A bullet (mass = m₁)is fired at speed V into a block of mass m₂ at rest.If the bullet escapes from the block with only a third of its original speed then the recoil speed of the block is given by A)m₁V/ 3m₂ B)2m₁V/ 3m₂ C)m₂V/ 3m₁ D)2m₂V/ 3m₁ E)4m₂V/ 9m₁

Question

Quizplus · Accepted Answer

The conservation of momentum principle states that the total momentum before an event must equal the total momentum after the event, assuming no external forces act on the system. Initially, the bullet has momentum $m_{1}V$ and the block is at rest, so its momentum is 0. After the collision, the bullet's speed is $V/3$, so its momentum is $m_{1}(V/3)$, and the block's momentum is unknown. Let's denote the recoil speed of the block as $V_{block}$.The total initial momentum is $m_{1}V$. The total final momentum is $m_{1}(V/3) + m_{2}V_{block}$.Setting the initial momentum equal to the final momentum gives:$$m_{1}V = m_{1}(V/3) + m_{2}V_{block}$$Solving for $V_{block}$ gives:$$V_{block} = \frac{m_{1}V - m_{1}(V/3)}{m_{2}} = \frac{2m_{1}V}{3m_{2}}$$Therefore, the correct answer is $2m_{1}V/ 3m_{2}$, which corresponds to choice B.

A Bullet (Mass = M1)is Fired at Speed V into a Block

A Bullet (Mass = M₁)is Fired at Speed V into a Block