Law of Conservation of Linear Momentum Questions

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The Law of conservation of linear momentum states that for an isolated system, the initial momentum of a system is equal to the final momentum of the system.

  • The total linear momentum of an isolated system remains constant according to the law of conservation of linear momentum
  • In other words, if the net external force acting on the system is zero, the total linear momentum of the system remains constant.
  • Momentum is the quantity used to describe a non-zero mass object's state of motion. 
  • It is the amount of movement contained within a body. It is represented numerically by the product of a body's mass and velocity.
  • Linear momentum is a vector quantity and its SI unit is kg m/s.
  • The law of linear momentum conservation is consistent with Newton's Third Law of Motion.

If m1 and m2 are the masses of the two bodies moving initially with velocities u1 and u2 respectively. Let these bodies collide and move with final velocity v1 and v2 respectively. Then according to the law of conservation of linear momentum,

Initial linear momentum = Final linear momentum

m1u, + m2u2 = m1v1 + m2v2


Very Short Answers Questions [1 Mark Questions]

Ques. Two bodies moving with constant velocities collide with each other. Which of the following quantities remain conserved?

  1. Speed
  2. Velocity
  3. Momentum
  4. Force

Ans. The correct answer is c. Momentum

Explanation: Because there is no external force involved in this case, the momentum will remain constant, and there will be no change in momentum.

Ques. A ball is moving with a constant velocity. After some time, it collides with a wall. Which one of the following remains conserved except momentum?

  1. Displacement
  2. Energy
  3. Force
  4. Power

Ans. The correct answer is b. Energy

Explanation: Apart from velocity, energy is also conserved during collision. In fact, the energy of a system often remains conserved.

Ques. What is the plural of the word momentum?

  1. Moments
  2. Momentums
  3. Moneta
  4. Momenta

Ans. The correct answer is d. Momenta

Explanation: Momentum is the amount of motion of a moving body, calculated as the product of its mass and velocity. The plural form of momentum is momenta.

Ques. Two bodies are accelerating toward each other. After some time they collide. Which one of the following statements is true about the system?

  1. The total momentum remains conserved at the instant of collision
  2. The total momentum does not remain conserved
  3. The total momentum remains conserved during the collision
  4. The total momentum remains conserved

Ans. The correct answer is a. The total momentum remains conserved at the instant of collision.

Explanation: Because they are accelerating, overall momentum is not conserved during the motion. However, at the point of contact, both bodies have the same velocity. As a result, the overall momentum will be conserved just before and after the collision.

Ques. When a body collides with a wall or the ground, what assumption do we make?

  1. The body is stationary
  2. The body is perfect
  3. The mass of the body is not negligible
  4. The mass of the body is negligible as compared to the mass of the wall or the ground

Ans. The correct answer is d. The mass of the body is negligible as compared to the mass of the wall or the ground.

Explanation: When a body collides with a wall or the ground, the mass of the body is considered small in comparison to the mass of the wall or the ground. We also make this assumption when a feather, which has just a small mass, collides with a truck, which has a large mass in comparison.


Short Answers Questions [2 Marks Questions]

Ques. What is the Law of conservation of linear momentum?

Ans. According to the law of conservation of linear momentum, the total linear momentum of an isolated system remains constant. In other words, the total linear momentum of the system remains constant if the net external force acting on this system is zero.

Ques. What is the formula for the law of conservation of linear momentum?

Ans. Let p1, p2, p3,.... be the linear momentum of n particles in a system, then according to the law of conservation of linear momentum, the total linear momentum of the system remains conserved i.e.

p1 + p2 + p3 +........+ pn = Constant

Ques. What if momentum is not conserved?

Ans. Momentum is not conserved when gravity, friction, or net force are present; net force refers to the overall amount of force. It demonstrates that if these quantities act on any object, its momentum will change. This should be noticeable because you are changing the object's velocity and momentum.

Ques. What is momentum?

Ans. Momentum is the quantity used to characterize the state of motion of a non-zero mass object. It is defined as the amount of motion contained within a body. Numerically, it is given by the product of the mass of a body and its velocity.

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Long Answers Questions [3 Marks Questions]

Ques. What are the applications of the law of conservation of linear momentum?

Ans. Linear momentum is important in rocket propulsion as well as the recoil of a gun. They are as follows:

  • Rocket propulsion: Chemicals in the rocket ignite and release gases at high speeds. These gases exit through the tail nozzle in a downward direction, and the rocket then rises upward to balance the momentum of the gases. Even if the amount of gas exhaust is low, the velocity adds up to momentum. Because the rocket receives equivalent momentum in the opposite direction, it travels upward at a high velocity despite its large mass.
  • The recoil of a gun: When a bullet is fired from a gun, to balance the momentum imparted by the bullet, the gun moves backward with a recoil velocity. Recoil velocity is the speed with which a gun goes backward after firing a bullet.
  • Motion of the motorboats: Motorboats work on the same principle, pushing the water back and forth to conserve momentum.

