## Chapter 2 Topic C Newtons Third Law Of Motion And Law Of Conservation Of Linear Momentum Synopsis:

**Newton’s Third Law Of Motion:**

To every action, there is an equal and opposite reaction.

Forces can be divided primarily into two types

- Contact forces and
- Non-contact force.

A forces that require being in contact with another object are contact forces.

**Examples:** Frictional force, tension force, etc.

A forces that can be exerted without requiring any contact with any object are non-contact forces.

**Examples:** Gravitational force, magnetic force, electrostatic force, etc.

**Read and Learn More WBBSE Solutions for Class 9 Physical Science and Environment**

**Law Of Conservation Of Linear Momentum: **

If no external force is applied on a system of particles, the total linear momentum of the system remains unchanged.

**Explanation:**

Suppose two bodies of masses m_{1} and m_{2} moving along a straight line with velocities u_{1} and u_{2} (u_{1} > u_{2}) respectively collide with each other.

After the collision, the velocities of these two bodies are v_{1} and v_{2} respectively.

The total linear momentum of the system before the collision is m_{1}u_{1} + m_{2}u_{2} and total linear momentum of the system after the collision is m_{1}v_{1} + m_{2}v_{2}.

According to the law of conservation of linear momentum,

m_{1}u_{1} +m_{2}u_{2} = m_{1}v_{1} +m_{2}v_{2}

## Chapter 2 Topic C Newtons Third Law Of Motion And Law Of Conservation Of Linear Momentum Short And Long Answer Type Questions

**Question 1. Write down Newton’s third law of motion and explain it.**

**Answer:**

To every action, there is an equal and opposite reaction.

Suppose, block P in a moving state applies a force \(F_{Q P}\) on the block Q.

At the same time, block Q also applies a force \(F_{P Q}\) on block P.

If force \(F_{Q P}\) is action, then \(F_{P Q}\) is reaction.

According to Newton’s third law of motion,

\(F_{Q P}\) = –\(F_{P Q}\)

As long as action exists, reaction also continues to exist.

If there is no action, there is also no reaction. This means that action and reaction take place simultaneously, it never happens that one occurs earlier and the other occurs later.

**Question 2. With the help of an experiment, show that action and reaction are equal and opposite.**

**Answer:**

Two spring balances are taken. The hook of one balance is attached to the hook of the other. Now the remaining two ends of the balances are pulled by two hands in such a way that there is no displacement of the two spring balances.

It is seen that the reading shown by the left spring balance is equal to the reading shown by the right spring balance.

Now, the right spring balance is fixed with the wall and the end of the left spring balance is pulled. In this case also, the reading of the left spring balance is the same as the reading of the right spring balance.

In the first case, if the force of the left hand is designated as action, then the force of the right hand is reaction.

In the second case also, if the force applied on the spring is action, then the force applied by the wall is reaction. Hence, it is found that action and reaction are equal and opposite.

**Question 3. Opposite poles of two magnets attract each other—explain this phenomenon with the help of Newton’s third law of motion.**

**Answer:**

If the north pole of a bar magnet is brought near the south pole of another bar magnet, it is seen that the two magnets move towards each other, that is, the two poles attract each other.

If the magnet on the left side attracts the magnet on the right side with force F_{1} and the magnet on the right side attracts the magnet on the left side with force F_{2}, then F_{1} = -F_{2}. Therefore, if force F_{1 }is called action, then F_{2} is the reaction.

Since action and reaction are equal and opposite, hence this type of attraction is called mutual attraction force.

**Question 4. When a shell is fired from a cannon, the cannon recoils. Explain this phenomenon with the help of Newton’s third law of motion.**

**Answer:**

When a shell is fired from a cannon, the shell moves forward with high velocity and the cannon recoils immediately. The force applied by the cannon on the shell is taken as action, then the shell moves forward due to this action. The equal and opposite reaction exerted by the shell on the cannon causes the cannon to recoil.

**Question 5. Give a scientific explanation of our walking with the help of Newton’s third law of motion.**

**Answer:**

When we walk, we obliquely exert a force (F) on the ground. The ground also exerts an equal reaction (R) on us in the opposite direction. The horizontal component of this reaction force (H) helps us to move forward.

**Question 6. Why is a boat pushed backward, if a passenger jumps from it towards the shore?**

**Answer:**

When a passenger jumps from a boat, he applies a force on the boat with his legs. Due to the influence of this force, the boat moves backward. At the same moment, the boat also applies a reaction force of equal magnitude, opposite to the direction of force applied by the passenger.

