Two bodies A and B of masses m and 2m respect | vedclass

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Question

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 sam

Vedclass · Accepted Answer

$\frac{{2m{v^2}_0}}{{3{x^2}_0}}$

Vedclass · Answer

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

$\frac{1}{2} m v_{0}^{2}=\frac{1}{2} m v^{2}+\frac{1}{2}(2 m) v^{2}+\frac{1}{2} k x_{0}^{2}$

$\frac{1}{2} m v_{0}^{2}=\frac{3}{2} m v^{2}+\frac{1}{2} k x_{0}^{2}$

$\frac{1}{2} m v_{0}^{2}=\frac{3}{2} m\left(\frac{v_{0}}{3}ight)^{2}+\frac{1}{2} k x_{0}^{2}$

$	herefore \frac{1}{2} k x_{0}^{2}=\frac{1}{2} m v_{0}^{2}-\frac{1}{6} m v_{0}^{2}$

Or $\frac{1}{2} k x_{0}^{2}=\frac{1}{3} m v_{0}^{2}$

$k=\frac{2}{3} \frac{m v_{0}^{2}}{x_{0}^{2}}$

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