Question

The figure below shows a block $C$ (of mass $3\, M .$ moving on a smooth horizontal surface at speed such that it makes an elastic head-on impact with a block ' $A$ ' (of mass $M$ ) initially at rest. ' $A$ ' is attached to an identical block ' $B$ ' (also at rest initially) through a massless ideal spring of stiffiness ' $k$ ' which is also at a relaxed state initially. If the collision occurs between $A$ and $C$ at time $t=0$, then the earliest time at which the speed of the block ' $B$ ' will be $\frac{3 v_{0}}{8}$ is at $t=\frac{\pi}{3} \sqrt{\frac{M}{x k}} .$ Find $x .$

Answer 1

Just after impact, at $t=0$ the velocities of $A$ and $B$ are $\frac{3 v_{0}}{2}$ and $0$ respectively and the CoM (of $A$ and $B$ ) is $\frac{3 v_{0}}{4} .$
Since the spring starts from relaxed state at $t=0$, the blocks $A$ and $B$ are in SHM (wrt their CoM) with time period $T=2 \pi \sqrt{\frac{M}{2 k}}$ where $M / 2$ is the "reduced mass" of the system.
So w.r.t CoM both $A$ and $B$ have a speed (max speed) of $\frac{3 v_{0}}{4}$ (towards CoM) initially, and therefore will have a speed of $\frac{3 v_{0}}{8}$ (half of max speed) at $t=\frac{T}{6}=\frac{\pi}{3} \sqrt{\frac{M}{2 k}}$.
That is when (w.r.t ground) $A$ and $B$ will have individual speeds of $\frac{9 v_{0}}{8}$ and $\frac{3 v_{0}}{8}$ respectively