~s}^{-2}\)

The velocity of car A after travelling 200 m is given by from the third equation of motion, \(v^2=u^2+2 an s\)

[for car A here, u=10 \(\mathrm{~ms}^{-1}, s=200 \mathrm{~m}]\)

⇒ \(v^2 =(10)^2+2 \times \frac{1}{10} \times 200\)

⇒ \(v^2 \)=100+40=140

v =\(\sqrt{140}=11.8 \mathrm{~ms}^{-1}\)

According to the question, after moving 200 m, the speed of car A, m =11.8 \(ms^{-1}\)

Just after the collision, the final speed of car A, \(v_1\)=0 before the collision, the initial speed of car B, \(w_2\)=0

From the conservation of linear momentum,

⇒ \(m_1 \mathrm{H}_1+m_2 \mathrm{H}_2=m_1 v_1+m_2 \mathrm{~N}_2\)

[Let just after the collision, the speed of the car B is \(v_2\)]

⇒ \(m_2 \times 11.8+m_2 \times 0=m_4 \times 0+m_2 \times v_2\)

⇒ \(11.8 m_2=m_2 v_2\)

⇒ \(118 m_2=m_1 v_2 \quad\left\{ m_1=m_2\right\}\)

∴ \(v_2=11.8 \mathrm{~m} / \mathrm{s}\)

Question 8. A car is moving with uniform velocities; 18 kmh^-1, 36 kmh^-1, 54 \(kmh{-1}\) and 72 \(kmh{-1}\) at some intervals. Find the KE of the boy of 40 kg sitting in the car at these velocities. Draw a graph between the KE and the velocities. Also, find the nature of the curve.

Answer:

At \(v_1=18 \mathrm{kmh}^{-1}=\frac{18 \times 1000}{3600}=5 \mathrm{~ms}^{-1}\),

⇒ \(KE_1=\frac{1}{2} m \mathrm{~m}^2=\frac{1}{2} \times 40 \times(5)^2=500 \mathrm{~J}\)

At \(v_2 =36 \mathrm{kmh}^{-1}=10 \mathrm{~ms}^{-1}\),

⇒ \(\mathrm{KE}_2 =\frac{1}{2} m v_2^2=\frac{1}{2} \times 40 \times(10)^2=2000 \mathrm{~J}\)

At \(v_3 =54 \mathrm{kmh}^{-1}=15 \mathrm{~ms}^{-1}\)

⇒ \(\mathrm{KE}_3 =\frac{1}{2} m v_3^2=\frac{1}{2} \times 40 \times(15)^2=4500 \mathrm{~J}\)

At \(v_4 =72 \mathrm{kmh}^{-1}=20 \mathrm{~ms}^{-1}\)

⇒ \(\mathrm{KE}_4 =\frac{1}{2} m v_4^2=\frac{1}{2} \times 40 \times(20)^2=8000 \mathrm{~J}\)

The graph between the KE and the velocities is shown below.

The graph is a parabolic curve because \(\mathrm{KE} \propto v^2\).

Class 9 Science Notes For Chapter 10  Work, Energy, And Power

Work:  Work is said to be done if by applying a force on an object, it is displaced from its position in the direction of force.

Scientific Conception of Work

From the point of view of science, the following two conditions need to be satisfied for work to be done.

1. A force should act on an object.
2. The object must be displaced.

If any one of the above conditions does not exist, work is not done.

For Example. A girl pulls a trolley and the trolley moves through a distance. In this way, she has exerted a force on the trolley and it is displaced. Hence, work is done.

Work Done by a Constant Force

‘Work done by a force on an object is equal to the magnitude of the force multiplied by the distance moved in the direction of force.’

Read and Learn  More Class 9 Science Notes

Let us assume if a constant force F acts on an object at point A, due to which the object gets displaced through a distance s in the direction of the force and reaches point B, then the work done (W) by force {F) on that object will be equal to the product of the force and displacement.

Work done = Force x Displacement in the direction of force or W = F x s

SI Unit of Work

1. If F = 1 N and s = l m. then the work clone hy the force will he 1 N-m.
2. The SI unit of work is newton-metre (N-m) which is also called joule (J)
3. Thus 1 J is the amount of work done on an object when a three of 1 N displaces it by 1 m along the line of action of the force.
4. 1 joule = 1 newton x 1 metre ⇒ 1J = lN-m
5. Work is a scalar quantity, it has only magnitude and no direction.

Example 1. A force of 10 N is acting on an object. The object is displaced through 5 m in the direction of force. What is the work done in this case?
Answer:

Given, force, F = 10N, displacement, s=5m

Work done, W = f x J = 10 Nx 5m = 50 N-m or 50J

Positive, Negative, and Zero Work

When the force F and displacement s are in the same direction (the angle between the direction of force and displacement is 0°), work done will be positive, i.e. work is done by the force.

For Example. A boy pulls an object towards himself.

W = + F xs

When the force F and displacement s are in opposite directions (the angle between the direction of force and displacement is 180°), work done will be negative, i.e. work is done against the force.

For Example. The frictional force acts in the direction opposite to the direction of displacement, so work done by friction will be negative. W=-Fxs

When the force and displacement are in the perpendicular direction (the angle between the direction of force and displacement is 90°), the work done is zero.

For Example. A coolie carrying a load on his head. In this case, the gravitational force is acting vertically downward (weight of load) and displacement is along the horizontal direction, i.e. force and displacement are perpendicular to each other. So, in this case, the work done by the gravitational force is zero. W= 0

Example 2. A crane lifts” a crate upwards through a height of 20 m. The lilting force provided by the crane is 5 kN. How much work Is done by the force?

Answer: 

Given, force, F = 5 kN = 5000 N Displacement, s = 20 m

Work done, W =?

We know that work done, W = F s

Here, force and displacement are in the same direction.

So, W = Fs ⇒ W =5000Nx 20m = 100000 J

So, the work done by the force is 100000 J or 100 kj

Energy

1. It is the ability to do work It is always essential for performing any mechanical work. An object having the capability to do work is said to possess energy.
2. The object that does the work, loses energy, and the object on which work is done, gains energy.
3. The energy of an object is measured in terms of its capacity to do work.
4. The SI unit of energy is the same as that of work, i.e. joule (J). 1 joule of energy is required to do 1 J of work. A larger unit of energy is kJ.
5. 1 kilo joule (kJ) =10³J
6. Work done against a force is therefore stored as energy.
7. For Example, When a fast-moving cricket ball hits a stationary wicket, the wicket is thrown away.
8. When a raised hammer falls on a nail placed on a piece of wood, it drives the nail into the wood.
9. Sun is the biggest natural source of energy for us. We can also get energy from the nuclei of atoms, the interior of the Earth, and the tides in the ocean.

Forms of Energy

Energy exists in various forms mechanical energy (the sum of potential energy and kinetic energy), heat energy, chemical energy, electrical energy, and light energy.

Kinetic Energy

1. The energy which is possessed by an object due to its motion is called kinetic energy.
2. Its SI unit is joule (J). The kinetic energy of a body moving with a certain velocity” is equal to the work done on it to make it acquire that velocity. The kinetic energy of an object increases with its speed.
3. Due to kinetic energy”, a bullet fired from a gun can pierce a target.
4. A moving hammer drives a nail into the wood. Due to its motion, it has kinetic energy or ability to do work.
5. The kinetic energy possessed by an object of mass m, moving with a uniform velocity v is given by,
6. \(\mathrm{KE}\) or \((E_K)=\frac{1}{2} m v^2\)

Calculation of Kinetic Energy

The kinetic energy of an object is measured by the amount of work, it can do before coming to rest. Consider an object of mass m moving with a uniform velocity u.

A force F is applied to it which displaces it through a distance s and it attains a velocity v.

Then, work is done to increase its velocity from u to v. W =Fs

According to the equation of motion,

⇒ \(v^2-u^2 =2 a y\)

s =\(\frac{v^2-u^2}{2 a}\)

where a is uniform acceleration, u is initial velocity and v is final velocity,

Also from, F=m a

Substituting the values of F and t in Eq. (1), we have

W = ma \(\cdot \frac{v^2-u^2}{2 a}\) of W=\(\frac{1}{2} m\left(v^2-u^2\right)\)

This is known as the work-energy theorem (i.e. total work is equal to the change in kinetic energy).

If initial velocity,

Then, u = 0

W =\(\frac{1}{2} m v^2\)

This work is equal to the kinetic energy of the object.

⇒ \(\mathrm{KE}\)(or \(E_K)=\frac{1}{2} m v^2\)

Some Important Results can be Derived from the Formula KE=\(\frac{1}{2} m v^2\)

These are given below :

1.  If the mass of an object is doubled, its kinetic energy also gets doubled.
2. If the mass of an object is halved, its kinetic energy also gets halved.
3.  If the speed of an object is doubled, its kinetic energy becomes four times.
4. If the speed of an object is halved, its kinetic energy becomes one-fourth.
5. Heavy objects moving with high speed have more kinetic energy than small objects moving with less speed.

Example 3. A bullet of mass 8 g is fired with a velocity of 80 \(\mathrm{~ms}^{-1}\). Calculate its kinetic energy.
Answer:

Given, mass, m=8 \(\mathrm{~g}=\frac{8}{1000} \mathrm{~kg}, velocity, v=80 \mathrm{~ms}^{-1}\)

KE of the bullet =\(\frac{1}{2} m v^2=\frac{1}{2} \times \frac{8}{1000} \times(80)^2 \)

= \(\frac{1}{2} \times \frac{8}{1000} \times 80 \times 80=25.6 \mathrm{~J}\)

Example 4. If a body of mass 5 kg is moving along a straight line with a velocity of 10 ms^-1 and acceleration of 20 ms^-2. Find its kinetic energy (KE) after 10 s.
Answer:

Given, the mass of the body, m = 5 kg

Initial velocity, u = 10 ms\(\mathrm{~ms}^{-1}\)

Acceleration, a = 20 \(\mathrm{~ms}^{-2}\)

Time, r = 10s Velocity, v = ?; KE = ?

