Question

A uniform rod $AB$ of length $L$ and mass $m$ is suspended freely at $A$. and hangs vertically at rest when a particle of same mass $m$ is fired horizontally with speed $v$ to strike the rod at its midpoint. If the particle is brought to rest after the impact. Then the impulsive reaction at $A$ in horizontal direction is

• A. $mv/4$
• B. $mv/2$
• C. $mv$
• D. $2mv$

Answer 1

The correct option is A $mv/4$

From conservation of angular momentum, just before and just after impact about point $O$ we have,
$L_l=L_f$ $\therefore mv\frac{L}{2}=\left(\frac{m{L}^2}{3}\right)\cdot \omega$
$\therefore \omega =\frac{3v}{2L}$
$v_C=\frac{1}{2}\cdot \omega =\frac{3}{4}v$
Impulse $J$ has changed the momentum of particle from mv to $O.$ Hence
$J=mv$ For rod $J_H+J=m{v}_C=\frac{3}{4}mv$
$\therefore J_H=\frac{3}{4}mv-J=\frac{3}{4}mv-mv$
$=-\frac{mv}{4}$ or $\left|J_H\right|=\frac{mv}{4}$