This force helps the passenger to reach the shore.

**Question 7. Though action and reaction forces are equal and work in opposite directions, why cannot they establish equilibrium?**

**Answer:**

Two equal and opposite forces applied on a body with the same line of action, keep the body in equilibrium. But though action and reaction are equal and in opposite directions, they always act on two different bodies.

Hence, action and reaction forces cannot establish equilibrium.

**Question 8. What do you mean by tension of a rope?**

**Answer:**

One end of a rope is firmly fixed and the other end is hanged with a wooden block attached to it.

The weight (W) of the block is acting downward in a perpendicular direction, so the block applies a force of equal amount of its weight on the rope which acts downward.

The rope also applies a force on the block equal to the weight of the body, acting in an opposite direction. This force is called tension of a rope.

**Question 9. According to the third law of motion, if a person pushes a book, then the book also pushes the person in the opposite direction. In which direction does the book move?**

**Answer: **

As the person pushes the book, then according to Newton’s third law of motion, the book also pushes the person in the opposite direction with equal magnitude of force.

The force applied by the person is action force whereas the force applied by the book is reaction force. Action and reaction are always of equal magnitude working in opposite directions but they always act on separate bodies.

That is why they cannot establish equilibrium. So the book moves in the direction of the force applied by the person. But the amount of force applied should be greater than the frictional force applied by the book to put it in motion.

**Question 10. What do you mean by attraction and repulsion?**

**Answer:**

If the opposite poles of two-rod magnets are brought nearby, it is seen that due to mutual action-reaction, they come close to each other. This type of force is called attraction.

Again, if the like poles of these two-rod magnets are brought nearby, it is seen that due to mutual action-reaction, they recede away. This type of force is called repulsion.

This means if two bodies come nearby due to mutual action-reaction, then that force is called attraction and if two bodies move away from each other, then that force is called repulsion.

**Question 11. In the following given cases, what is the acting force?**

- A stone is attached to a string and the system is suspended.
- A rolling iron ball strikes another ball.
- A man is standing on the floor.
- An iron nail kept in front of a magnet moves towards the magnet.
- When two rod magnets are kept In front of each other, they repulse each other and move away In opposite directions.
- A book when pushed on a table, traverses some distance and then stops.

**Answer:**

Incidents and acting forces are mentioned below:

**Question 12. Among the forces mentioned below, Which are responsible for acceleration, deceleration, and establishment of equilibrium:**

- Tension,
- Force due to collision,
- Normal force,
- Attraction,
- Repulsion,
- Friction.

**Answer:**

**Tension:**force that establishes equilibrium.**Force due to collision:**Force that creates acceleration or deceleration.**Normal force:**Force that establishes equilibrium.**Attraction:**Force that creates acceleration or deceleration.**Repulsion:**Force that creates acceleration.**Friction:**Force that creates deceleration.

**Question 13. Give an example to show that attraction force may create acceleration as well as deceleration.**

**Answer:**

If a body is thrown upwards from the earth’s surface, its velocity gradually decreases due to force of gravity acting on it. This means it decelerates. Here, gravity is the attraction force.

Again, when the body moves downwards after reaching the highest point, its velocity gradually increases. This means it accelerates. This acceleration also takes place due to gravity.

Hence, attraction force may create acceleration as well as deceleration.

**Question 14. A sand clock is weighed twice using a sensitive common balance, once when sand particles are slowly falling from upper chamber to lower chamber and the other time when all particles have accumulated in the lower chamber. Are the two weights equal?**

**Answer:**

In the first case, when some sand particles keep falling downward from the upper chamber, they do not apply any force on the common balance.

But in the second case, when all the particles have accumulated in the lower chamber, all the sand particles apply force on the common balance. So, reading in the first case is somewhat lower than the reading in the second case.

**Question 15. What do you mean by thrust?**

**Answer:**

Suppose, a man is sitting on a floor. He applies a force on the floor acting downward and equal to his own weight and the floor also applies an equal reaction force on the man, but in the opposite direction. This type of force is called thrust.

In other words, if a body is kept on another body, then the mutual action-reaction force applied by them is called thrust.