First, we use the equation of meeting, N=N+ at to hinsl p, Then, we see \(KI-\frac{1}{2} m v^7\) to find kinetic energy.

r=\(\Delta+\Delta f\)

r=\((v+a t) m^{-1}\)

As we know, kinetic erg, KE =\(\frac{1}{2} m w^2\)

So. \(\mathrm{KE}=\frac{1}{2} m \times(w+a t)^2\)

KI=\(\frac{1}{2} \times 5 \times(10+20 \times 10)^2\)

∴ \(\mathrm{KE}=\frac{1}{2} \times 5 \times 210 \times 210=110250 \mathrm{j}\)

Example 5. What is the work to be done to increase the velocity of a van from 10 m/s to 20 m/s, if the mass of the is 2000 kg?
Answer:

Given,m=2000 kg, \(v_1=10 ms^{-1}, p_2=20 \mathrm{~ms}^{-1}\)

The initial Kinetic energy of the van

⇒ \(\mathrm{KE}_1.=\frac{1}{2} m \nu_1^2\) \([\mathrm{KE}=\frac{1}{2} m \nu^2]\)

= \(\frac{1}{2} \times 2000 \mathrm{~kg} \times\left(10 m s^{-1}\right)^2=100000 \mathrm{~J}=100 \mathrm{~kJ}\)

The final kinetic energy of the van

⇒ \(\mathrm{KE}_2=\frac{1}{2} m{ }_2^2=\frac{1}{2} \times 2000 \mathrm{~kg} \times\left(20 \mathrm{~ms}^{-1}\right)^2\)

= 400000J = 400 kJ

The work done = Change in kinetic energy = 400 kJ -100 kJ = 300 kJ

So, the kinetic energy of the van increases by 300 kJ when it speeds up from 10 \(m s^{-1}\) to 20 \(m s^{-1}\).

Potential Energy

1. The energy possessed by a body due to its change in position or shape is called potential energy. Its SI unit is joule (J).
2. We can say that the potential energy possessed by a body is the energy present in it by its position or configuration,
3. For Example. a stretched rubber band, spring, string on the bow, etc. Now, we can say that a body possesses energy even when it is not in motion.

Examples of potential energy are

Water stored in a dam has potential energy due to its position at the height.

A stone lying on the roof of the building has potential energy due to its height.

A wound spring of a watch has potential energy due to the change in its shape.

Potential Energy of an Object at a Height

1. When an object is raised through a certain height above the ground, its energy increases. This is because work is done on it against gravity while it is being raised.
2. The energy present in such an object is the gravitational potential energy. The gravitational potential energy of an object at a point above the ground is defined as the work done in raising it from the ground to that point against gravity.

Expression for Potential Energy

• Consider an object of mass m, lying at point A on the Earth’s surface. Here, its potential energy is zero and its weight mg acts vertically downwards.
• To lift the object to another position B at a height of h, we have to apply a minimum force that is equal to mg in the upward direction. So, work is done on the body against the force of gravity. Therefore,
• Work done = Force x Displacement or W = F x s
• As, F = mg [weight of the body]
• Here, s = h
• Therefore, W = mg x h = mg i.e. PE = mg
• This work is equal to the gain in energy of the body. This is the potential energy (PE) of the body.
• The potential energy of an object at a height depends on the ground level or the zero level you choose.
• An object in a given position can have a certain potential energy concerning one level and a different value of potential energy concerning another level.
• The work done by gravity depends on the difference in vertical heights of the initial and final positions of the objects and not on the path along which the object is moved.
• It is clear from the given
• In both the above situations, the work done on the object is much.

Example 6. Suppose you have a body of mass 1 kg in your hand. Tb what height will you raise it, so that it may acquire a gravitational potential energy of 1 J? (Take, g = 10 ms^-2)
Answer:

Given, PE =1J, mass, m = 1 kg,

Acceleration due to gravity, g = 10 ms^-2, h =?

We know that, PE = mg or 1 = 1 x 10 x h

Height, b = \(\frac{1}{1 \times 10}\)=0.1 m = 10 cm

Example 7. A boy weighing 40 kg climbs up a vertical height of 200 m. Calculate the amount of work done by him. How much potential energy does he gain? (Take, g =9.8 ms^-2)
Answer:

Given that, mass, m = 40 kg

Acceleration due to gravity, g = 9.8 ms^-2, height, h = 200 m

Work done by the body = mgt = 40 x 9.8 x 200 = 78400J = 7.84 x 10⁴ J

Gain in PE = Work done = 7.84 x 10⁴J

Example 8. Suppose two bodies A and B having equal masses are kept at heights of h and 3 h, respectively. Find the ratio of their potential energies.
Answer:

Let the mass of each body be m.

PE of body A= mg PE of body B=m g \(\times\) 3 h

Ratio of their potential energies =\(\frac{m g b}{m g \times 3 b}=\frac{1}{3}\)=1: 3

Law of Conservation of Energy

The law of conservation of energy states that energy can neither be created nor be destroyed, it can only be transformed from one form to another. The total energy before and after transformation always remains constant.

Conservation of Energy During the Free Fall of a Body

Consider an object of mass m, lying at position B. It is made to fall freely from a height (h) above the ground

At point B: At the start, the potential energy is mgh and kinetic energy is zero (as its velocity is zero),

i.e. PE = mg

KE = 0

Total energy, TE = PE + KE = mgh

At point A: As it falls, its potential energy will change into kinetic energy. If v is the velocity of the object at a given instant, its kinetic energy would be \(\frac{1}{2} m v^2\).

PE = mg(h – x)

From Newton’s third equation of motion, v

⇒ \(v^2=u^2+2 g x\)

⇒ \(v^2\)=2 g x[u=0]

⇒ \(\mathrm{KE}=\frac{1}{2} m v^2=\frac{1}{2} m \times 2 g\) x=m g x [\(v^2=2 g x]\)

Total energy, TE = mg

At point C: As the fall of the object continues, the potential energy would decrease while the kinetic energy would increase. When the object is about to reach the ground, h =0 and v will be the highest.

PE =0

KE =\(\frac{1}{2} m v^2=\frac{1}{2} m\left(2 g^h\right)\)=m g h [ \(v^2=2 g^h\)]

Total energy, TE = mg

Thus, the sum of the potential energy and kinetic energy of the object would be the same at all points,

PE + KE = Constant Or mgh+\(\frac{1}{2} m v^2\)= constant

This verifies the law of conservation of energy.

Example 9. An object of mass 10 kg is dropped from a height of 5 m. Fill in the blanks by computing the potential energy and kinetic energy in each case.(Take, g = 10 ms^-2]
Answer:

Given, mass, m =10 kg Height, h = 5 m

Acceleration due to gravity, g = 10 ms^-2

At height b = 5 m,

KE = 0, as v = 0

PE = mgt =10x 10x 5 = 500J

Total energy (KE + PE) at height 5 m = 500 J

At height h = 4m,

KE =\(\frac{1}{2} m v^2=\frac{1}{2} m \times 2 g y [ v^2=2 g]\)

Distance covered, s = 5-4 = 1 m

KE =\(\frac{1}{2} \times 10 \times 2 \times 10 \times 1=100 \mathrm{~J}\)

PE = mgt =10 \(\times 10 \times 4=400 \mathrm{~J}\)

Total energy \((\mathrm{KE}+\mathrm{PE})\) at height 4 \(\mathrm{~m}=(100+400)=500 \mathrm{~J}\)

At height b=3 m,

s =5-3=2 m

KE =\(\frac{1}{2} m \nu^2=\frac{1}{2} m \times 2 g r=m g s=10 \times 10 \times 2=200 \mathrm{~J}\)

Therefore, PE =\(\mathrm{mgh}=10 \times 10 \times 3=300 \mathrm{~J}\)

Total enetgy (KE + PE) at height 3 m =(200+300)=500 J

At height b=2 m,

s =5-2=3 m

KE=m g y=10 \(\times 10 \times 3=300 \mathrm{~J}\)

PE = mgh =10 \(\times 10 \times 2=200 \mathrm{~J}\)

Tatal encrgy (KE + PE) at height 2 m = 300+200=500 J

At height k=1 m,

s = 5-1 = 4 m

KE = \(\frac{1}{2} m v^2=\frac{1}{2} \times m \times 2 \mathrm{~g}\)

= mgr =10 \(\times 10 \times\) 4=400 J

PE = mgt =10 \(\times 10 \times 1=100 \mathrm{~J}\)

Total energy( KE + PE ) at height 1 m =(400+100)=500 J

At just above the ground, h=0,

s=5-0=5 m

KE=\(\frac{1}{2} m v^2=m_S=10 \times 10 \times 5=500 J\)

PE = m \(\mathrm{mg}^{\prime}=10 \times 10 \times\) 0=0

Total energy (KE + PE) at just above the ground

=(500+0)=500 J

Thus, total mechanical energy remains constant at each height, which proves that energy is always conserved.

Example 10. A man is moving with a high velocity of 30 ms^-1. Determine the total mechanical energy of the man weighing 60 kg, if he is on a height of 50 m at this speed. [Take, g =10 ms^-2)

Answer:

Given, the mass of man, m=60 kg

Velocity, v=30 \(ms^{-1}\), height, b=50 m

The total energy (TE) of the man at a height of 50 m is given by

TE = PE + KE

where, PE = potential energy ( = mg )

and KE = kinetic energy \(\left(=\frac{1}{2} m y^2\right)\)

TE =m g h+\(\frac{1}{2} m v^2\) [from Eq. (1)]

=60 \(\times 10 \times 50+\frac{1}{2} \times 60 \times(30)^2\)

=30000+27000

=57000 J

Transformation of Energy (Are Various Energy Forms Interconvertible?)

1. One form of energy can be converted into another form of energy and this phenomenon is called transformation of energy.
2. When an object is dropped from some height, its potential energy continuously converts into kinetic energy.
3. When an object is thrown upwards, its kinetic energy continuously converts into potential energy.

For Example:

1. Green plants prepare their food (stored in the form of chemical energy) by using solar energy through the process of photosynthesis.
2. When we throw a ball, the muscular energy which is stored in our body gets converted into the kinetic energy of the ball.
3. The wound spring in the toy car possesses potential energy. As the spring is released, its potential energy changes into kinetic energy due to which, the toy car moves.
4. In a stretched bow, potential energy is stored. As it is released, the potential energy of the stretched bow gets converted into the kinetic energy of the arrow which moves in the forward direction with large velocity.

