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The 'e' Equation

A spherical b...

Question

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

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

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

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

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

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Solution

The correct option is A $\frac{4 m_{1} m_{2}}{(m_{1} + m_{2})^{2}}$
According to momentum conservation, we get
$m_{1} v_{1 i} = m_{1} v_{1 f} = m_{2} v_{2 f} . . . . (i)$
whered $v_{1 i}$ 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
$\frac{1}{2} m_{1} v_{1 i}^{2} = \frac{1}{2} m_{1} v_{1 f}^{2} + \frac{1}{2} m_{2} v_{2 f}^{2}$
$m_{1} v_{1 i}^{2} = m_{1} v_{1 f}^{2} + m_{2} v_{2 f}^{2} . . . . . (i i)$
From Eqs. (i) and (ii), it follows that
$m_{1} v_{1 i} (v_{2 f} - v_{1 i}) = m_{1} v_{1 f} (v_{2 f} - v_{1 f})$
or $v_{2 f} (v_{1 i} - v_{1 f}) = v_{1}^{2} i - V_{1 f}^{2} = (v_{1 i} - v_{1 f}) (v_{1 i} + v_{1 f})$
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
$K_{1 i} = \frac{1}{2} m_{1} v_{1 i}^{2}$
The final kinetic energy of the mass $m_{1}$ is
$K_{1 f} = \frac{1}{2} m_{1} v_{1 f}^{2} = \frac{1}{2} m_{1} {(\frac{m_{1} - m_{2}}{m_{1} + m_{2}})}^{2} v_{1 i}^{2}$ (Using (iii))
The fraction of kinetic energy lost by $m_{1}$ is
$f = \frac{K_{1 i} - K_{1 f}}{K_{1 i}} = \frac{\frac{1}{2} m_{1} v_{1 i}^{2} - \frac{1}{2} m_{1} {(\frac{m_{1} - m_{2}}{m_{1} + m_{2}})}^{2} v_{1 f}^{2}}{\frac{1}{2} m_{1} v_{1 i}^{2}}$
$= 1 - {(\frac{m_{1} - m_{2}}{m_{1} + m_{2}})}^{2} = \frac{4 m_{1} m_{2}}{(m_{1} + m_{2})^{2}}$ .

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