Question

Ball 'A' of mass 'm' moving along x-axis with Kinetic energy of 'k' & momentum 'p'.it undergoes elastic collision with a stationary 'B' of mass 'm',after the collision the ball 'A' moves along negative x-axis with Kinetic energy K/9.Find the momentum of 'B'??

Answer 1

Ball A of mass $m$ is moving along x-axis with kinetic energy of $K$ and momentum $p$. It undergoes elastic collision with a stationary ball B of mass $m$, after the collision the ball A moves along negative x-axis with kinetic energy $K/9$. Find the momentum of B.

Solution: Kinetic energy of the ball A before the collision was $K = \frac{m \cdot v_0^2}{2}$, where $v_0$ is the speed of the ball before the collision. After the collision it became equal to $K' = \frac{m \cdot v_A^2}{2}$, where $v_A$ is the speed of the ball after the collision. From the initial conditions we see, that $K' = \frac{K}{9}$, $\frac{K}{K'} = 9$, $\left(\frac{v_0}{v_A}\right)^2 = 9$, $v_A = \frac{1}{3} v_0$.

The momentum of the ball A before the collision was $p_{A0} = m \cdot v_0 = p$, and after the collision it became equal to $p_A = -m \cdot v_A = -m \cdot \frac{1}{3} \cdot v_0 = -\frac{1}{3} p_{A0}$ (minus sign means, that after the collision ball changed the direction of the movement to opposite). According to the law of conservation of the momentum, total momentum of balls A and B before the collision should be equal to total momentum after the collision: $p_{A0} + p_{B0} = p_A + p_B$. Ball B didn't move before the collision, it means that $p_{B0} = 0$.

Then, momentum of the ball B after the collision is: $p_{B} = p_{A0} - p_{A} = p_{A0} - \left(-\frac{1}{3} p_{A0}\right) = \frac{4}{3} p_{A0} = \frac{4}{3} p$.

Answer: $\frac{4}{3} p$.