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Acceleration

A body mass ...

Question

A body mass $m_{1}$ strikes a stationary body of $m_{2}$ . If the collision is elastic, the fraction of kinetic energy transmitted by the first body to second bodu is

A

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

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B

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

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C

\frac{4 m_{1} m_{2}}{(m_{1} + m_{2})^{2}}

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D

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

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Solution

The correct option is C $\frac{4 m_{1} m_{2}}{(m_{1} + m_{2})^{2}}$
Let $u_{1}$ be the initial velocity of mass $m_{1}$ and $v_{1}$ be the final velocity of the mass $m_{1}$ .
As it perfect elastic collision , from law of conservation of momentum, we get
$m_{1} u_{1} = m_{1} v_{1} + m_{2} v_{2}$
$m_{2} v_{2} = m_{1} u_{1} - m_{1} v_{1}$
$m_{2} v_{2} = m_{1} (u_{1} - v_{1})$ ----------(1)
Again from law of conservation of kinetic energy we get,
$\frac{1}{2} m_{1} u_{1}^{2} = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}$
$m_{1} u_{1}^{2} = m_{1} v_{1}^{2} + m_{2} v_{2}^{2}$ -------------(2)
Dividing 2 and 1 we get,
$v_{2} = u_{1} + v_{1}$ -------(3)
Putting 3 in 1 we get,
$\frac{m_{1} u_{1} - m_{1} v_{1}}{m_{2}} = u_{1} + v_{1}$
$m_{1} u_{1} - m_{1} v_{1} = m_{2} u_{1} + m_{2} v_{1}$
$u_{1} (m_{1} - m_{2}) = v_{1} (m_{2} + m_{1})$
Thus,
$\frac{{K . E}_{i} - {K . E}_{f}}{{K . E}_{i}} = 1 - \frac{(1 / 2) m_{1} v_{1}^{2}}{(1 / 2) m_{1} u_{1}^{2}} = 1 - {\frac{(v_{1}}{u_{1}})}^{2}$
$1 - [{\frac{m_{1} - m_{2}}{m_{1} + m_{2}}]}^{2} = \frac{4 m_{1} m_{2}}{[{m_{1} + m_{2}]}^{2}}$

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