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
(i) 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

Given: 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

Solution:

Let C be centre of mass of the rod of mass M and length L. Consider the rod and the particle together as a system. Let v be velocity of C and $\omega$ be angular velocity of the rod just after collision. The linear momentum of the system just before and just after the collision is,

$p_i=m{v}_0,p_f=Mv$

where $p_i{,}_f$ are the initial and final linear momentum of the system.

There is no external force on the system in x direction. Hence, linear momentum in x-direction is conserved i.e., $p_i=p_f$, which gives,

$Mv=m{v}_0⟹v_0=\frac{Mv}{m}........\left(i\right)$.

The angular momentum of the system about C just before and just after the collision is,

$L_i=m{v}_0\frac{L}{2},L_f=\omega I_c=M\omega \frac{L^2}{12}$.

There is no external torque on the system about C. Hence, angular momentum of the system about C is conserved i.e., $L_i=L_f$, which gives,

$m{v}_0=M\omega \frac{L}{6}........\left(ii\right)$.

The kinetic energy before and after the collision is,

$K_i=\frac{1}{2}m{v}_0^2,K_f=\frac{1}{2}M{v}^2+\frac{1}{2}{I}_c{\omega }^2$.

Since kinetic energy is conserved in elastic collision, $K_i=K_f$, i.e.,

$m{v}_0^2=M{v}^2+\left(\frac{M{L}^2}{12}\right){\omega }^2.......\left(iii\right)$.

Solve above equations we get

$\frac{m}{M}=\frac{1}{4},v=\frac{v_0}{4},\omega =\frac{3v_0}{2L}$.

The velocity of a point P with position vector $\overrightarrow{r_{PC}}$ from C is given by $\overrightarrow{v_P}=\overrightarrow{v_C}+\overrightarrow{\omega }\times \overrightarrow{r_{PC}}$. Just after the collision $\overrightarrow{r_{PC}}=y\hat{j}$. Thus, P is at rest if,

$\overrightarrow{v_P}=\frac{v_0}{4}\hat{i}+\left(\frac{3v_0}{2L}\hat{k}\right)\times \left(y\hat{j}\right)=\frac{v_0}{4}-\frac{3y{v}_0}{2L}=0$,

which gives $y=\frac{L}{6}$.

The distance $AP=AC+CP=\frac{L}{2}+\frac{L}{6}=\frac{2L}{3}$.

After the collision, C keeps moving with $\overrightarrow{v_C}=\frac{v_0}{4}\hat{i}$ and angular velocity of the rod remains $\overrightarrow{\omega }=\frac{3v_0}{2L}\hat{k}$.

The angular displacement of the rod in time $t=\frac{\pi L}{3v_0}$ is $\omega t=\frac{\pi }{2}$ and hence after time t position vector of P w.r.t. C is $\overrightarrow{r_{PC}}=-\frac{L}{6}\hat{i}+0\hat{j}$.

The velocity of P is,

$\overrightarrow{v_P}=\overrightarrow{v_C}+\overrightarrow{\omega }\times \overrightarrow{r_{PC}}=\frac{v_0}{4}\hat{i}+\left(\frac{3v_0}{2L}\hat{k}\right)\times (-\frac{L}{6}\hat{i})=\frac{v_0}{4}\hat{i}-\frac{v_0}{4}\hat{j}$,

and its magnitude is $|\overrightarrow{v_P}|=\frac{v_0}{2\sqrt{2}}$