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Velocity of Separation and Approach

A body of mas...

Question

A body of mass $m_{1}$ moving at a constant speed undergoes an elastic collision with a body of mass $m_{2}$ initially at rest. The ratio of the kinetic energy of mass $m_{1}$ after the collision to that before the collision is

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

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

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

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

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Solution

The correct option is D ${(\frac{m_{1} - m_{2}}{m_{1} + m_{2}})}^{2}$
Let $u_{1}, u_{2}$ be the speeds of masses $m_{1}$ and $m_{2}$ before the collision and $v_{1}, v_{2}$ be the speeds of those masses after the collision.
Here $u_{2} = 0$ .

From the conservation of linear momentum,
$m_{1} u_{1} = m_{1} v_{1} + m_{2} v_{2}$

From the equation of restitution
$e = \frac{v_{2} - v_{1}}{u_{1} - u_{2}}$

$1 = \frac{v_{2} - v_{1}}{u_{1}}$

After the solving the above equations, we get

$v_{1} = (\frac{m_{1} - m_{2}}{m_{1} + m_{2}}) u_{1}$

and $v_{2} = (\frac{2 m_{1}}{m_{1} + m_{2}}) u_{1}$

$∴ K E$ of $m_{1}$ after collision $= \frac{1}{2} m_{1} v_{1}^{2}$

$K E_{f} = \frac{1}{2} m_{1} {(\frac{m_{1} - m_{2}}{m_{1} + m_{2}})}^{2} u_{1}^{2}$

$K E$ of $m_{1}$ before collision $= \frac{1}{2} m_{1} u_{1}^{2}$

The ratio of the two is ${(\frac{m_{1} - m_{2}}{m_{1} + m_{2}})}^{2}$ .

Hence, the correct choice is $(A)$ .

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