Rate of Doing Work: Power

1. The rate of doing work or the rate at which energy is transferred used or transformed to other forms is called power.
2. If work W is done in time t, then
3. Power, P=\(\frac{\text { Work }}{\text { Time }}\)
4. P=\(\frac{\mathbb{W}}{t}\)
5. The SI unit of power is watt in honor ofJarr.c: Wait having the symbol W. We express a larger rate of energy transfer in kilowatt (kW).
6. 1 W=1 \(\mathrm{Js}^{-1}\) or 1 kW=1000 W=1000 \(\mathrm{Js}^{-1}\)
7. 1 MW = \(10^6\) W, 1 (horsepower) HP =746 W

Average Power

1. Average power is defined as the ratio of total work done to the total time taken. An agent may perform work at different rates at different intervals of time.
2. In such a situation, the average power is considered by dividing the total energy consumed by the total time taken.
3. Average power =\(\frac{\text { Tocal energy consumed }}{\text { Total time raken }}\)

Example 11. A boy does 400 J of work in 20 s and then he does 100 J or work in 2s. Find the ratio of the power delivered by the boy in two cases.
Answer:

Case 1 Work done by the boy. \(W_1\)=400 J

Time taken, \(z_1\)=20 s

power, \(P_1\)=z

Case 2 Work done by the bor, \(W_2\)=100 J

Time taken, \(s_2\)=2 s,

power, \(R_2\)= ?

Power, P=\(\frac{\text { Work dont }(W)}{\text { Time taken }(t)}\)

⇒ \(R_1=\frac{W}{t_1}=\frac{400}{20}=20 W\)

and \(R_2=\frac{W_2}{t_2}=\frac{100}{2}=50 W\)

Time raken,\(s_2=2 s\),

Power, P=\(\frac{\text { Work done }(W)}{\text { Time taken }(t)}\)

and \(P_2=\frac{W_2}{t_2}=\frac{100}{2}=50 W\)

So, \(\frac{B_1}{P_2}=\frac{20}{50}=2: 5\)

Example 12. A boy of mass of 55 kg runs up a staircase of 50 steps in 10 s. If the height of each step is 10 cm, find his power. Take g =10 m/s².
Answer:

Weight of the boy = mg

= 55×10 = 550 N

Height of the stains, h=\(\frac{50 \times 10}{100}\)=5 m

Time taken to dim =10 s

Power, p =\(\frac{\text { Work done }}{\text { Time taken }}\)

=\(\frac{\text { mgt }}{t}=\frac{550 \times 5}{10}=275 W\)

Question 13. The heart does 1.2 J of work in each _g Hp heartbeat. How many times per minute does it beat, if its power is 2 W?
Answer:

Here, work done in each heartbeat =1.2 J

t=1 min =60 s, power, P = 2W = 2 \(Js^{-1}\)

Total work done =P x r = 2 x 60=120 J

Number of times heart beats per minute

= \(\frac{\text { Total work done }}{\text { Work done in each beart beat }}\)

= \(\frac{120}{1.2}=100 \text { times }\)

Example 14. A horse exerts a pull on a cart of 500 N so that the horse cart system moves with a uniform velocity of 36 km^-1. What is the power developed by the horse in watts as well as in horsepower?
Answer:

Given, F=500 N

v=\(36 \mathrm{kmh}^{-1}=\frac{36 \times 1000}{3600} \mathrm{~ms}^{-1}=10 \mathrm{~ms}^{-1}\)

As, P=\(\frac{W}{t}=\frac{F s}{t}\)

\(p^2 =\frac{W}{t}=\frac{F_s}{t}\) [ W=\(F_s\) and v = t/t]

=F v=500 x 10=5000 W

In horsepower.

P=\(\frac{5000}{746}\)=6.70 HP

Activity Zone

Activity 1

Objective: To understand why we use the term work differently in science.

Procedure

We come across several activities which are normally considered as work in our day-to-day life.

For each of these activities, the teacher should ask the following questions

1.  What is the work being done?
2. What is happening to the object?
3. Who (what) is doing the work?

Conclusion

1.  When an object moves a distance by applying force on it, work is said to be done.
2. The object moves under the influence of the applied force.
3. The agency that exerts the force is doing work.

Question 1. Work done by the force depends on.
Answer:

Work done by the force depends on the path or displacement of the body.

Question 2. Net work during the motion of the Earth around the Sun Is zero. Why?
Answer:

Net work during the motion of the Earth around the Sun is zero because the angle between force and displacement is 90°.

Question 3. What Is the SI unit of work?
Answer:

The SI unit of work is N-m or joule (J).

Question 4. Give mathematical expression for work done.
Answer:

The mathematical expression for work done (W) is as follows: W = Fs cos θ

where F is force, s is displacement, and θ is the angle between F and s.

Question 5. During a circular motion, what will be the work done?
Answer:

During a circular motion, force is perpendicular to the displacement, so the net work done is zero.

Activity 2

Objective: To understand how work is done.

Procedure:

1. Think of some situations from your daily life involving work.
2.  List them.
3.  Discuss with your friends whether work is being done in each situation or not.
4. Try to reason out your response by taking an example of an engine pulling a train.

Conclusion

The engine exerts a force on the train. Work is being done on the train due to which the train begins to move.

Question 1. 5 N of force Is applied to an object, but the object does not move. Then, how much work Is being done?
Answer:

If there is no displacement due to the application of force, then the net work done will be zero.

Question 2. What are the two conditions that need to be satisfied during work?
Answer:

The two conditions that need to be satisfied during work are

1. A force should act on an object.
2. There must be displacement due to applied force.

Question 3. When an engine pulls a train, who exerts force on It?
Answer:

When an engine pulls a train, the engine exerts a force on it

Question 4. 2 N of force displaces a body by 2 m, the work done will be.
Answer:

Work done by the given force = Force x Displacement =2 x2 = 4 J

Question 5. N-m Is the SI unit of which physical quantity?
Answer:

Ans N-m is the SI unit of work done.

Activity 3

Objective: To understand how work is not done.

Procedure:

1. Think of some situations when the object is not displaced despite a force acting on it Also, think of situations when an object gets displaced in the absence of a force acting on it
2. List all the situations that you can think of for each.
3. Discuss with your friends whether work is done in these situations. Suppose, we work hard to push a huge rock.
4. The rock does not move despite all the effort and we get completely exhausted. However, we have not done any work on the rock as there is no displacement of the rock.
5. Suppose a car is moving at a constant speed. The car moves without any external force acting on it However, there is no work done as there is no external force acting on the car.

Conclusion

1. When a force is applied to an object and it is not displaced from its position, the work done will be zero.
2. When an object gets displaced in the absence of a force acting on it, the work done will be zero.

Question 1. When a bullock pulls a cart, who does work?
Answer:

When a bullock pulls a cart, work is being done by the bullock.

Question 2. We pick a suitcase from the Earth to a height of h. Net work done by us will be (where, mass of suitcase =m, acceleration due to gravity =g).
Answer:

Net work done during the pickup of a suitcase to a height h is W = Fxs = mg

Question 3. Net work done during motion in a circular path is
Answer:

Net work done during motion in a circular path is zero as the angle between force and displacement is 90°.

Question 4. If a body does not move by the application of force, the net work done will be
Answer:

If a body does not move by the application of force, then the work done will be zero because of zero displacement.

Question 5. A waiter is carrying a tray to the table, the net work done by the waiter will be
Answer:

A waiter is carrying a tray to the table, the net work done by him will be zero as the weight of the tray is acting downwards and he is moving horizontally. The angle between force and displacement is 90°.

Hence, work done = 0.

Activity 4

Objective: To understand that the work done by a force can be either positive or negative.

Procedure

1. Lift an object. Work is done by the force exerted by you on the object. The object moves upwards.
2. The force exerted by you is in the direction of displacement. However/there is the force of gravity acting on the object

Conclusion

1. The work done by the force applied by you on the object is positive because it acts in the direction of displacement of the object
2. As the object moves up the force of gravity acts in the downward direction, i.e. in the opposite direction of the displacement of the object.
3. Hence, the work done by the force of gravity on the object is negative.

Question 1. What is the direction of force and displacement, if net work done is zero?
Answer:

If the net work done is zero, it means the direction of force is perpendicular to the direction of displacement.

Question 2. Give the work done in the given diagram.

Answer:

Work done when force and displacement are in opposite direction will be negative, i.e. W = -Fs

Question 3. Give the work done in case of a given diagram

Answer:

Work done in the given diagram is (W) = Fs, as the angle between F and s is 0°.

Question 4. What is the angle between force and displacement for maximum work?
Answer:

Work done will be maximum, if

cos θ = 1 or 0 = 0°, i.e. W = Fs cosθ°=1

Question 5. What is the work done by frictional force when the body is dragged along a rough surface?
Answer:

Work done is negative because the displacement of the body and frictional force is in the opposite direction

Activity 5

Objective: To understand that a falling body possesses kinetic energy due to its motion.

Procedure

1. Take a heavy ball. Drop it on a thick bed of sand. A wet bed of sand would be a better way than any other material.
2. Drop the ball on the sand bed from a height of about 25 cm. The ball creates a depression.
3. Repeat this activity from heights of 50 cm, lm, and 1.5 m.
4. Ensure that all the depressions are distinctly visible.
5. Mark the depressions to indicate the height from which the ball was dropped. Compare their depths.

Observation

1. The depression is deepest when the ball is dropped from 1.5 m height.
2. The depression is shallowest when the ball is dropped from 25 cm height, as it is the minimum height from which, the ball is dropped.

Conclusion

The larger the height from which the ball is dropped, the larger the kinetic energy gained by the ball on reaching the ground and more will be its capability to do work.

Question 1. A stone Is dropped from some height. What kind of energy is present in stone on reaching the ground?
Answer:

On reaching the ground, the stone has kinetic energy.

Question 2. A moving object has velocity due to which it has which type of energy?
Answer:

A moving object has velocity due to which it has kinetic energy.

Question 3. If two balls A and 8 are dropped from the same height. Which one has maximum kinetic energy?
Answer:

If two balls are dropped from the same height, then the heavier ball has greater kinetic energy w.r.t. the lighter one.

Question 4. If two bodies A and 8 of the same mass fall from height hJ and h, respectively, on the sand, where h₂ >hv Which body has more energy?
Answer:

Body B falling from height h₂ has more energy.

Question 5. A ball Is allowed to fall freely from a tower. Which energy Is gained at every point during the fall?
Answer:

Kinetic energy.

Activity 6

Objective: To understand the conversion of energy from one form to another.

Procedure

1. Many of the human activities and the gadgets we use involve the conversion of energy from one form to another.
2. Make a list of such activities and gadgets.
3. Identify in each activity/gadget the kind of energy conversion that takes place.

