Question

A pendulum consists of a wooden bob of mass $M$ and length $l$. A bullet of mass $m$ is fired towards the pendulum with a speed $v$. The bullet emerges immediately out of the bob from the other side with a speed of $v/2$ and the bob starts rising. Assume no loss of mass of bob takes place due to penetration. What is the momentum transferred to the bob by the bullet?

• A. $mv$
• B. $\frac{mv}{2}$
• C. $\frac{Mv}{2}$
• D. $Mv$

Answer 1

The correct option is B $\frac{mv}{2}$

Velocities of the bullet before and after the collision are $v$ and $\frac{v}{2}$ respectively.

Let the momentum of the wooden bob after the collision be $P$.

Initial momentum of the bob is $Zero$.

Applying conservation of momentum before and after the collision :

$mv+0=P+m\times \frac{v}{2}$

$⟹P=\frac{mv}{2}$

Net momentum transferred to the bob $=P=\frac{mv}{2}$