Ques. How much net horizontal force is required to accelerate a 500 kg car at 3 m/s2?

Ans. Given

  • The mass of the body, m = 500 kg
  • Acceleration of the body, a = 3 m/s2

According to Newton’s second law of motion, we have

Force, F = ma

⇒ F = 500 x 3

⇒ F = 1500 N

Hence the net horizontal force required to accelerate the body with acceleration of 3 m/s2 is 1500 N.

Ques. Give the applications of Newton’s second law.

Ans. The amount of force necessary to move or stop an object may be calculated using Newton's second law of motion.

Some applications of Newton’s second law of motion are

  • Kicking a ball: When we kick a ball, we apply force in one direction alone. The harder we kick the ball, the more force we apply to it, and the further it will travel.
  • Pushing a cart: In a supermarket, an empty cart is simpler to push than a filled one, because more mass requires more acceleration.
  • Two people walking: If one of the two persons walking is heavier than the other, the larger person will walk slower since the acceleration of the lighter person is higher.

Very Long Answers Questions [5 Marks Questions]

Ques. Two objects each of mass 5 kg are moving in a straight line but in opposite directions towards each other with the same speed of 3 m/s. They stick together after the collision. What will be the velocity of the combined object after collision?

Ans. Let m1 and m2 be the mass and u1 and u2 be the initial velocity of the first and second objects respectively.

Given

  • m1 = m2 = 5 kg
  • u1 = 3 m/s
  • u2 = – 3 m/s

Here initial velocity of the second object is taken to be negative because it is moving in the opposite direction to that of the first body.

Total momentum before collision,

Pi = m1u1 + m2u2

⇒ Pi = (5 x 3) + (5 x -3) = 0

After the collision, both bodies stick together and move as one body with velocity V

Total momentum after the collision,

Pf = m1V + m2V

⇒ Pf = (m1 + m2)V

⇒ Pf = (5 + 5)V

⇒ Pf = 10V

According to the conservation of linear momentum, total momentum before collision is equal to the total momentum after collision i.e.

Pi = Pf

⇒ 0 = 10 V

⇒ V = 0

Hence the velocity of the combined object after collision is zero.

Ques. A bullet of mass 50 g is fired from a gun with an initial velocity of 35 m/s. If the mass of the gun is 4 kg, then calculate the recoil velocity of the gun.

Ans. The velocity with which a gun moves backward after firing a bullet is known as recoil velocity.

Let

  • v1 be the velocity of the bullet after firing
  • v2 be the recoil velocity of the gun i.e. velocity of the gun after firing
  • m1 be the mass of the bullet
  • m2 be the mass of the gun
  • u1 and u2 are the velocities of the bullet and the gun before firing

Given

  • m1 = 50 g = 50 x 10-3 kg
  • m2 = 4 kg
  • v1 = 35 m/s

Before firing, u1 = u2 = 0

Total linear momentum before firing,

Pi = m1u1 + m2u2

⇒ Pi = (50 x 10-3 x 0) + (4 x 0)

⇒ Pi = 0

Total linear momentum after firing,

Pf = m1v1 + m2v2

⇒ Pf = (50 x 10-3 x 35) + (4 x v2)

⇒ Pf = 1.75 + 4v2

According to the conservation of linear momentum, total momentum before collision is equal to the total momentum after collision i.e.

Pi = Pf

⇒ 0 = 1.75 + 4v2

⇒ v2 = - 1.75/4

⇒ v2 = - 0.44 m/s

Hence the recoil velocity of the gun is 0.44 m/s. The negative sign indicates that the direction of the recoil velocity is opposite to the direction of the bullet.

Ques. A boy of mass 50 kg is running with a velocity of 2 m/s. He jumps over a stationary cart of 2 kg while running. Find the velocity of the cart after the jumping of the boy.

Ans. After jumping the boy over the stationary cart, the cart and the boy move with the same velocity.

Let

  • u1 = 2 m/s, be the initial velocity of the boy
  • u2 = 0, be the initial velocity of the cart
  • m1 = 50 kg, be the mass of the boy
  • m2 = 2 kg, be the mass of the cart
  • V is the final velocity of the cart and the boy

Total linear momentum before the jump,

Pi = m1u1 + m2u2

⇒ Pi = (50 x 2) + (2 x 0)

⇒ Pi = 100 kg m/s

Total linear momentum after the jump,

Pf = m1v1 + m2v2

⇒ Pf = (50 x V) + (2 x V)

⇒ Pf = 52 V

According to the conservation of linear momentum, total momentum before collision is equal to the total momentum after collision i.e.

Pi = Pf

⇒ 100 = 52V

⇒ V = 100/52

⇒ V = 1.92 m/s

Hence the velocity of the cart after the jumping of the boy is 1.92 m/s.


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