Observation and Conclusion

1. A list of such activities and gadgets is as follows:
2. Electric cell → Chemical energy changes into electrical energy.
3. Electric motor  → Electrical energy changes into mechanical energy.
4. Electric heater → Electrical energy changes into heat energy.
5. Dynamo → Mechanical energy changes into electrical energy.
6. Headphone → Electrical energy changes into sound energy.
7. Hydroelectric power station → Mechanical energy changes into electrical energy.
8. Microphone → Sound energy changes into electrical energy.
9. Steam engine → Heat energy changes into mechanical energy.
10. Photoelectric cell → Light energy changes into electrical energy.

Question 1. In a photocell, which type of energy conversion takes place?
Answer:

In photocell, light energy is converted into electrical energy.

Question 2. In solar cells, light energy Is converted to what?
Answer:

In solar cells, Sun energy is converted into electrical energy.

Question 3. Give the sequence of energy changes during the production of electricity In the dam.
Answer:

In the production of electricity in the dam, firstly stored potential energy of water is converted into mechanical energy to move the turbine and this mechanical energy gets converted into electricity or electrical energy by a dynamo.

Question 4. Give the energy change In the electric motor.
Answer:

In an electric motor, electrical energy is converted into mechanical energy (to move the body).

Question 5. In the headphones, which type of energy gets converted into sound energy?
Answer:

In the headphones, electrical energy gets converted into sound energy.

Activity 7

Objective: To understand that the sum of kinetic energy and potential energy of an object is its total mechanical energy.

Procedure

An object of mass 20 kg is dropped from a height of 4 m. Fill in the blanks in the following table by computing the potential energy and kinetic energy in each case.

For simplifying the calculations, take the value of g as 10 ms^-2.

Calculation:

At height h=4 m ,

⇒ \(E_P\)=m g h=20 x 10 x 4=800 J

V = 0

⇒ \(E_K=\frac{1}{2} m v^2\)=0

Total energy =\(E_p+E_K\)=800+0=800 J

At height h=3 m,

⇒ \(E_P=m g t=20 \times 10 \times 3\)=600 J

Distance covered, s=4-3=1 m

From \(v^2-w^2\)=2 g s,

⇒ \(v^2 =u^2+2 g s \Rightarrow v^2\)=0+2 g s=2 g s

⇒ \(E_K =\frac{1}{2} m v^2=\frac{1}{2} m \times\) 2 g s=m g s

=20 \(\times 10 \times 1=200 \mathrm{~J}\)

Total energy =\(E_P+E_K\)=600+200=800 J

At height h=2 m,

⇒ \(E_P =m g h=20 \times 10 \times 2\)=400 J

s =4-2=2 m

⇒ \(E_E =m g r=20 \times 10 \times\) 2=400 J

Total energy =\(E_P+E_K\)=400+400=800 J

At height h =1 m,

⇒ \(E_P =m g h=20 \times 10 \times 1\)=200 J

s =4-1=3 m

⇒ \(E_K =m g s=20 \times 10 \times\) 3=600 J

Total energy =\(E_{\Gamma}+E_K\)=200+600=800 J

At just above the ground h=0,

s =4-0=4 m

⇒ \(E_r\)=m g h=0

⇒ \(E_K\) =m g s=20 x 10 x 4=800 J

Total energy =\(E_R+E_K\)

=0+800=800 J

Therefore, the table will be as follows:

Conclusion:

The total mechanical energy remains constant at each height, hence energy is conserved.

Question 1. When a body falls from a certain height, which energy transformation takes place?
Answer:

When a body falls suddenly, then its potential energy gradually gets converted into kinetic energy. On just reaching the ground, the whole potential energy of the body gets converted into kinetic energy.

Question 2. A car accelerates up a hill. What happens to its kinetic energy and its potential energy?
Answer:

The speed of the car is increasing so its kinetic energy increases. Going up the hill implies a gain in the vertical height so its potential energy increases.

Question 3. Give the variation of KE with height (h) graphically.
Answer:

Question 4. Give the variation of PE with height (h) graphically.
Answer:

Question 5. A ball Is thrown upwards from point A. It reaches up to the highest point 8 and then returns what Is Its PE at 8?
Answer:

From the law of conservation of energy, PE at point B = KE at point A.

Activity 8

Objective: To understand the agents that transfer energy to do work at different rates.

Procedure

1. Consider two children, say A and B. Let us say, they have the same weight. Both start climbing up a rope separately and both reach a height of 8 m.
2. For this A takes 15 s while B takes 20 s to accomplish the task.
3. Work done by each child is mgh, where mg is the weight of each child.

Observation

The same amount of work is being done by both but A takes less time (15 s) while B takes more time (20 s) to do the same work.

Conclusion

1. So, A has done more work in 1 s.
2. Hence, A has more power than B.

Question 1. Give the relation between work and power.
Answer:

The relation between work and power is as follows:

Power, P=\(\frac{\text { Work }}{\text { Time }}=\frac{W}{t}
Time t\)

Question 2. A body takes 2 s to do 10 J of work. What is its power?
Answer:

Power =\(\frac{\text { Work done }}{\text { Time taken }}=\frac{10}{2}=5 \mathrm{~W}\)

Question 3., Boy A does 400 J of work In 10 min her boy 8 does 500 J of work in 20 min. Who expends more power A or 8?
Answer:

Power expended by A=\(\frac{W}{t}=\frac{400}{10 \times 60}=0.67 W[ 1 \mathrm{~min}=60 \mathrm{~s}]\)

and power expended by B=\(\frac{W}{t}=\frac{500}{20 \times 60}\)=0.42 W

A expends more power.

Question 4. If a force F acts on the body and moves it with constant velocity v. Then, the power of the body will be
Answer: 

The power of a body moving with velocity v by a force F is given by

Power, P = Force x Velocity ⇒ P = Fv

Question 5. The unit kWh Is used for which physical quantity?
Answer:

kWh is the unit of energy. It is used for commercial purposes.

Work, Energy, And Power Question And Answers

Question 1. A force of 7 N acts on an object. The displacement is say 8 m, in the direction of the force. Let us take it that the force acts on the object through the displacement. What is the work done in this case?

Answer:

As work done, W = Fs[Work done is positive because force and displacement are in the same direction] W=7×8=56J

Question 2. When do we say that work is done?

Answer:

If a force acting on a body causes some displacement, then we can say that work is being done by the force on the body that is displaced.

Question 3. Write an expression for the work done when a force is acting on an object in the direction of its displacement.
Answer:

Expression for the work done is given by

Work done, W = + Fs [F and s are in the same direction] where, F = force, s = displacement,

So, W = Fs

Question 4. Define 1 J of work.

Answer:

1 J of work is defined as the amount of work done on an object when a force of 1 N displaces it by 1 m along the line of action of the force.

Question 5. A pair of bullocks exerts a force of 140 N on a plow. The field being ploughed is 15 m long. How much work is done in plowing the length of the field?

Answer:

Given, force, F =140 N, displacement, s =15 m

Work done, W = + F . s [Fand s are in the same direction]

So, W =F-s = 140×15 =2100J

Question 6. What is the kinetic energy of an object? kinetic energy at the surface of the earth 

Answer:

The kinetic energy of an object is defined as the energy due to its motion.

Question 7. Write an expression for the kinetic energy of an object.

Answer:

The expression for kinetic energy for an object is given by KE=\(\frac{1}{2} m v^2\)

where, KE = kinetic energy, w = mass of the body, v – velocity of the body.

Question 8. The kinetic energy of an object of mass m moving with a velocity of 5 ms^-1 is 25 J. What will be its kinetic energy when its velocity is increased three times?

Answer:

As for the question,

Kinetic energy \(\left(\mathrm{KE}_i\right)\) initially is given by \(\frac{1}{2} m v_i^2\) where, m= mass of the body, v_i= initial velocity

⇒ \(\mathrm{KE}_l \left.=\frac{1}{2} m v_i^2\left[\text { given, } \mathrm{KE}_i=25\right], w_i=5 \mathrm{~ms}^{-1}\right]\)

25 =\(\frac{1}{2} m\left(5^2\right)\)

m =\(\frac{25 \times 2}{5 \times 5}\)

m =2 kg

Now, as from the question, final velocity \(\left(v_f\right)\) becomes 3 times its initial velocity, i.e.

⇒ \(v_f=3 v_i \Rightarrow v_f=3 \times 5=15 \mathrm{~ms}^{-1}\)

Now, kinetic energy, KE =\(\frac{1}{2} m v_f^2=\frac{1}{2} \times 2 \times 15 \times 15\)

=225

Question 9. What is power?

Answer:

Power is defined as the rate of doing work. If the work done by an object in time t is W Then, power, P = \(\frac{W}{t}\)

Its unit is Js^-1 or watt.

Question 10. Define 1 watt of power.

Answer:

Power, P=\(\frac{W}{t}\), If W =1 J, t=1 s

Then, P=\(\frac{1 \mathrm{~J}}{1 \mathrm{~s}}=1 \mathrm{~W}\)

The power of an object is said to be 1 watt if it does 1 J of work in 1 s.

Question 11. A lamp consumes 1000 J of electrical energy in 10 seconds. What is the power?

Answer:

Given energy = 1000

i.e. wark done, W=1000

Time, r=10 s

Power of lamp, P=\(\frac{\mathrm{W}}{t}=\frac{1000}{10}\)=100 W

Question 12. Define average power.
Answer:

Average power is defined as the ratio of total work done to the total time taken.

Exercises

Question 1. Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term ‘work.’

1. Seema is swimming in a pond.
2. A donkey is carrying a load on its back.
3. A windmill is lifting water from a well.
4. A green plant is carrying out photosynthesis.
5. An engine is pulling a train.
6. Food grains are getting dried in the Sun.
7. A sailboat is moving due to wind energy.

Answer:

1. Work is being done by Seema because she displaces the water by applying force.
2. No work is being done by the gravitational force because the direction of force, i.e. load is vertically downward and displacement is horizontally. If displacement and force are perpendicular, then no work is done.
3. Work is done because the mill is lifting the water, i.e. it is changing the position of water.
4. No work is done because there is no force and displacement.
5. Work is done because the engine is changing the position of the train.
6. No work is done because there is no force and no displacement.
7. Work is done because of the force acting on the boat, it starts moving.

Question 2. An object thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and the final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object?

