A spherical ball of mass $m_{1}$ collides head on with another ball of mass $m_{2}$ at rest. The collision is elastic. The fraction of kinetic energy lost by $m_{1}$ is

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

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

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

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

answer is A.

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Detailed Solution

According to momentum conservation, we get

$m_{1} v_{1 i} = m_{1} v_{1 f} + m_{2} v_{2 f} . . . (i)$

where $v_{1 i}$ is the initial velocity of spherical ball of mass $m_{1}$ before collision and $v_{1 f}$ and $v_{2 f}$ are the final velocities of the balls of masses $m_{1}$ and $m_{2}$ after collision.

According to kinetic energy conservation, we get

$\frac{1}{2} m_{1} {v^{2}}_{1 i} = \frac{1}{2} m_{1} {v^{2}}_{1 f} + m_{1} {v^{2}}_{1 f} m_{1} {v^{2}}_{1 i} = m_{1} {v^{2}}_{1 f} + m_{2} {v^{2}}_{2 f} (i i)$

From Eqr. $(i)$ and $(i i)$ ,it follows that

$m_{1} v_{1 i} (v_{2 f} - v_{1 i}) = m_{1} v_{1 f} (v_{2 f} - v_{1 f}) o r v_{2 f} = v_{1 i} + v_{1 f}$

Substituting this in Eq. $(i)$ ,we get

$v_{1 f} = \frac{(m_{1} - m_{2})}{m_{1} + m_{2}} v_{1 i} . . . (i i i)$

The initial kinetic energy of the mass $m_{1}$ is

$k_{1 i} = \frac{1}{2} m_{1} v_{2^{} 1 i}$

The final kinetic energy of the mass $m_{1}$ is

$k_{1 f} = \frac{1}{2} m_{1} {v^{2}}_{1 f} = \frac{1}{2} m_{1} {(\frac{m_{1} - m_{2}}{m_{1} - m_{2}})}^{2} {v^{2}}_{1 i} (u \sin g (i i i))$

The fraction of kinetic energy lost by $m_{1}$ is

$f = \frac{k_{1 i} - k_{1 f}}{k_{1 i}} = \frac{\frac{1}{2} m_{1} {v^{2}}_{1 i} - \frac{1}{2} m_{1} {(\frac{m_{1} - m_{2}}{m_{1} + m_{2}})}^{2} {v^{2}}_{1 i}}{\frac{1}{2} m_{1} {v^{2}}_{1 i}} = 1 - {(\frac{m_{1} - m_{2}}{m_{1} + m_{2}})}^{2} = \frac{4 m_{1 m_{2}}}{{(m_{1} + m_{2})}^{2}}$