Question

Two identical balls each of mass $m$ interconnected by a small light inextensible string, are kept on a smooth horizontal surface
One of the balls is pushed towards the other, with a speed $v_0$ find the loss of K.E of the ball. What will be the final velocities of the balls? $(e=1)$

Answer 1

We know that, when two identical bodies collide elastically, they transfer their momenta. That means, just after collision the stationary ball starts moving with a speed $v_0$ till the string is taut. Then both the balls will continue to move with same speed, say $v$. Since no external force is acting on the system of the balls and string (tension in the string is internal force), the system conservess its linear momentum. Conserving the momentum of the system just before and after the string is taut obtain,
$m{v}_0=mv+mv\Rightarrow v=\frac{v_0}{2}$
(a) Now the loss in KE of the ball
$\Delta KE=\frac{1}{2}m{v}^2+\frac{1}{2}m{v}^2-\frac{1}{2}m{v}_0^2$
Putting $v=\frac{v_0}{2}$, value of $\Delta KE$ can be found