Answer:

As we know work done is the product of force and displacement and here in this case, displacement in the direction of gravitational force (change in height) is zero, so work done by the force of gravity on the object is zero.

Question 3. A battery lights a bulb. Describe the energy changes involved in the process.

Answer:

A battery contains chemicals and supplies electrical energy. So, a battery converts chemical energy into electrical energy.

In an electrical bulb, the electrical energy is first converted into heat energy. This heat energy causes the filament of the bulb to become white-hot and produce light energy.

Thus, the energy changes are Chemical energy →  Electrical energy → Heat energy → Light energy

Question 4. Certain force acting on a mass 20 kg mass changes its velocity from 5 ms^-1 to 2 ms^-1. Calculate the work done by the force.

Answer:

Given, mass, m = 20 kg Initial velocity, u – 5 ms 1 Final velocity, v = 2 ms^-1

Work done by the force = Change in kinetic energy = Final kinetic energy – Initial kinetic energy

= \(\frac{1}{2} m v^2-\frac{1}{2} m u^2\)

= \(\frac{1}{2} m\left(y^2-v^2\right)=\frac{1}{2} \times 20\left[(2)^2-(5)^2\right]\)

= 10 (4-25)=10 x (- 2,1) = – 210 J

Question 5. A mass of 10 kg is at point A on the table. It is moved to a point. If the line joining A and 6 is horizontal, what is the work done on the object by the gravitational force? Explain your Solwer.

Answer:

Here, both the initial and final positions are on the same horizontal line. So, there is no difference in height,

i.e. h =0. where, h = difference in the heights of initial and final positions of the object.

We know that work done by gravitational force, W = mg
Work done, W = mg x 0 = 0

Question 6. The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?

Answer:

Indeed, the potential energy of a freely falling object decreases progressively. But as the object falls Ground^ down, its speed increases, i.e. the kinetic energy of the object increases progressively (kinetic energy will increase with the increase in speed).

Now, we can say that the law of conservation of energy is not violated, because the decrease in potential energy results in the increase of kinetic energy.

Question 7. What are the various energy transformations that occur when you are riding a bicycle?

Answer:

In the case of riding a bicycle, the muscular energy is converted into the kinetic energy of the bicycle.

Question 8. Does the transfer of energy take place when you push a huge rock with all your might and fail to move it? Where is the energy you spend going?

Answer:

When we push a huge rock, then the rock also exerts a large force on us (according to Newton’s third law of motion). The muscular energy spent by us in the process is used to oppose the huge force acting on us due to the rock.

Question 9. An object of mass 40 kg is raised to a height of 5 m above the ground. What is its potential energy, if the object is allowed to fall? Find its kinetic energy when it is halfway down.

Answer:

Given, mass, m = 40 kg, height, h = 5 m As potential energy is given by PE = mgh

So, PE= 40 x 10 x 5 =2000J [v =10 ms^-2]

When it is allowed to fall, its PE gets converted into kinetic energy KE. So, when it reaches to half-way, half of its PE gets converted to KE.

So, mg \(\frac{b}{2}=\frac{1}{2} m v^2\) [where, v= velocity at the bottom]

So,KE =m g \(\frac{b}{2}\)

=40 \(\times 10 \times \frac{5}{2}\)=1000 J

Question 10. What is the work done by the force of gravity on a satellite moving around the Earth? Justify your answer.

Answer:

Work done by the force of gravity on a satellite moving around the Earth is zero. Because of the angle between force (centripetal) and displacement in the case of circular motion.

So, work done, W = 0

Question 11. Can there be displacement of an object in the absence of any force acting on it? Think, and discuss this question with your friends and teacher.

Answer:

Yes, if an object moves with a constant velocity, i.e. there is no acceleration, then no force acts on it. As the object is moving, so it gets displaced from one position to another position.

Question 12. A person holds a bundle of hay over his head for 30 min and gets tired. Has he done some work or not? Justify your answer.

Answer:

On holding a bundle of hay over the head, the work done by the person is zero because there is no displacement.

Question 13. Illustrate the law of s conservation of energy by discussing the energy changes which occur when we draw pendulum bob to one side and allow it to oscillate. Why does the bob eventually come to rest? What happens to its energy eventually? Is it a violation of the law of conservation of energy?

Answer:

1. Let a simple pendulum be suspended from a rigid support S and OS be the equilibrium position of the pendulum. Let the pendulum be displaced to a position P, where it is at rest.
2. At position P, the pendulum has potential energy {mgh). When the pendulum is released from position P, it begins to move towards position O.
3. The speed of the pendulum increases and its height decreases which means the potential energy is converted into kinetic energy.
4. At position O, the whole of the potential energy of the pendulum is converted into its kinetic energy.
5. Then, the pendulum swings to the other side due to the inertia of motion. As the pendulum begins to move towards position Q, the speed of the pendulum decreases and height increases which means kinetic energy is converted into potential energy.
6. At point Q, the whole of the kinetic energy is converted into potential energy. Thus, we find that the potential energy is converted into kinetic energy and vice-versa during the motion of the pendulum. But the total energy remains constant.
7. When the pendulum oscillates in air, the air friction opposes its motion. So, some part of the kinetic energy of the pendulum is used to overcome this friction.
8. With time, the energy of the pendudecreasessing and finally becomes zero.
9. The energy of the pendulum is transferred to the atmosphere. So, energy is being transferred, i.e. is converted from one form to another. So, no violation of the law of conservation of energy takes place.

Question 14. An object of mass m is moving with a constant velocity How much work should be done on the object to bring the object to rest?

Answer:

Concept Change in kinetic energy (KE) =Work done

Given, mass = m, initial velocity, u = v

Final velocity, v = 0

So, W=\(\frac{1}{2} m v^2-\frac{1}{2} m s^2\)

=\(\frac{1}{2} m(0)^2-\frac{1}{2} m v^2\)

W=-\(\frac{1}{2} m v^2\)

Hence, the work that should be done to bring the object to rest is \(\frac{1}{2} m v^2\).

Question 15. Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km-1.

Answer:

Concept Change in kinetic energy is equal to the work done W.

Given, initial velocity, u=60 \(\mathrm{kmh}^{-1}\)

=60 \(\times \frac{5}{18}=\frac{50}{3} \mathrm{~ms}^{-1} {\left[1 \mathrm{kmh}^{-1}=5 / 18 \mathrm{~ms}^{-1}\right] }\)

Final velocity v=0

So, the magnitude of change in kinetic energy =W

=\(\frac{1}{2} m v^2-\frac{1}{2} m v^2\)

W =\(\frac{1}{2} m\left(v^2-s^2\right)\)

=\(\frac{1}{2} \times 1500 \times\left(\frac{-50 \times 50}{9}\right)\)

=-\(\frac{1}{2} \times \frac{1500 \times 50 \times 50}{9}\)

W\(\left.=-\frac{625000}{3}=-208333.3\right)\)

Hence, the work required to be done to stop a car is 208333.3 J.

Question 16. In each of the following, a force F is acting on an object of mass m. The direction of displacement is from West to East shown by the longer arrow. Observe the figure and state whether the work done by the force is negative, positive, or zero.

Answer:

In given. (1), the angle between F and s is 90°, so the work done is zero.

In given. (2), the angle between F and s is 0°, so the work done is positive.

In given. (3), the angle between F and s is 180°, so the work done is negative.

Question 17. Soni says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with her? Why?

Answer:

Yes, I agree with Soni, the acceleration of an object can be zero even when several forces are actonng onacting the resultant of all the forces acting on an object is zero.

Question 18. A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?

Answer:

When a freely falling object strikes the Earth, some sound and heat are produced. So, the kinetic energy of the object converts into sound energy and heat energy.

Summary

Work is said to bo done in a physical activity involving n form and movement in the direction of force and the process an equal amount of energy is used up.

Two conditions need to be satisfied for work to be done.

1. A force should act on the object.
2. The object must be displaced.

Work done by a force on an object is equal to the magnitude of the force multiplied by the distance moved in the direction of force, i.e. U’ = F xs.

• The SI unit of work is joule (J).
• 1 J is the amount of work done on an object when a force of 1 X displaces it by 1 m along the line of action of force.
• Energy is the ability to do work.
• The SI unit of energy is the same as that of work, i.e. joule.
• A larger unit of energy is kilojoule (kJ).
• Relation between Joule and Kilojoule
• 1 kilojoule = 1000 joule.
• The SI unit of work and energy is named after British Physicist, James Prescott Joule.
• Then energy that is possessed by an object due to its motion is called kinetic energy.
• The SI unit of kinetic energy is joule (J).
• The kinetic energy possessed by an object of mass m, moving
  noth a uniform velocity v is KE = \(\frac{1}{2} m v^2\)
• The work-energy theorem is given asW = \(\frac{1}{2} m\left(v^2-u^2\right)\). If initial velocity u = 0, then W = \(\frac{1}{2} m v^2\)
• The energy possessed by a body due to its position or change in its configuration (i.e. shape or position) is known as potential energy.
• The SI unit of potential energy is the joule (J).
• Tho potential energy of an object can be given ns PE = mg.
• According to the law of conservation of energy, energy can neither be created nor be destroyed but it can be transformed from one form to another.
• During the free fall of a body, its energy is always conserved. The sum of the potential energy and kinetic energy of an object is the same at all points.
• One form of energy can be converted into other forms of energy, this phenomenon is known as transformation of energy.

Some Energy Transformations

1. Electric Motor Electrical energy into mechanical energy.
2. Electric Generator Mechanical energy into electrical energy.
3. Steam Engine Heat energy into kinetic energy.
4. Electric Bulb Electrical energy into light energy.
5. Dry Cell Chemical energy into electrical energy.
6. Solar Cell Light energy into electrical energy.
7. The rate of doing work or the rate at which the energy is
   transformed is known as power (P).
8. Power =\(\frac{\text { Work done }}{\text { Time }}\)
9. P =\(\frac{W}{t}\).
10. The SI unit of power is watt (W).
11. The relation between watt and horsepower can be given as 1 (horsepower) HP = 746 W.
12. Average power is defined as the ratio of total work done to total time taken.

Class 9 Science Chapter 11  Sound Very Short Answer Type Questions

Question 1. What is the name of the strings which vibrate in our voice box when we talk?
Answer:

Vocal cord.

Question 2. How would you communicate in an atmos-phereless region?
Answer:

We can communicate through wireless sets using radio waves because radio waves can travel without a medium.

