Question

The figure below shows a block $C$ (of mass $3M.$ 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 $\frac{3v_0}{8}$ is at $t=\frac{\pi }{3}\sqrt{\frac{M}{xk}}.$ Find $x.$

Answer 1

Answer: 2

Just after impact, at $t=0$ the velocities of $A$ and $B$ are $\frac{3v_0}{2}$ and $0$ respectively and the CoM (of $A$ and $B$ ) is $\frac{3v_0}{4}.$
Since the spring starts from relaxed state at $t=0$, the blocks $A$ and $B$ are in SHM (wrt their CoM) with time period $T=2\pi \sqrt{\frac{M}{2k}}$ 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 $\frac{3v_0}{4}$ (towards CoM) initially, and therefore will have a speed of $\frac{3v_0}{8}$ (half of max speed) at $t=\frac{T}{6}=\frac{\pi }{3}\sqrt{\frac{M}{2k}}$.
That is when (w.r.t ground) $A$ and $B$ will have individual speeds of $\frac{9v_0}{8}$ and $\frac{3v_0}{8}$ respectively