**Question 16. What do you mean by push?**

**Answer:**

A tennis ball is thrown toward the wall. When the ball touches the wall, it applies a force on the wall and the wall also applies an equal and opposite force on the tennis ball as reaction. As a result, the tennis ball and the wall try to move away from each other. This type of action-reaction is called push.

In other words, if two bodies want to move away from each other during the period of their contact due to mutual action-reaction, then this action- reaction is called push.

**Question 17. Establish the law of conservation of third law of motion.**

**Answer:**

Suppose two bodies of masses m_{1} and m_{2} are moving in the same straight line with velocities u_{1} and u_{2}. There is a collision between these two bodies if u_{1} > u_{2}.

After collision, these bodies move with velocities v_{1} and v_{2} in the same straight line. During collision, force applied by the body with mass m_{1} on the body with mass m_{2} is F_{1} and the force applied by the body with mass m_{2} on the body with mass m_{1} is F_{2}.

Clearly, F_{1} and F_{2} are action and reaction,

∴ according to Newton’s third law of motion,

F_{1} = -F_{2 }……(1)

Now, force F_{1} is applied on the body with mass m_{2}.

∴ \(F_1=\frac{m_2 v_2-m_2 u_2}{t}\) [t= time of collision]

Again, force F_{2} is applied on the body with mass m_{1}.

∴ \(F_2=\frac{m_1 v_1-m_1 u_1}{t}\)

So from equation (1), we get

\(\frac{m_2 v_2-m_2 u_2}{t}=-\frac{m_1 v_1-m_1 u_1}{t}\)or, \(m_2 v_2-m_2 u_2=-m_1 v_1+m_1 u_1\)

or, \(m_1 u_1+m_2 u_2=m_1 v_1+m_2 v_2\)

∴ total linear momentum remains unchanged after collision of two bodies, unless an external force acts on the system.

**Question 18. Establish Newton’s third law of motion with the help of law of conservation of linear momentum.**

**Answer:**

Suppose two bodies of masses m_{1} and m_{2} are moving in the same straight line with velocities u_{1} and u_{2} (u_{1 }> u_{2}). After collision, these bodies will move with velocities v_{1} and v_{2} in the same straight line.

From the law of conservation of linear momentum, we get

\(m_1 u_1+m_2 u_2=m_1 v_1+m_2 v_2\)or, \(m_2 u_2-m_2 v_2=m_1 v_1-m_1 u_1\)

or, \(-m_2\left(v_2-u_2\right)=m_1\left(v_1-u_1\right)\)

or, \(\frac{m_2\left(v_2-u_2\right)}{t}=-\frac{m_1\left(v_1-u_1\right)}{t}\) [t= time of collision]

Now, it is known that \(F_1=\frac{m_2\left(v_2-u_2\right)}{t}\) is the force applied on the second body by the first body and \(F_2=\frac{m_1\left(v_1-u_1\right)}{t}\) is force applied on the first body by the second body.

∴ F_{1} = -F_{2}

Hence, if F_{1} is action, F_{2} is reaction. Further, F_{1} and F_{2} are equal and act in opposite directions. This is Newton’s third law of motion.

**Question 19. If a bullet is fired from a gun, explain this phenomenon with the help of the law of conservation of linear momentum.**

**Answer:**

When a bullet is fired from a gun, the bullet moves forward with high velocity, and the gun also recoils or kicks back immediately. We can explain this phenomenon with the help of the law of conservation of linear momentum.

Suppose, a bullet of mass m is fired with velocity u in the forward direction from the gun of mass M. As a result, the velocity of the gun becomes V.

Just before firing the bullet, total linear momentum of the system = 0, and the moment after the firing the bullet, total linear momentum = MV+ mu.

Since no external force is acting here. So from the law of conservation of linear momentum, we get 0 = MV + mu or,V = –\(\frac{m}{M}\)u…..(1)

Negative sign in equation (1) indicates that directions of velocities of bullet and gun are opposite, i.e., if the bullet moves forward, the gun recoils backward.

**Question 20. A body is at rest in a floor. After the explosion, it divided into two parts of equal mass. A parts moves towards north direction with velocity u. In which direction and with what velocity the other part of the body will move?**

**Answer:**

Let, mass of the each parts of the body be m and the second part moves with velocity v.

Now, according to conservation of linear momentum we can write, (m + m) x 0 = mu + m • V m

or, \(v=-\frac{m}{m} \cdot v=-u\)

∴ The second part of the body will move with velocity u and in the direction opposite to the first part i.e., the second part will move towards south direction with the velocity u.