Question 3. If a freely suspended vertical spring is pulled in a downward direction and then released, which type of waves are produced in the spring?
Answer:

In this situation, longitudinal waves will be produced in spring.

Question 4. If a stone is dropped on the surface of water in a pond. Which type of waves are produced in water?
Answer:

In this situation, transverse waves will be produced in water.

Question 5. Give one example of a transverse and longitudinal wave.
Answer:

Transverse wave – Light Longitudinal wave – Sound

Question 6. A baby recognises her mother by her voice. Name the characteristic of sound involved.
Answer:

Quality of the sound is that characteristic of sound due to which distinction can be made between two sounds.

Question 7. On what factor does the quality of the sound depend?
Answer:

The quality of the sound depends on the shape of the sound wave.

Question 8. Why do we hear the loud sound of the horn of an approaching car before the car reaches us?
Answer:

This is because the velocity of the soul is much greater than that of the car.

Read and Learn More Class 9 Science Solutions

Question 9. Among air, water and steel, in which medium, the sound wave will travel faster?
Answer:

The sound wave will travel faster in steel because the speed of sound is the fastest in the solids.

Question 10. How does temperature affect the speed of sound?
Answer:

The speed of sound increases with the increase in temperature.

Question 11. Give the correct picture of the reflection of sound marked with the angle of incidence, angle of reflection and the normal.
Answer:

1. \(\angle i\)= angle of incidence,
2. \(\angle r\)=angle of reflection

Question 12. From which type of surface, the reflection of sound is better?
Answer:

The hard surface is a better reflector of sound.

Question 13. Give examples of multiple echoes.
Answer:

1.  Rolling of thunder,
2. Whispering galleries.

Question 14. At 20°C, what should be the minimum distance of a person from a sound-reflecting surface to hear an echo?
Answer:

The minimum distance from the sound-reflecting surface to hear an echo should be at least 17.2 m.

Question 15. Why are roofs and walls of an auditorium/hall generally covered with sound-absorbent materials?
Answer:

To reduce reverberation, the roofs and walls of an auditorium or hall are generally covered with sound-absorbent materials.

Question 16. Name the phenomenon responsible for the rolling of thunder.
Answer:

It is due to multiple reflections of sound from objects like clouds, land, etc.

Question 17. Why should the curved soundboard be placed behind the stage?
Answer:

It is because, after reflection from the curved sound hoard, sound waves spread evenly apart the width of the hall.

Question 18. What kind of waves are produced in an earthquake before the main shock wave begins?
Answer:

Infrasonic waves are produced in an earthquake before the main shock wave begins.

Question 19. State two important uses of ultrasounds for medical purposes.
Answer:

1. Diagnosing the tumours in the human body.
2. To analyse the development of the foetus.

Question 20. What is the frequency of the wave with a period of 0.025 s?
Answer:

Given, T=0.025 s

v=\(\frac{1}{T}=\frac{1}{0.025}\)=40 Hz

Question 21. A human heart beats 72 times in a minute. Calculate its frequency.
Answer:

Frequency =\(\frac{\text { Number of beats }}{\text { Total time }(\text { in second) }}\)

= \(\frac{72}{60}=1.2 \mathrm{~Hz}[1 \mathrm{~min}=60 \mathrm{~s}]\)

Class 9 Science Chapter 11  Sound Short Answer Type Questions

Question 1. Why do astronauts talk to each other through radiotelephone?
Answer:

As sound waves cannot travel through a vacuum in space. So, they use electromagnetic waves such as radio waves to communicate as it does not require a medium for propagation.

Question 2. Why are the longitudinal waves also called pressure waves?
Answer:

It is because longitudinal waves travel in a medium as a series of alternate compressions and rarefactions, i.e. they travel as variations in pressure. Therefore, they are called pressure waves.

Question 3. When the wire of a guitar is plucked, what types of waves are produced in

1. air
2. wire

Give reasons in support of your answer.

Answer:

1. When the wire of a guitar is plucked, a longitudinal wave is produced in the air due to the to-and-fro motion of the string of the guitar.
2. In the wire of a guitar, the transverse wave is produced as the particle vibrates perpendicular to the direction of motion.

Question 4. Through what type of medium, can

1. the transverse waves and
2. the longitudinal waves be transmitted? Explain in brief.

Answer:

(1) Since transverse waves travel in the form of crests and troughs, the)’ involves changes in the shape of the medium.

So, they can be transmitted through a medium having elasticity of shape. As solids and liquids have elasticity of shape, hence transverse waves can be transmitted through solids and liquids.

(2) Since longitudinal waves travel in the form of compressions and rarefactions, they involve changes in the volume and density of the medium.

All media, i.e. solids, liquids and gases have elasticity of volume. Hence, these waves (longitudinal waves) can be transmitted through all three types of media.

Question 5. Name the types of waves and two examples associated with

1. compressions and rarefactions
2. crests and troughs.

Answer:

1. The waves which travel along a slinky when it is pushed and pulled at one end, are the longitudinal waves. The following figure shows longitudinal waves.
2. The waves produced by moving one end of a long rope (or spring) up and down rapidly whose other end is fixed, are transverse waves as shown

Question 6. When the wire of a sitar is plucked, what kind of waves are produced in

1.  the wire and
2.  air

Answer:

1. Transverse waves are produced in the wire.
2. Longitudinal waves are produced in the air.

Question 7. Why the waves produced by a motorboat sailing in the sea, are both longitudinal and transverse?
Answer:

Longitudinal waves are produced due to vibrations of the rudder inside the sea water but on the surface of the water, transverse waves are generated. Hence, waves produced by motorboats are both longitudinal and transverse

Question 8. At what frequency, a source produces 500 sound waves per second?
Answer:

The frequency of sound waves is called the number of waves, produced in 1 s.

Here, several waves are produced in 1 s = 500 So, the frequency of this sound wave is 500 Hz.

Question 9. What happens to the wavelength, velocity and frequency of sound waves, when it travels from one medium to another?
Answer:

When a sound wave travels from one medium to another medium, its wavelength as well as velocity may change. But frequency does not change.

Question 10. Some animals get disturbed before the earthquakes. Give reason.
Answer:

During earthquakes, some animals get disturbed because earthquake produces low-frequency infrasound before the main shock waves, which is audible to some animals.

Question 11. Which of the two graphs (1) and (2) representing the human voice is likely to be the male voice? Give a reason for your answer

Answer:

Graph (1) represents the male voice. Since the pitch and frequency of the male voice are lower than the pitch of the female voice vibration of the graph (2) represents a higher frequency and higher pitch.

Question 12. Waves of frequency 100 Hz are produced in a string as shown. Give its

1. Amplitude
2. Wavelength
3. Velocity
4. Nature

Answer:

1. Amplitude = 5 cm
2. Wavelength – 20 cm
3. Velocity, v=\(v \lambda=100 \times 20 \times 10^{-2}=20 \mathrm{~ms}^{-1}\)
4. Nature It is a transverse wave.

Question 13. What is wave motion? Write any four characteristics of wave motion.
Answer:

• Wave motion is nothing but a mode of transfer of energy from place to place periodically without material transport.
• Four characteristics of wave motion are:
• It is the disturbance which travels forward through the medium but not the particles of the medium.
• Each particle receives vibrations a little later than its preceding particle.
• The wave velocity is different from the velocity of the particles with which they vibrate about their mean positions.
• The wave velocity remains constant in a given medium, whereas particle velocity changes continuously during its vibrations about the mean position.

Question 14. The graph shows a trace of a sound wave which is produced by a particular tuning fork.

1. Draw a trace of the sound wave which has a higher frequency than that shown
2. Draw a trace of the sound wave which has a larger amplitude than that shown

Answer:

1. The property which leads to the formation of echoes is the reflection of sound. When a sound is repeatedly reflected and if the minimum distance between the sound source and the reflecting surface is more than 17.2 m, then we hear echoes.
2. If the reverberation time in a big hall is too long, then the sound becomes distorted, blurred and confusing.

Question 15. Give the basic difference between wave velocity and particle velocity.

Answer:

The wave velocity is constant for a given medium and is expressed by v = vA, while the particle velocity changes with time. It is maximum at the mean position and zero at the extreme position.

Question 16 (1) Which property of sound leads to the formation of echoes? Briefly explain.

(2) What will happen, if the reverberation time in a big hall is too long?

Answer:

The property which leads to the formation of echoes is the reflection of sound. When a sound is repeatedly reflected and if the minimum distance between the sound source and reflecting surface is more than 17.2 m, then we hear echoes.

If the reverberation time in a big hall is too long, then the sound becomes distorted, blurred and confusing.

Question 17. What is meant by the loudness of sound? State the factor on which it depends. Draw figures to illustrate

1. soft sound and
2. loud sounds.

Answer:

1. Loudness is the measure of sound energy reaching the ear per second. The greater the sound energy reaching our ears per second, the louder the sound will appear to be.
2. It depends on the amplitude of the sound wave, which depends upon the force with which an object is made to vibrate.

Question 18. A nail was gently touched by the hammer and then was hit harder.

1. When will be the sound created louder?
2. Which characteristic of sound here is responsible for the change in sound?

Answer:

1. Sound will be produced when we beat hard on the nails.
2. The amplitude of the vibrating body is responsible for the change in sound.

Question 19. State the conditions required to hear an echo.

Answer:

The conditions to hear an echo are

1. The time interval between source sound and reflected sound must be at least 0.1s.
2. The minimum distance between the obstacle and the source of sound should be at least 17.2 m.

Question 20. When can we distinctly hear the echo of a sharp sound? Why cannot we hear an echo in a small hall?

Answer:

1. As we know, to listen to echo for a sound wave whose speed in air is 344 ms¹ and persistence of sound is 0.1 s, a minimum distance of 17.2 m between the observer and reflecting surface is required.
2. The condition of the minimum distance between the die source and the reflecting surface is not satisfied, so in a small hall, we do not hear echoes.

Question 21. Explain the rolling action of thunder.

Answer:

The rolling of thunder is due to multiple reflections of the sound of thunder from a number of reflecting surfaces such as clouds and land as several echoes may be heard.

Question 22. Distinguish between the terms

1. music and noise
2. tone and note.

Answer:

(1) Music The sound which is pleasant to the ears is called music. It is produced by regular periodic vibrations. There is no sudden change in loudness, e.g. Sound produced from a tabla.

Noise The sound which is unpleasant to the cars is called noise. It is produced at irregular intervals. There is a sudden change in its loudness, For Example. The sound produced in a market and the sound produced by an explosion.

