Question

A block $m_{1}$ strikes a stationary block $m_{3}$inelstically. Another block $m_{2}$ is kept on $m_{3}$. Neglecting the friction between all contacting surfaces, the fractional decrease of K. E. of the system in collision is

• A. $\frac{m_{1}}{m_{1} + m_{2} + m_{3}}$
• B. $\frac{m_{1}}{m_{2} + m_{3}}$
• C. $\frac{m_{3}}{m_{1} + m_{3}}$
• D. $\frac{m_{2} + m_{3}}{m_{1} + m_{2} + m_{3}}$

Answer 1

Answer: (C)

$\frac{m_{3}}{m_{1} + m_{3}}$

Answer 2

Since the impact between $m_{1}$and $m_{3}$ is inelastic, $m_{1}$and $m_{3}$ will move together towards right and $m_{2}$does not move due to the absence of friction.

The velocity of the combined mass = $v' = \frac{m_{1}v}{m_{1} + m_{3}}$

⇒ $\frac{|\Delta KE|}{KE} = \frac{\frac{1}{2}m_{1}v^{2} - \frac{1}{2}\left( m_{1} + m_{3} \right)v'^{2}}{\frac{1}{2}m_{1}v^{2}} = 1 - \left( \frac{m_{1} + m_{3}}{m_{1}} \right)\left( \frac{v'^{2}}{v} \right)$

$1 - \left( \frac{m_{1} + m_{3}}{m_{1}} \right)\left( \frac{m_{1}v}{m_{1} + m_{3}} \right)^{2} = 1 - \frac{m_{1}}{m_{1} + m_{3}} = \frac{m_{3}}{m_{1} + m_{3}}$