Question

A rod AB of mass M and length L is lying on a horizontal frictionless surface. A particle of mass m travelling along the surface hits the end A of the rod with a velocity $v_0$ in a direction perpendicular to AB. The collision is elastic. After the collision the particle comes to rest.
(a) Find the ratio m/M.
(b) A point P on the rod is at rest immediately after collision. Find the distance AP.
(c) Find the linear speed of the point P a time $\pi L/3v_0$ after the collision.

Answer 1

(a) Let just after collision, velocity of CM of rod is v and angular velocity about CM is $\omega$. Applying following three laws.
(1) External force on the system (rod + mass) in horizontal plane along x-axis is zero .
$\therefore$ Applying conservation of linear momentum in x-direction.
$m{v}_0=mv$...(i)
(2) Net torque on the system about CM of rod is zero
$\therefore$ Applying conservation of angular momentum about
CM of rod, we get $m{v}_0\left(\frac{L}{2}\right)=I\omega$
or $m{v}_0\frac{L}{2}=\frac{M{L}^2}{12}\omega$
or $m{v}_0=\frac{ML\omega }{6}$ ...(ii)
(3) Since, the collision is elastic, kinetic energy is also conserved.
$\therefore \frac{1}{2}m{v}_0^2=\frac{1}{2}M{v}^2+\frac{1}{2}I{\omega }^2$
or $m{v}_0^2=M{v}^2+\frac{M{L}^2}{12}{\omega }^2$ ...(iii)
From Eqs. (i), (ii) and (iii), we get the following results
$\frac{m}{M}=\frac{1}{4}$
$v=\frac{m{v}_0}{M}$ and $\omega =\frac{6m{v}_0}{ML}$
(b) Point P will be at rest if x $\omega =v$
or $x=\frac{v}{\omega }=\frac{m{v}_0/M}{6m{v}_0/ML}$
or $x=L/6$
$\therefore AP=\frac{L}{2}+\frac{L}{6}$
or $AP=\frac{2}{3}L$
(c) After time $t=\frac{\pi L}{3v_0}$
angle rotated by rod, $\theta =\omega t=\frac{6m{v}_0}{ML}.\frac{\pi L}{3v_0}$
$2\pi \left(\frac{m}{M}\right)=2\pi \left(\frac{1}{4}\right)\therefore \theta =\frac{\pi }{2}$
Therefore, situation will be as shown below
$\therefore$ Resultant velocity of point P will be
$|v_P|=\sqrt{2}v=\sqrt{2}\left(\frac{m}{M}\right)v_0$
$\frac{\sqrt{2}}{4}{v}_0=\frac{v_0}{2\sqrt{2}}$ or $|v_P|=\frac{v_0}{2\sqrt{2}}$