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 cord with a velocity $v_0$ in a direction perpendicular to $AB$ and comes to rest. The collision is completely elastic.
(a) Find the ratio $m/M$
(b) A point $P$ on the cord is at rest immediately after the collision. Find the distance $AP$
(c) Find the linear speed of the point $P$ at a time $\pi L/\left(3v_0\right)$ after the collision.

Answer 1

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,$$\begin{array}{ccc}{p}_i=m{v}_0,& & p_f=Mv.\end{array}$$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,$$\begin{array}{c}Mv=m{v}_0.\end{array}$$The angular momentum of the system about C just before and just after the collision is,$$\begin{array}{ccc}{L}_i=m{v}_{0}L/2,& & L_f=\omega I_c=M\omega L^2/12.\end{array}$$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,$$\begin{array}{c}m{v}_0=M\omega L/6.\end{array}$$The kinetic energy before and after the collision is,$$\begin{array}{ccc}& K_i=\frac{1}{2}m{v}_0^2,& K_f=\frac{1}{2}M{v}^2+\frac{1}{2}{I}_c{\omega }^2.\end{array}$$Since kinetic energy is conserved in elastic collision, $K_i=K_f$, i.e.,$$\begin{array}{c}m{v}_0^2=M{v}^2+\left(M{L}^2/12\right){\omega }^2.\end{array}$$Solve above equations to get $m/M=1/4$, $v=v_0/4$, and $\omega =3v_0/\left(2L\right)$.
The velocity of a point P with position vector ${\overrightarrow{r}}_{\text{PC}}$ from C is given by ${\overrightarrow{v}}_P={\overrightarrow{v}}_C+\overrightarrow{\omega }\times {\overrightarrow{r}}_{\text{PC}}$. Just after the collision ${\overrightarrow{r}}_{\text{PC}}=y\hat{ȷ}$. Thus, P is at rest if,$$\begin{array}{c}{\overrightarrow{v}}_P=\frac{v_0}{4}\hat{ı}+\left(\frac{3v_0}{2L}\hat{k}\right)\times \left(y\hat{ȷ}\right)=\frac{v_0}{4}-\frac{3y{v}_0}{2L}=0,\end{array}$$which gives $y=L/6$. The distance $AP=AC+CP=L/2+L/6=2L/3$. After the collision, C keeps moving with ${\overrightarrow{v}}_C=v_0/4\hat{ı}$ and angular velocity of the rod remains $\overrightarrow{\omega }=3v_0/\left(2L\right)\hat{k}$. The angular displacement of the rod in time $t=\pi L/\left(3v_0\right)$ is $\omega t=\pi /2$ and hence after time $t$ position vector of P w.r.t. C is ${\overrightarrow{r}}_{\text{PC}}=-L/6\hat{ı}+0\hat{ȷ}$. The velocity of P is,$$\begin{array}{cc}& {\overrightarrow{v}}_P={\overrightarrow{v}}_C+\overrightarrow{\omega }\times {\overrightarrow{r}}_{\text{PC}}=\frac{v_0}{4}\hat{ı}+\left(\frac{3v_0}{2L}\hat{k}\right)\times (-\frac{L}{6}\hat{ı})=\frac{v_0}{4}\hat{ı}-\frac{v_0}{4}\hat{ȷ},\end{array}$$and its magnitude is $|{\overrightarrow{v}}_P|=\frac{v_0}{2\sqrt{2}}$