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De Broglie's Hypothesis

A particle of...

Question

A particle of mass $m_{1}$ experienced a perfectly elastic collision with a stationary particle of mass $m_{2}$ . What fraction of the kinetic energy does the striking particle lose, if the collision is a head-on one?

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

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

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

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

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Solution

The correct option is A $η = \frac{4 m_{1} m_{2}}{{(m_{1} + m_{2})}^{2}}$
In head-on collision both momentum and kinetic energy are conserved hence, by conservation of momentum
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
$m_{1} u_{1} + m_{2} (0) = m_{1} v_{1} + m_{2} v_{2}$
$m_{1} u_{1} = m_{1} v_{1} + m_{2} v_{2}$ ............(1)
And conservation of kinetic energy gives
$\frac{1}{2} m_{1} (u_{1})^{2} = \frac{1}{2} m_{1} (v_{1})^{2} + \frac{1}{2} m_{2} (v_{2})^{2}$ ...........(2)
From equation (1) and (2)
$m_{1} (u_{1} v_{1}) = m_{2} v_{2}$ ...........(3)
$m_{1} [(u_{1})^{2} (v_{1})^{2}] = m_{2} v_{2}$ ...........(4)
Dividing equation (4) by (3) we get
$u_{1} + v_{1} = v_{2}$
Using this value in (3)
$m_{1} (u_{1} v_{1}) = m_{2} (u_{1} + v_{1})$
$m_{1} u_{1} m_{2} u_{2} = m_{2} v_{2} + m_{1} v_{1}$
$u_{1} (m_{1} - m_{2}) = v_{1} (m_{1} + m_{2})$
$\Rightarrow v_{1} = \frac{u_{1} (m_{1} - m_{2})}{(m_{1} + m_{2})}$
Hence, fractional loss in KE of striking particle is ratio of change in KE to initial KE.
$η = \frac{E_{i} E_{f}}{E_{i}} = 1 - \frac{E_{f}}{E_{i}}$
$\Rightarrow η = 1 - (\frac{v_{1}}{u_{1}})^{2}$
$\Rightarrow η = 1 (\frac{(m_{1} - m_{2})}{(m_{1} + m_{2})})^{2}$
$\Rightarrow η = \frac{(m_{1} + m_{2})^{2} (m_{1} - m_{2})^{2}}{m_{1} + m_{2})^{2}}$
$\Rightarrow η = \frac{4 m_{1} m_{2}}{(m_{1} + m_{2})^{2}}$

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