Question

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

• A. $\frac{4m_{1}m_{2}}{(m_{1} + m_{2})^{2}}$
• B. $\frac{m_{1}}{m_{1} + m_{2}}$
• C. $\frac{m_{2}}{m_{1} + m_{2}}$
• D. $\frac{m_{1}m_{2}}{(m_{1} + m_{2})}$

Answer 1

Answer: (A)

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

Answer 2

According to momentum conservation, we get

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

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

According to kinetic energy conservation, we get

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

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

Form Egs. (i) and (ii), it follows that

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

Or $v_{2f}(v_{1i} - v_{1f}) = v_{1}^{2}i - v_{1f}^{2} = (v_{1i} - v_{1f})(v_{1i} + v_{1f})$

$\therefore v_{2f} = v_{1i} + v_{1f}$

Substituting this in Eq. (i) we get

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

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

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

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

$k_{1f} = \frac{1}{2}m_{1}v_{1f}^{2} = \frac{1}{2}m_{1}\left( \frac{m_{1} - m_{2}}{m_{1} + m_{2}} \right)^{2}v_{1i}^{2}$ (Using (iii))

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

$f = \frac{k_{1i} - k_{1f}}{k_{1i}}$

$= \frac{\frac{1}{2}m_{1}v_{1i}^{2} - \frac{1}{2}m_{1}\left( \frac{m_{1} - m_{2}}{m_{1} + m_{2}} \right)^{2}v_{1i}^{2}}{\frac{1}{2}m_{1}v_{1i}^{2}}$

$= 1 - \left( \frac{m_{1} - m_{2}}{m_{1} + m_{2}} \right)^{2} = \frac{4m_{1}m_{2}}{(m_{1} + m_{2})^{2}}$