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A block $C$ of mass $m$ is moving with velocity $v_0$ and collides elastically with block $A$ of mass $m$ which connected to another block $B$ of mass $2\,m$ through a spring of spring constant $k$. What is $k$ if $x_0$ is the compression of spring when velocity of $A$ and $B$ is same?
$\frac {mv_0^2}{x_0^2}$
$\frac {mv_0^2}{2x_0^2}$
$\frac {3}{2} \frac {mv_0^2}{x_0^2}$
$\frac {2}{3} \frac {mv_0^2}{x_0^2}$
Solution
Using conservation of linear momentum, we have
$m v_{0}=m v+2 m v$
$\Rightarrow v=\frac{v_{0}}{3}$
Using conservation of energy, we have
$\frac{1}{2} m v_{0}^{2}=\frac{1}{2} k x_{0}^{2}+\frac{1}{2}(3 m) v^{2}$
where $x_{0}$ is compression in the string.
$\therefore m v_{0}^{2}=k x_{0}^{2}+(3 m) \frac{v_{0}^{2}}{g}$
$\Rightarrow k x_{0}^{2}=m v_{0}^{2}-\frac{m v_{0}^{2}}{3}$
$\Rightarrow k x_{0}^{2}=\frac{2 m v_{0}^{2}}{3}$
$\therefore k=\frac{2 m v_{0}^{2}}{3 x_{0}^{2}}$
