Question

A spring-block system is resting on a frictionless floor as shown in the figure. The spring constant is $2.0N{m}^{-1}$ and the mass of the block is $2.0kg$. Ignore the mass of the spring. Initially, the spring is in an unstretched condition. Another block of mass $1.0kg$ moving with a speed of $2.0m{s}^{-1}$ collides elastically with the first block. The collision is such that the $2.0kg$ block does not hit the wall. The distance in metres between the two blocks when the spring returns to its unstretched position for the first time after the collision is

Answer 1

Let $V_1$ be the velocity of 1 kg mass and $V_2$ be the velocity of 2 kg mass.
After collision,
$\begin{array}{}{V}_1+2V_2=2& \text{(1)}\end{array}$
Also coefficient of restitution, $\begin{array}{}e=\frac{V_2-V_1}{2}& \text{(2)}\end{array}$
Solving (1) and (2)
$V_1=-\frac{2}{3}$ and $V_2=\frac{4}{3}$
$T=2\pi \sqrt{\frac{m}{k}}=2\pi \sec$

block returns to original position in $\frac{T}{2}=\pi \sec$
Distance between the blocks after collision when spring returns to unstretched position,$d=\frac{2}{3}\left(\pi \right)=\frac{2}{3}\left(3.14\right)=2.0933m$
$d=2m$