# Collision of Point Masses with Rigid Bodies

If a torque $\vec\tau$ is applied to a body for a short time $\Delta t$ then angular impulse imparted to the body is given by \begin{align} \Delta \vec{J}=\vec{\tau}\Delta t. \nonumber \end{align} The angular impulse imparted to a body in a time interval $t$ is given by \begin{align} \vec{J}=\int_0^{t} \vec{\tau}\,\mathrm{d} t. \nonumber \end{align} Impact or collision is a short duration phenomenon. The impulsive forces on the body are very large in comparison to the other forces. The effect of small (and/or slowly varying) forces like gravity is neglected during collision.

The kinetic energy of the system is conserved in an elastic collision. The conservation of (i) linear momentum, (ii) angular momentum and (iii) energy, should be explored to solve the problems involving collision.

In an elastic collision, the velocity of approach at the impact point is equal to the velocity of separation at this point.

## Solved Problems from IIT JEE

### Problem from IIT JEE 2000

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 velocity $v_0$ in the direction perpendicular to AB. The collision is elastic. After the collision the particle comes to rest.

1. Find the ratio $m/M$.
2. A point P on the rod is at rest immediately after collision. Find the distance AP.
3. Find the linear speed of the point P a time $\frac{\pi L}{3v_0}$ after the collision.

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, \begin{align} p_i&=mv_0,\\ p_f&=Mv.\nonumber \end{align} 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{align} \label{hpb:eqn:1} Mv=mv_0. \end{align} The angular momentum of the system about C just before and just after the collision is, \begin{align} L_i&=mv_0 L/2,\\ L_f&=\omega I_c=M\omega L^2/12.\nonumber \end{align} 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{align} \label{hpb:eqn:2} mv_0=M\omega L/6. \end{align} The kinetic energy before and after the collision is, \begin{align} K_i&=\tfrac{1}{2}mv_0^2,\\ K_f&=\tfrac{1}{2}Mv^2+\tfrac{1}{2}I_c\omega^2.\nonumber \end{align} Since kinetic energy is conserved in elastic collision, $K_i=K_f$, i.e., \begin{align} \label{hpb:eqn:3} mv_0^2=Mv^2+(ML^2/12)\omega^2. \end{align} Solve above equations to get ${m}/{M}={1}/{4}$, $v={v_0}/{4}$, and $\omega={3v_0}/{(2L)}$.

The velocity of a point P with position vector $\vec{r}_\text{PC}$ from C is given by $\vec{v}_P=\vec{v}_C+\vec{\omega}\times\vec{r}_\text{PC}$. Just after the collision $\vec{r}_\text{PC}=y\,\hat\jmath$. Thus, P is at rest if, \begin{align} \vec{v}_P&=\frac{v_0}{4}\;\hat\imath+\left(\frac{3v_0}{2L}\,\hat{k}\right) \times \left(y\,\hat\jmath\right)\\ &=\frac{v_0}{4}-\frac{3yv_0}{2L}=0, \end{align} which gives $y={L}/{6}$. The distance $\mathrm{AP}=\mathrm{AC}+\mathrm{CP}={L}/{2}+{L}/{6}={2L}/{3}$. After the collision, C keeps moving with $\vec{v}_C={v_0}/{4}\,\hat\imath$ and angular velocity of the rod remains $\vec{\omega}={3v_0}/{(2L)}\,\hat{k}$. The angular displacement of the rod in time $t={\pi L}/{(3v_0)}$ is $\omega t={\pi}/{2}$ and hence after time $t$ position vector of P w.r.t. C is $\vec{r}_\text{PC}=-{L}/{6}\,\hat\imath+0\,\hat\jmath$. The velocity of P is, \begin{align} \vec{v}_P&=\vec{v}_C+\vec{\omega}\times\vec{r}_\text{PC}\\ &=\frac{v_0}{4}\hat\imath+\left(\frac{3v_0}{2L} \,\hat{k}\right)\times \left(-\frac{L}{6}\,\hat\imath\right)\\ &=\frac{v_0}{4}\,\hat\imath-\frac{v_0}{4}\,\hat\jmath,\nonumber \end{align} and its magnitude is $|\vec{v}_{P}|=\frac{v_0}{2\sqrt{2}}$.