Question

A uniform rod of mass $m$ and length $l$ is free to rotate about a fixed horizontal axis passing through hinge. A small ball of mass $m$ moving horizontally with velocity $v_o$ strikes the rod at depth $y$ from hinge as shown in figure. If angular velocity of rod and velocity of ball just after the collision, be $\omega$ and $v$ respectively, then which of the following statements are wrong?

• A. If $y=\frac{l}{2}$, then in case of elastic collision, $v=\frac{v_o}{2}$ and $\omega =\frac{v_o}{l}$
• B. If $y=\frac{l}{2}$, then in case of elastic collision,momentum of ball-rod system remains conserved
• C. If momentum of ball-rod system remains conserved then $y<\frac{l}{2}$
• D. If $y=\frac{l}{2}$, then in case of elastic collision, magnitude of $v=\frac{v_o}{7}$ and $\omega =\frac{12v_o}{7l}$

Answer 1

Answer: (A), (B), (C)

If momentum of ball-rod system remains conserved then $y<\frac{l}{2}$

Conservation of angular momentum about $O$
$m{v}_{0}y=mvy+I\omega$
$v_{0}y=vy+\frac{l^2\omega }{3}......1$

Using Newtons law of collision
$e=\frac{v_0}{\omega y-v}......2$

For $y=l/2,e=1$
We get
$v_0=\frac{\omega l}{2}-v.....3$
$v_0=v+\frac{2\omega l}{3}.....4$
Adding $3,4$
$2v_0=\frac{7\omega l}{6}$
$\omega =\frac{12v_0}{7l}$
$v=\frac{-v_0}{7}$

Momentum of ball rod system is not conserved as reaction force at hinge acts as external force.