(2) Tone The sound of a single frequency is called a tone.

Note The sound which is a mixture of several frequencies is called a note.

Question 23. When a workman hammers to one end of the long iron pipeline, an observer places his ear on the other end of the pipeline. How he can distinctly hear two sounds? Justify your answer.

Answer:

Due to the propagation of sound through solids such as iron, the sound of hammering to one end will be heard by the observer on the other end of the pipeline.

Also, the sound of hammering will be propagated through air to reach the observer.

As we know, sound travels faster in iron than in air. So, the observer hears two sounds. The first one, travelling through the iron pipeline and the second travelling through air.

Question 24. (1) Define

1.  infrasonic wave
2. ultrasonic wave.

(2) Name two species of animals which can produce and detect

1.  infrasonic waves
2. ultrasonic waves.

Answer: 

(1) (1) Infrasonic wave The sound of frequency lower than 20 Hz is known ai infrasonic sound (or wave).

(2) Ultrasonic wave The sound of frequency higher than 20,000 Hz arc is called ultrasonic wave.

(2) (A) Animals like whales, elephants and rhinoceroses produce infrasonic sound of frequency 5 Hz,

(B) Dogs can hear and detect ultrasonic waves of frequency up to 50 Hz.

Question 25. (1) Which has a shorter wavelength infrasonic or ultrasonic?

(2) Can dolphins detect ultrasonic waves?

(3) Name any one living organism which can detect infrasonic waves.

Answer:

1. The ultrasonic wave has shorter wavelengths.
2. Yes, dolphins can detect ultrasonic waves.
3. Rhinoceros can detect infrasonic waves.

Question 26. A key of a piano is struck gently and then struck again but much harder this time. What will happen in the second case?

Answer:

1. In the second case, the key is struck harder, so the amplitude of vibration of the string increases and hence loudness increases.
2. Also, frequency and hence pitch increase with an increase in force or tension in the string.

Question 27. A student went to a hill station early in the morning, he could hear the echo of his clap after 0.1 s. When he went to the same place in the afternoon he could not hear the echo at all. Explain the reason for his changed observation.

Answer:

There is a rise in temperature in the afternoon, so the speed of sound increases with an increase in temperature, As speed increases, the time taken by reflected sound will be less, which may be less than 0.1 s in the afternoon.

That is why the student could not hear the echo at all in the afternoon.

Question 28. A girl is sitting in the middle of a park of dimension 12 m x 12 m. On the left side of it, there is a building adjoining the park and on the right side of the park, there is a road adjoining the park. A sound is produced on the road by a cracker. Is it possible for the girl to hear the echo of this sound? Explain your answer.

Answer:

No, the girl can’t hear the echo of this sound because, the distance between the girl and the obstacle (building) is only 6 m approx but an echo is heard only, if the minimum distance between the observer at the source of sound and the obstacle is 11.3 m

Question 29. The wavelength of a sound wave was measured by Rohit as 8 m. The frequency of a sound wave is given as 40 Hz. What is the speed of sound as calculated by Rohit?

Answer:

Given, \(\lambda\)=8 m and v=40 Hz

As speed of sound, v= v \(\lambda=40 \times 8=320 \mathrm{~ms}^{-1}\)

Question 30. How are ultrasonic waves different from ordinary sound waves? State two applications of ultrasound.

Answer:

Ultrasonic waves have greater frequency (more than 20,000 Hz). Ordinary sound has a lower frequency than ultrasonic waves.

Due to their high frequencies,

1.  they have high power.
2. they can penetrate anywhere to a large extent.
3. they are able to travel along well-defined straight paths, even in the presence of obstacles.

Applications of ultrasound are

1. To deter the flaw or defect in metal.
2. In the diagnosis of diseases.

Question 31. An echo is heard on a day when the temperature is about 22°C. Will echo be heard sooner or later, if the temperature increases to 40°C?

Answer:

An echo will be heard sooner than the echo heard when the temperature is 22°C because the speed of sound increases with an increase in temperature.

Also, speed of sound in air = \(\frac{\text { distance }}{\text { time }}\)

i.e. If the speed of sound increases, then the time after which the echo will be heard decreases.

Question 32. (1) Write two main properties of ultrasound.

(2) Mention one application of ultrasound in (A) industries (B) the medical field.

Answer:

(1) The two main properties of ultrasound are

•  High frequency
• These travel along well-defined paths even in the presence of obstacles.

(2) (A) In industries, ultrasound is used to detect cracks and flaws in metal blocks.

(B) In the medical field, ultrasound is used to break stones in the gall bladder and kidney.

Question 33. How is it Mint, hots or able to fly at night without colliding with other objects?

Answer:

1. Bats search out prey and fly at night by emitting and detecting reflections of ultrasonic waves.
2. Bats emit high-frequency ultrasonic squeaks while flying and listen to the echoes produced by the reflection of their squeaks from the obstacle i.e. prey or object in their path.
3. From the time taken by the echo to be heard, bats can determine the distance of the object and can avoid the object by changing the direction without colliding with it.

Question 34. Find the wavelength of sound for frequencies up to 120 kHz at which a bat can hear it. Take the speed of sound in air as 344 99 \(\mathrm{~ms}^{-1}\).

Answer:

Given, speed of sound in air, v=340 \(\mathrm{~ms}^{-1}\)

Time taken to hear echo, t=3 s

As we know that, distance, s= speed (v) \(\times time (t)\)

=340 \(\times\) 3=1020 m

As in 3s, sound has to travel twice the distance between man and the cliff. So, the distance between man and the cliff is

=\(\frac{s}{2}=\frac{1020}{2}\)=510 m

Question 35. Find the wavelength of the tone produced by a body vibrating with a frequency of 4 kHz. [Given the speed of sound in air is 344 m/s]

Answer:

Given,v=4 \(\mathrm{kHz}=4000 \mathrm{~Hz}\) and v=344 \(\mathrm{~m} / \mathrm{s}\)

As, v=v \(\lambda \Rightarrow \lambda=\frac{v}{v}=\frac{344}{4000}=0.086 \mathrm{~m}\)

Question 36. A man fires a rifle in front of a cliff and hears the echo after 3 seconds. Calculate the distance of man from the cliff, if the velocity of sound in air is 340 ms-^-1.

Answer:

Given, speed of sound in air, v = 340 ms^-1 Time taken to hear echo, t = 3 s As we know that, distance, s=speed (v) x time (t)

= 340 x 3 =1020m

In 3 s, sound has to travel twice the distance between man and the cliff. So, the distance between man and the cliff is

=\(\frac{s}{2}=\frac{1020}{2}=510 \mathrm{~m}\)

Question 37. A sound wave has a frequency of 3 kHz and a wavelength of 45 cm. How long will it take to travel 1.8 km?

Answer:

Given, frequency,v=3 kHz=3 \(\times 10^3 \)Hz

Wavelength, \(\lambda=45 \mathrm{~cm}\)=45 \(\times 10^{-2} \mathrm{~m}\)

From the relation, speed of wave, v=v \(\lambda\)

=3 \(\times 10^3 \times 45 \times 10^{-2}=1350 \mathrm{~ms}^{-1}\)

Distance to be covered =\(1.8 \mathrm{~km}=1.8 \times 1000\)

=1800 m

Time taken by the wave to cover the distance of 1.8 km

= \(\frac{\text { Distance }}{\text { Speed }}=\frac{1800}{1350}=1.33 \mathrm{~s}\)

Question 38. Ocean waves of time period of 0.01 s have a speed of 15 ms-1. Calculate the wavelength of these waves. Find the distance between a crest and an adjoining trough.

Answer:

Given, time period of the waves, r=0.01 s

Speed of the waves, v=15 \(\mathrm{~ms}^{-1}\)

Wavelength of these waves, \(\lambda=\frac{v}{v} [ v=v \lambda]\)

or \(\lambda=\nu \times T=15 \times 0.01=0.15 \mathrm{~m}\)

The distance between the crest and adjoining trough

= \(\lambda / 2=\frac{0.15}{2}\)

= 0.075 m

Question 39. In a ripple tank, 14 full ripples are produced in 1 s. If the distance between a crest and the next trough is 12 cm, calculate (1) wavelength and (2) velocity of the wave.

Answer:

(1) Given, the distance between a crest and the next trough

=\(\frac{\lambda}{2}=12 \mathrm{~cm}\)

The wavelength of the wave,

⇒ \(\lambda=12 \times 2=24 \mathrm{~cm}\)=0.24 m

(2) Number of ripples produced in 1 s=14 Hz

Frequency of wave, v=14 Hz

As velocity, \(\nu=\mathrm{v} \lambda\)

v=14 \(\times 0.24=3.36 \mathrm{~ms}^{-1}\)

Question 40. The sound produced by a thunderstorm is heard 10 seconds after the lightning is seen. Calculate the approximate distance of the thundercloud. (Given, the speed of sound = 340 ms^-1)

Answer:

Given, time, t =10 s and speed, v =340 ms^-1

We know that, distance = speed x time

= 340 x 10 = 3400 m = \(\frac{3400}{1000} \mathrm{~km}\)

= 3.4 km

Question 41. Compare the frequencies of notes A and C played during a musical concert when the speed of sound is 340 ms^-1 and the wavelength of A and C are 1.5 m and 1.33 m, respectively.

Answer:

As we know that, from the relation, v =v\(\lambda\)

Frequency of the sound wave (v)

Speed of sound (y)

Wavelength of the sound wave (k)

=\(\frac{\text { Speed of sound }(v)}{\text { Wavelength of the sound wave }(\lambda)}\)

So, frequency of note A, \(v_A=\frac{340}{1.5}=226.66 \mathrm{~Hz}\)

Again, frequency of note C, \(v_C=\frac{340}{1.33}\)=255.63 Hz

So, the frequency of note C is more than the frequency of note A.

Question 42. Aditi clapped her hands near a cliff and heard the echo after 4 seconds. What is the distance of the cliff from her, if the speed of sound is taken as 346 ms^-1?

Answer:

Given, the speed of sound, v =346 ms

Time taken to hear the echo,

t = 4 s

Distance travelled by the sound = v x t

= 346 x 4 = 1384m

In 4 s, sound has to travel twice the distance between the cliff and Aditi. Therefore, the distance between the cliff and Aditi is \(\frac{1384}{2}\)=692 m

Question 43. A sound wave travels at a speed of 340 ms¹. If its wavelength is 1.5 cm. What is the frequency of the wave? Will it be audible?