**Question 21. Explain the working principle of a rocket with the help of law of conservation of linear momentum.**

**Answer:**

The working principle of a rocket can be explained with the help of law of conservation of linear momentum. Due to the burning of fuel in the combustion chamber of a rocket, gas produced is emitted out through the nozzle at one end with high velocity, producing thrust.

Gas emitted through the nozzle has a high value of linear momentum. As a result, the rocket also attains a linear momentum of equal value in opposite direction and moves forward with high velocity.

**Question 22. Five balls of equal masses are kept in a straight line, touching each other. The balls are suspended from separate strings. Now, the ball at the extreme left is displaced a little and then let loose. It is seen that when this ball strikes the next one, only one ball of the other end is budged (displaced). Again, if two balls of the left side strike in the same way, two balls of the right side are displaced. What is the reason behind these phenomena?**

**Answer:**

According to the question, the balls have the ass. When one of them is displaced and released, it strikes the next ball. Now according to the law of conservation of momentum, the total linear momentum of the system should remain conserved.

Consequently, only one ball of the other end is deflected with the same velocity (the velocity with which the left ball struck).

If two balls of the left side strike in the same way, then two balls from the right side are deflected with the same velocity so that the total linear momentum of the system remains conserved.

**Question 23. Why is it difficult to hold a hose pipe properly when water emits through the pipe with high velocity?**

**Answer:**

When water emits out with high velocity through a hose pipe, the emitted water has a definite momentum in the forward direction. So, according to the law of conservation of linear momentum, the pipe would also have a momentum of equal value but in the opposite direction.

For this reason, force has to be applied to hold the pipe, making it a bit difficult.

**Question 24. Why a cricket player lowers his hand while catching a cricket ball? Explain.**

**Answer:**

When a cricketer catches a ball, its momentum reduces to zero in his hands. If initial momentum of the cricket ball be P, t be the time duration in which the ball stops in his hand and F be the average force applied by the cricketer to stops the ball, then,

\(F=\frac{\text { change of momentum }}{\text { time }}=\frac{0-P}{t}=\frac{P}{t}\)(-ve sign shows that force is applied against the motion of the ball).

By lowering his hands while catching the ball, the player allows a longer time for change in momentum and stops the ball slowly in hands, and protects his hands from getting injury.

**Question 25. A wooden block of mass m is kept on a table. What is its normal reaction?**

**Answer:**

A is a wooden block of mass m kept on a table. The earth extras downward force on the block which is its weight, w = mg.

The table also extras an equal and opposite reaction force of magnitude W = mg on the block. This is the normal reaction force (N) on the block that directed vertically upward.

**Question 26. A body is placed on an inclined plane such that it is in rest without slipping. In which direction normal reaction force acts? Explain with proper diagram.**

**Answer:**

If a body is kept at rest in an inclined plane then normal reaction force acts on the body in upward direction perpendicular to the inclined plane.

[If mass of the body be m and angle of inclination of the inclined plane be θ, then weight of the body W = mg, component of weight perpendicular to the plane = W cosθ.

Therefore normal reaction force is R = Wcosθ.]

## Chapter 2 Topic C Newtons Third Law Of Motion And Law Of Conservation Of Linear Momentum Very Short Answer Type Questions Choose The Correct Answer

Question 1. The basis of working principle of a jet engine is

- Principle of conservation of mass
- Principle of conservation of energy
- Principle of conservation of linear momentum
- Principle of conservation of angular momentum

Answer: 3. Principle of conservation of linear momentum

Question 2. The motion of a rocket is established on the principle of conservation of

- Energy
- Kinetic energy
- Linear momentum
- Mass

Answer: 3. Linear momentum

Question 3. Action and reaction

- Are applied on the same body
- Are applied on different bodies
- Are equal and unidirectional
- Are in opposite directions but their values are not

Answer: 2. Are applied on different bodies

Question 4. A gun recoils if a bullet is fired from it

- According to Newton’s second’law of motion
- According to Newton’s first law of motion
- Due to reaction force
- Due to action force

Answer: 3. Due to reaction force

Question 5. The principle of conservation of linear momentum states that the linear momentum of a system

- Can not be changed
- Can not remain constant
- Can be changed only if internal forces act
- Can be changed only if external forces acts

Answer: 4. Can be changed only if external forces acts

Question 6. A boy of mass 40 kg jumps from a boat of mass 200 kg at a velocity 10 m/s. What is the velocity of the boat?

