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Two bodies $A$ and $B$ of masses $m$ and $2m$ respectively are placed on a smooth floor. They are connected by a spring. A third body $C$ of mass $m$ moves with velocity $V_0$ along the line joining $A$ and $B$ and collides elastically with $A$ as shown in fig. At a certain instant of time $t_0$ after collision, it is found that instantaneous velocities of $A$ and $B$ are the same. Further at this instant the compression of the spring is found to be $x_0$. Determine the spring constant
$\frac{{2m{v^2}_0}}{{3{x^2}_0}}$
$\frac{1}{3}\frac{{m{v^2}_0}}{{{x^2}_0}}$
$\frac{1}{4}\frac{{m{v^2}_0}}{{{x^2}}}$
$\frac{4}{5}\frac{{m{v^2}_0}}{{{x^2}_0}}$
Solution
Initially, the blocks $A$ and $B$ are at rest and $C$ is moving with velocity $v_{0}$ to the right. As masses of $\mathrm{C}$ and $\mathrm{A}$ are same and the collision is elastic the body $C$ treansfers its whole momentum $m v_{0}$ to body $A$ and as a result the body $\mathrm{C}$ stops and $\mathrm{A}$ starts moving with velocity $v_{0}$ to the right. As this instant the spring is uncompressed and the body $B$ is still at rest.
The momentum of the system at this instant $=m v_{0}$
Now, the spring is compressed and the body $B$ comes in motion. After time $t_{0},$ the compression of the spring is $x_{0}$ and common velocity of $A$ and $B$ is $\mathrm{v}(\mathrm{say})$
As external force on the system is zero, the law of conservation of linear momentum gives
$m v_{0}=m v+(2 m) v$ or $v=\frac{v_{0}}{3}$
The law of conservation of energy gives
$\frac{1}{2} m v_{0}^{2}=\frac{1}{2} m v^{2}+\frac{1}{2}(2 m) v^{2}+\frac{1}{2} k x_{0}^{2}$
$\frac{1}{2} m v_{0}^{2}=\frac{3}{2} m v^{2}+\frac{1}{2} k x_{0}^{2}$
$\frac{1}{2} m v_{0}^{2}=\frac{3}{2} m\left(\frac{v_{0}}{3}\right)^{2}+\frac{1}{2} k x_{0}^{2}$
$\therefore \frac{1}{2} k x_{0}^{2}=\frac{1}{2} m v_{0}^{2}-\frac{1}{6} m v_{0}^{2}$
Or $\frac{1}{2} k x_{0}^{2}=\frac{1}{3} m v_{0}^{2}$
$k=\frac{2}{3} \frac{m v_{0}^{2}}{x_{0}^{2}}$
