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Two bodies $A$ and $B$ of mass $m$ and $2\, m$ respectively are placed on a smooth floor. They are connected by a spring of negligible mass. $A$ third body $C$ of mass $m$ is placed on the floor. The body $C$ moves with a velocity $v_0$ along the line joining $A$ and $B$ and collides elastically with $A$. At a certain time after the collision it is found that the instantaneous velocities of $A$ and $B$ are same and the compression of the spring is $x_0$. The spring constant $k$ will be
$m\,\frac{{v_0^2}}{{x_0^2}}$
$m\,\frac{{{v_0}}}{{2{x_0}}}$
$2m\,\frac{{{v_0}}}{{{x_0}}}$
$\frac{2}{3}m\,{\left( {\frac{{{v_0}}}{{{x_0}}}} \right)^2}$
Solution
Initial momentum of the system block $(C)$
$ = m{v_0}$ .After striking with $A$, the block $C$ comes to rest and now both block $A$ and $B$ moves with velocity $v$ when compression in spring is ${x_0}.$
By the law of conseravtion of linear momentum
$m{v_0} = \left( {m + 2m} \right)v \Rightarrow \frac{{{v_0}}}{3}$
By the law of conservation of energy
$K.E. \,of\, block \,C = K.E.\ of\, system + P.E. of system$
$\begin{array}{l}
\frac{1}{2}mv_0^2 = \frac{1}{2}\left( {3m} \right){\left( {\frac{{{v_0}}}{3}} \right)^2} + \frac{1}{2}kx_0^2\\
\Rightarrow \,\frac{1}{2}mv_0^2 = \frac{1}{6}mv_0^2 + \frac{1}{2}kx_0^2\\
\Rightarrow \,\frac{1}{2}kx_0^2 = \frac{1}{2}mv_0^2 – \frac{1}{6}mv_0^2 = \frac{{mv_0^2}}{3}\\
\therefore \,\,\,\,\,\,\,k = \frac{2}{3}m{\left( {\frac{{{v_0}}}{{{x_0}}}} \right)^2}
\end{array}$