- 2 m/s
- 1 m/s
- 4 m/s
- 5 m/s

Answer: 1. 2 m/s

Question 7. A bullet of mass m hits a wooden block of mass M with velocity v and attached with it. Velocity of the system after collision is

- \(\frac{M}{M+m} v\)
- \(\frac{M-m}{M} v\)
- \(\frac{m}{M+m} v\)
- \(\frac{M+m}{M} v\)

Answer: 3. \(\frac{m}{M+m} v\)

Question 8. A gun fires N bullets per second, each of mass m with velocity v. The force exerted by the bullet on the gun is

- \(\frac{m v}{N}\)
- \(\frac{m v^2}{N}\)
- mNv
- mNv
^{2}

Answer: 3. mNv

Question 9. A nucleus disintegrates into two nuclear parts which have their velocities in the ratio of 3:2. The ratio of their mass will be

- 3:2
- 2:3
- 9:4
- 4:9

Answer: 2. 2:3

Question 10. The difference between the ,nature of a rocket and a jet is

- Rockets can only move in air but jets can not
- Rockets can move out side the atmosphere but jets can not
- Rockets obtain oxygen from air but jets carries its own oxygen
- Both of them carries own oxygen

Answer: 2. Rockets can move out side the atmosphere but jets can not

Question 11. If 10 N force acts on a body, then reaction force will be

- 10 N along the direction applied force
- 20 N along the direction of applied force
- 10 N opposite to the direction of the applied force
- 20 N opposite to the direction of the applied force

Answer: 3. 10 N opposite to the direction of the applied force

Question 12. A and B are two spring balances of negligible small weight. Mass M is hang from lower end of B. Reading of any of the two spring balances is

- M kgf
- Mg kgf
- 2M kgf
- 2 Mg kgf

Answer: 1. M kgf

Question 13. Conservation of linear momentum is applicable when external force will be

- Very small in magnitude
- Zero
- Very large in magnitude
- Constant in magnitude

Answer: 2. Zero

Question 14. An iron ball collides with another iron ball at rest. Here the force comes into play is

- Friction force
- Normal reaction force
- Tension force
- Collision force

Answer: 4. Collision force

Question 15. When a book lying on a table is pushed, it moves some distance and then stops. The external force which affects the motion of the body is

- Collision force
- Tension
- Friction
- Normal reaction

Answer: 3. Friction

Question 16. In the case of a game of tug-of-war, if both the parties exert a force T from either side, what will be the tension in the rope?

- T
- 2T
- T/2
- 3

Answer: 1. T

Question 17. According to Newton’s third law of motion, the angle between action force and its reaction force

- 0°
- 180°
- 90°
- 360°

Answer: 2. 180°

Question 18. Which is not a contact force?

- Friction
- Gravitational force
- Tensile force
- Collision force

Answer: 2. Gravitational force

Question 19. Recoil velocity of a gun

- Less than the velocity of the bullet
- Grater then the velocity of the bullet
- Equal to the velocity of the bullet
- A and B Both

Answer: 1. Less than the velocity of the bullet

## Chapter 2 Topic C Newtons Third Law Of Motion And Law Of Conservation Of Linear Momentum Answer In Brief

**Question 1. What is the force due to which it is difficult to push a body on the ground?**

**Answer:** It is difficult to push a body on the gro due to the force of friction between ground and the body.

**Question 2. “Action-reaction can establish equilibrium” is the statement, true or false?**

**Answer:** The statement is false.

**Question 3. The working principle of a rocket depends on which basic principle?**

**Answer:** The working principle of a rocket depends on the principle of conservation of linear momentum.

**Question 4. What is repulsion?**

**Answer:** Due to action-reaction, if two bodies move away from each other, then it is called repulsion.

**Question 5. Write down the law of conservation of linear momentum.**

**Answer:** The law states that if no external force is applied, the total linear momentum of a body or system of bodies always remains unchanged.

**Question 6. What do you mean by friction?**

**Answer:** When a body moves or tends to move on another body or on any surface, then the opposing force that acts against this movement or tendency of movement is called friction.

