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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Centre of Mass, Linear momentum, Collision

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Centre of Mass, Linear momentum, Collision

A bullet of mass m travelling at a speed v strikes a ball of mass M...

A bullet of mass mm travelling at a speed vv strikes a ball of mass MM that is initially at rest. A small part with mass m′m' breaks off from the ball and sticks to the bullet. The remaining ball is observed to travel at a speed v1v_1 in the bullet's direction. Determine the velocity of the bullet following the collision.

Options

  1. Amv−Mv1m+m′\dfrac{mv - Mv_1}{m + m'} in the initial direction
  2. Bmv−(M−m′)v1m+m′\dfrac{mv - (M - m')v_1}{m + m'} in the initial direction
  3. Cmv+(M−m′)v1m+m′\dfrac{mv + (M - m')v_1}{m + m'} in the initial direction
  4. Dmv−(M−m′)v1m\dfrac{mv - (M - m')v_1}{m} in the initial direction

Correct answer

B. mv−(M−m′)v1m+m′\dfrac{mv - (M - m')v_1}{m + m'} in the initial direction

Step-by-step solution

By the principle of conservation of linear momentum, the total initial momentum of the system is equal to the total final momentum.

Initial momentum of the system is pi=mv+M(0)=mvp_i = mv + M(0) = mv.

After the collision, a part of mass m′m' breaks off from the ball and sticks to the bullet. The new mass of the bullet becomes m+m′m + m'. Let its final velocity be v2v_2.

The remaining mass of the ball is M−m′M - m', and it moves with a velocity v1v_1 in the initial direction of the bullet.

Final momentum of the system is pf=(m+m′)v2+(M−m′)v1p_f = (m + m')v_2 + (M - m')v_1.

Equating initial and final momentum:

mv=(m+m′)v2+(M−m′)v1mv = (m + m')v_2 + (M - m')v_1

(m+m′)v2=mv−(M−m′)v1(m + m')v_2 = mv - (M - m')v_1

v2=mv−(M−m′)v1m+m′v_2 = \dfrac{mv - (M - m')v_1}{m + m'}

The velocity of the bullet following the collision is mv−(M−m′)v1m+m′\dfrac{mv - (M - m')v_1}{m + m'} in the initial direction.
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