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Work Done on & by the Object

Two blocks of...

Question

Two blocks of masses $m_{1}$ and $m_{2}$ are connected by spring of constant K. The spring is initially compressed and the system is released from rest at t $=$ 0 second. The work done by spring on the blocks $m_{1}$ and $m_{2}$ be $W_{1}$ and $W_{2}$ respectively by time $t$ . The speeds of both the blocks at time t are non zero. Then the value of $\frac{W_{1}}{W_{2}}$ equals to:

A

\frac{m_{1}}{m_{2}}

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B

\frac{m_{2}}{m_{1}}

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C

{(\frac{m_{1}}{m_{2}})}^{2}

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D

{(\frac{m_{2}}{m_{1}})}^{2}

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Solution

The correct option is B $\frac{m_{2}}{m_{1}}$
Let work done by spring on the block of mass $m_{1}$ be $W_{1} =$ change in kinetic energy
$= \frac{1}{2} m_{1} {v_{1}}^{2} - \frac{1}{2} m_{1} {(0)}^{2} = \frac{1}{2} m_{1} {v_{1}}^{2}$
Similarly, $W_{2} = \frac{1}{2} m_{2} {v_{2}}^{2}$
$\Rightarrow \frac{W_{1}}{W_{2}} = \frac{\frac{1}{2} m_{1} v_{1}^{2}}{\frac{1}{2} m_{2} v_{2}^{2}}$
From conservation of momentum,
$m_{1} v_{1} = m_{2} v_{2}$
$∴ \frac{W_{1}}{W_{2}} = \frac{v_{1}}{v_{2}} = \frac{m_{2}}{m_{1}}$

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