**Question 7. What change in total linear momentum does occur in the case of collision between two bodies?**

**Answer:** If no external force is applied total linear momentum of the system will remain unchanged.

**Question 8. Give one example of contact force.**

**Answer:** Force of friction or simply friction is an example of contact force.

**Question 9. Why it is difficult to move a body over the earth’s surface?**

**Answer:** Due to the arise of frictional force it is difficult to move a body over the earth’s surface.

**Question 10. Give one example of a non contact force.**

**Answer:** Gravitational force of attraction is an example of non contact force.

**Question 11. What do you mean by friction?**

**Answer:** When a body moves or tends to move on another body or any surface, then the opposing force that acts against this movement or tendency of movement is called friction.

**Question 12. Why do not the forces of action and reaction cancel each other?**

**Answer:** Action and reaction forces act on different bodies which is why they do not cancel each other.

**Question 13. Name the principle on which a rocket works.**

**Answer:** Conservation of linear momentum is the principle on which a rocket works.

**Question 14. A bucket full of water is suddenly pushed. State whether the water splashes forward or backward?**

**Answer:** Water splashes backward.

## Chapter 2 Topic C Newtons Third Law Of Motion And Law Of Conservation Of Linear Momentum Fill In The Blanks

Question 1. Every action has an _________ and opposite _________

Answer: Equal, reaction

Question 2. Rocket propulsion is possible even in an __________ place.

Answer: Air-free

Question 3. Action-reaction cannot ________ mutually.

Answer: Balance

Question 4. The theory of flight of a jet plane can be explained by Newton’s _______ law of motion.

Answer: Third

Question 5. Action and reaction forces acts on ________ bodies.

Answer: Different

Question 6. If a body A slips over another body B the force comes into play that resist this motion is called ________

Answer: Friction

Question 7. The linear momentum of a system remains constant if no _______ force acts on it.

Answer: External

Question 8. Normal reaction force is a ________ force.

Answer: Contact

Question 9. Magnetic force is an example of _______ force.

Answer: Non-contact

## Chapter 2 Topic C Newtons Third Law Of Motion And Law Of Conservation Of Linear Momentum State Whether True Or False

Question 1. The total linear momentum of a system of bodies can be changed without the application of an external force.

Answer: False

Question 2. A spring can pull an object as well as push an object.

Answer: True

Question 3. A ball moving on a horizontal surface stops because of the force of friction.

Answer: True

Question 4. Any pair of equal and opposite forces forms an action-reaction pair.

Answer: False

Question 5. Magnetic force is contact force.

Answer: True

Question 6. Birds can fly in air-free space.

Answer: False

Question 7. If the magnitude of action force be 15 N then the magnitude of the reaction force will be 15 N.

Answer: True

Question 8. A boat is floating on still water. A man walks from one end of the boat moves backwards.

Answer: True

## Chapter 2 Topic C Newtons Third Law Of Motion And Law Of Conservation Of Linear Momentum Numerical Examples

**Useful relations**

Two bodies of masses m_{1} and m_{2} respectively, moves in a straight line with velocities u_{1 }and u_{2} (u_{1} > u_{2}) respectively, collides with each other.

- If after collision these two bodies moves along the same straight line with velocity v
_{1 }and u_{2}respectively, then, m_{1}u_{1}+ m_{2}u_{2}= m_{1}v_{1}+ m_{2}v_{2} - If after collision they coalesce with each other and moves with velocity V in a straight line then,

or, \(v=\frac{m_1 u_1+m_2 u_2}{m_1+m_2}\)

If a bullet of mass m is fired with a velocity u from a gun of mass M and the recoil velocity of the gun be V, then,

(M + m) x 0 = mv + M • V

or, V = -(\(\frac{m}{M}\))v

**Question 1. A bullet of mass 20 g is fired from a gun of mass 5 kg with a velocity of 400 m/s. What is the recoil velocity of the gun?**

**Answer:**

Mass of the gun, M = 5 kg; mass of the bullet, m = 20 g = 0.02 kg; velocity of the bullet, u = 400 m/s

Let the velocity of the gun be V.

So, from the law of conservation of linear momentum, we get

mu + MV = 0 or, MV = -mu

or, \(V=-\frac{m u}{M}=\frac{0.02 \times 400}{5}=-1.6 \mathrm{~m} / \mathrm{s}\)

∴ recoil velocity of the gun = 1.6 m/s .

**Question 2. A bullet of mass 20 g moving with a velocity of 500 m/s strikes a stationary wooden block of mass 3.98 kg. Calculate the velocity of the wooden block.**

**Answer:**

Mass of the bullet, m = 20 g = 0.02 kg; velocity of the bullet, u = 500 m/s; mass of the wooden block, M = 3.98 kg

Let us assume that the bullet is stuck in the wooden block and both, as a single system, move with a velocity V.

