## College Physics (4th Edition)

The ratio of the kinetic energy just after impact to the kinetic energy just before impact is $\frac{1}{3}$
Let $M$ be the mass of bob A. Then $2M$ is the mass of bob B. Let $v_A$ be bob A's speed just before the collision. We can use conservation of momentum to find the speed $v_f$ just after the collision: $m_f~v_f = m_A~v_A$ $(M+2M)~v_f = M~v_A$ $v_f = \frac{v_A}{3}$ We can find the initial kinetic energy: $KE_0 = \frac{1}{2}Mv_A^2$ We can find the final kinetic energy: $KE_f = \frac{1}{2}(3M)(\frac{v_A}{3})^2$ $KE_f = \frac{1}{3}\times \frac{1}{2}Mv_A^2$ $KE_f = \frac{1}{3}\times KE_0$ The ratio of the kinetic energy just after impact to the kinetic energy just before impact is $\frac{1}{3}$.