Answer:

Given that, speed of sound, v=340 \mathrm{~ms}^{-1}

Wavelength, \(\lambda=1.5 \mathrm{~cm}=\frac{1.5}{100} \mathrm{~m}\). Frequency, v=?

Using the formula, v=\(\mathrm{v} \lambda\)

= \(\frac{\nu}{\lambda}=\frac{340}{\left(\frac{1.5}{100}\right)} \mathrm{ms}^{-1}\)

= \(\frac{340 \times 100}{1.5}=22666.6 \mathrm{~Hz}\)

We know that the audible range of frequencies is 20 Hz to 20000 Hz.

So, the given frequency (i.e. 22666.6 Hz) is not audible.

Question 44. If the velocity of sound in air is 330 ms-1, then express the audible range of frequencies in terms of time period.

Answer:

The audible range of frequencies is V1 =20 Hz,

v₂ = 20000 Hz and the velocity, v = 330 ms^-1

So, time period will become,

⇒ \(T_1=\frac{1}{\mathrm{v}_1}=\frac{1}{20}=5 \times 10^{-2} \mathrm{~s}\)

and time period will become,

⇒ \(T_2=\frac{1}{v_2}=\frac{1}{20000}=5 \times 10^{-5} \mathrm{~s}\)

Thus, the audible range in terms of time period is from 5 \(\times 10^{-2} \mathrm{~s}\) to 5 \(\times 10^{-5} \mathrm{~s}\).

Question 45. A ship sends out an ultrasound that returns from the sea bed and is detected after 1.71 seconds. If the speed of ultrasound through seawater is 1531 ms^-1, what is the distance of the sea bed from the ship?

Answer:

Given, the time between transmission and detection, t =1.71 s

Speed of ultrasound through seawater, v =1531 ms^-1

Distance travelled by the ultrasound

= 2 x Depth of the sea = 2d

or 2d = 1 x r =1531 x 1.71 = 2618.01m

d=\(\frac{2618.01}{2}=1309 \mathrm{~m}\)

Hence, the distance of the sea bed from the ship is 1309 m.

Question 46. For hearing the loudest ticking sound heard by the ear, find the angle x in the given figure.

Answer:

We know that in the laws of reflection, the angle of incidence (x) is always equal to the angle of reflection (x). Since AOB is a straight line.

⇒ \(\angle A O B\) =\(180^{\circ}\)

⇒ \(50^{\circ}+x+x+50^{\circ}=180^{\circ}\)

[sum of all angles lies on the same side of a line is \(180^{\circ}\) ]

⇒ \(2 x+100^{\circ}=180^{\circ}\)

⇒ \(2 x=180^{\circ}-100^{\circ} \Rightarrow 2 x=80^{\circ}\)

x=\(\frac{80^{\circ}}{2} \Rightarrow x=40^{\circ}\)

Hence, the value of x is 40°.

Question 47. A boat moving with a velocity of 20 ms¹ in a sea is rocked by waves. If its crests are 80 m apart, then at what time does the boat bounce up?

Answer:

Given that, wavelength, \(\lambda\) = 80 m

(Distance between two consecutive crests and troughs is equal to the wavelength.)

Velocity, v = 20 ms^-1, time, T = ?

As we know that, T=\(\frac{\lambda}{v}=\frac{80}{20}=4 \mathrm{~s}\)

Question 48. The given graph shows the displacement versus time relation for a disturbance travelling with a velocity of 1500 ms^-1. Calculate the wavelength of the disturbance.

Answer:

Given, velocity, v=1500 \(\mathrm{~ms}^{-1}\)

Time taken in one complete cycle is 2 \(\mu \mathrm{s}\)

Time, T=2 \(\mu \mathrm{s}=2 \times 10^{-6} \mathrm{~s}\left[ 1 \mu \mathrm{s}=10^{-6} \mathrm{~s}\right]\)

We know that,v=\(\mathrm{v} \lambda\left[\mathrm{v}=\frac{1}{T}\right]\)

So, v=\(\frac{\lambda}{T} \Rightarrow \lambda\)=v T

where, \(\lambda\)= wavelength, v= frequency and T= time period

So,\(\lambda =1500 \times 2 \times 10^{-6}=3000 \times 10^{-6}\)

=3 \(\times 10^{+3} \times 10^{-6}=3 \times 10^{-3} \mathrm{~m}\)

Question 49. A construction worker’s helmet slips and falls when he is 78.4 m above the ground. He hears the sound of the helmet hitting the ground 4.23 s after it slipped. Find the speed of sound in the air.

Answer:

Here, s = distance travelled by the helmet to reach the ground =78.4 m

t = total time taken to hear the sound of the helmet hitting the ground =4.23 s

Let it be the time taken by the helmet to reach the ground.

So, we know that, according to the equation of motion, where, u = initial velocity g = acceleration due to gravity.

s=u t+\(\frac{1}{2} g t^2\left[\begin{array}{l}
\text { where, } u=\text { initial velocity } \\
g=\text { acceleration due to gravity. }\end{array}\right]\)

78.4=0 \(\times t+\frac{1}{2} \times 9.8 \times t^2 \Rightarrow 78.4=\frac{1}{2} \times 9.8 \times t^2\)

⇒ \(t^2=16 \Rightarrow\) t=4 s

So, time is taken by the sound wave to travel distance.

78.4 \(\mathrm{~m}=4.23 \mathrm{~s}-4 \mathrm{~s}=0.23 \mathrm{~s}\)

So, speed of sound in air =\(\frac{78.4 \mathrm{~m}}{0.23 \mathrm{~s}}\)=340.86 \(\mathrm{~ms}^{-1}\)

Question 50. A sound wave has a frequency of 2 kHz and a wavelength of 45 cm. It takes 4 s to travel. Calculate the distance it travels.

Answer:

Here, the frequency of sound waves,

v = 2 kHz = 2x 10 3 Hz

Wavelength of the wave, \(lambda\) = 45 cm =0.45 m

Velocity of the wave, v = v \(\lambda\)

= 2 x 103 x 0.45 =900 ms^-1

Time to travel = 4 s

Distance travelled by the wave

= vt =900 x 4 =3600 m

Class 9 Science Chapter 11  Sound Long Answer Type Questions

Question 1. (1) What is meant by frequency of sound
waves?

(2) Give the range of frequencies of sound waves that an average human ear can detect.

(3) A source of wave produces 20 crests and 20 troughs in 0.2 s. The distance between a crest and next trough is 50 cm. Find the

• wavelength
•  frequency
• period of the wave.

Answer:

(1) Frequency The number of waves produced per second is called the frequency of the wave.

(2) 20 Hz to 20 kHz

(3)(A) Since the distance between a crest and the next trough is \(\frac{\lambda}{2}\).

Therefore, \(\frac{\lambda}{2}\)=50 cm (given)

⇒ \(\lambda\)=100 cm or 1 m

(B) Distance covered in 20 crests =20 \(\times\) 1=20 m

Velocity of the wave, \(\nu=v \lambda=\frac{\lambda}{t}\)

v \(\times 1=\frac{20}{0.2}\)[ t=0.2 s]

v=100 Hz or

1 \(\text { crest and } 1 \text { trough }=1 \text { wave }\)

20 \(\text { crest and } 20 \text { trough }=20 \text { waves }\)

Frequency =\(\frac{\text { Number of waves }}{\text { Time }}=\frac{20}{0.2}=100 \mathrm{~Hz}\)

1 crest and 1 trough =1 wave

20 crest and 20 troughs =20 waves

Frequency =\(\frac{\text { Number of waves }}{\text { Time }}=\frac{20}{0.2}=100 \mathrm{~Hz}\)

(C)Time period, T=\(\frac{1}{\mathrm{v}}=\frac{1}{100}\)=0.01 s

Question 2. Define frequency and wavelength with reference to sound. Explain, what is echo. Give two applications of ultrasound.

Answer:

Frequency The number of complete sound waves (or oscillations) produced in one second is called the frequency of the sound wave.

Wavelength The minimum distance in which a sound wave repeats itself.

Echo The repeddon of sound caused by the reflection of sound waves is called an echo.

Two applications of ultrasound are

1. Ultrasound is used in industries to detect flaws in metal blocks without damaging them.
2. Ultrasound is used to investigate internal organs of the human body such as the liver, gall bladder, pancreas, kidneys, etc.

Question 3. Establish the relationship between the speed of sound, its wavelength and frequency. If the velocity of sound in air is 340 ms^-1. Calculate

1. wavelength when the frequency is 256 Hz.
2. frequency when the wavelength is 0.85 m.

Answer:

The speed of sound is defined as the distance at which a point on a wave, such as a compression or a rarefaction, travels per unit of time.

We know that, speed, v=\(\frac{\text { distance }}{\text { time }}=\frac{\lambda}{T}\)

Here, \(\lambda\) is the wavelength of the sound wave. It is the distance travelled by the sound wave in one time period (T) of the wave.

⇒ \(\nu=\frac{\lambda}{T}=\lambda \times \frac{1}{T}\)

We know that, v=\(\frac{1}{T}\) [where, v= frequency]

⇒ \(\nu=\lambda v\)

i.e. Speed = Wavelength x Frequency

Given, speed of sound in air, v=340 \(\mathrm{~ms}^{-1}\) and frequency, v=256 Hz

(1) Speed = Wavelength x Frequency

⇒ \(\Rightarrow 340=\lambda \times 256 \Rightarrow \lambda=\frac{340}{256}=133 \mathrm{~m}\)

(2) Again, given wavelength, \(\lambda\)=0.85 m

Then, frequency of sound in air, v=\(\frac{\text { Speed }}{\text { Wavelength }}\)

⇒ {Speed= Wavelength x Frequency}

v=\( \frac{340}{0.85}=\frac{340 \times 100}{85}=400 \mathrm{~Hz}\)

Thus, the frequency of sound is 400 Hz.

Question 4. Draw a curve showing density or pressure variations with respect to distance for a disturbance produced by sound. Mark the position of compression and rarefaction on this curve. Also, define wavelengths and time periods using this curve.

Answer:

We have a curve showing density or pressure variations with respect to distance for a disturbance produced by sound.

Wavelength can be defined as the distance between two successive compressions or rarefactions. It is denoted by λ.

The time taken by the waves to complete one full cycle so that its particles are in the same phase is called time period. It is denoted by T.