So, from the law of conservation of linear momentum, we get mu = (M + m)V

∴ \(V=\frac{m u}{(M+m)}=\frac{0.02 \times 500}{(3.98+0.02)}=2.5 \mathrm{~m} / \mathrm{s}\)

**Question 3. A piece of stone of mass m is sliding over a smooth surface of ice with a velocity v. The stone is collected by a boy of mass M who is standing on the surface. What is the gain in velocity of the boy standing on the surface of ice?**

**Answer:**

Since the surface of the ice is smooth, it is frictionless. So, no external resistant force is applied on the boy or the stone.

Then, according to the law of conservation of momentum, the sum of initial velocity of the moving stone and initial momentum of the boy is equal to the final velocity of the boy with stone in his hand.

So, mv + 0 x M = (m + M) x V where V = velocity attained by the boy.

∴ \(V=\frac{m v}{(m+M)}\)

**Question 4. A body (A) of mass 100 g and another body (B) of mass 400 g are approaching each other with velocities 100 cm/s and 10 cm/s, respectively. After a head on collision, they coalesce with each other. This coalesced mass will move in which direction after the collision and how much distance will it cover in 10s?**

**Answer:**

As no external force is being applied, so the law of conservation of momentum is applicable here. Suppose, velocity of the coalesced mass after collision = V.

So, sum of individual momentums of A and B before collision is equal to the total momentum after collision.

So, 100 X 100 + 400 X (-10) = (100 + 400) x V

[As velocities of the two masses are in opposite directions, if the velocity of one body is taken as positive, the other must be negative]

So, 10000 – 4000 = 500 V

Positive sign of the velocity of 12 cm/s signifies that after collision, the coalesced mass will move in the direction of the mass of 100 g.

∴ distance traversed by this coalesced mass in 10s, s = vt = 12 x 10 = 120 cm .

**Question 5. A rocket consumes fuel at the rate of 100 kg • s ^{-1}. Gas ejects out of it with a velocity of 5 x 10^{3} m/s. Find the force experienced by the rocket.**

**Answer:**

Here, rate of consumption of fuel is = 100 kg/s.

Velocity of ejected gas = 5 x 10^{3} m/s.

∴ Upward impulsive force experienced by the rocket is

= rate of consumption of fuel x velocity of ejected has = 100 x 5 x 10^{3} N = 5 x 10^{5} N

**Question 6. A body of mass 5 kg while moving with velocity 15 m/s explodes in two parts of mass 3 kg and 2 kg. if the first part moves with velocity 5 m/s find velocity of the second.**

**Answer:**

Initial mass of the body M = 5 kg and initial velocity V = 15 m/s.

Mass of the first part m_{1} = 3 kg and mass of the second part m_{2} = 2 kg.

Velocity of the first part v_{1} = 5 m/s

Let, velocity of the second part be v^{2} m/s.

∴ From conservation of linear momentum,

M • V = m_{1}v_{1} + m_{2}v_{2}

or, 5 x 15 = 3 x 5 + 2 x v_{2} or, 2v_{2} = 75 – 15

∴ v_{2 }= 60/2 = 30 m/s

∴ Velocity of the second part is 30 m/s.

**Question 7. From a gun of mass M, a bullet of mass m is fired with a velocity v. If recoil of the gun stops after time t second by applying an average force F, show that F = \(\frac{m}{v}{t}\).**

**Answer:**

Here, initial mass of the gun bullet system is = M and initial velocity of the system = 0.

Velocity of the bullet = v and mass of the bullet = m.

Let, recoil velocity of the gun be V

∴ According to conservation of linear momentum,

0 = MV + mv

∴ MV = -mv

∴ Average force applied on the gun to stop its recoil

\(F =\frac{\text { change of momentum of the gun }}{\text { time }}\)= \(\frac{0-M V}{t}=-\frac{M V}{t}=\frac{m v}{t}\) (proved)

## Chapter 2 Topic C Newtons Third Law Of Motion And Law Of Conservation Of Linear Momentum Miscellaneous Type Questions

** Match The Columns**

**Question 1.**

**Answer: **1. B, 2. A, 3. D, 4. C

**Question 2. **

**Answer: **1. C, 2. A, 3. D, 4. B