Question

A block m₁ strikes a stationary block m₃inelastically. Another block m₂ is kept on m₃. 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 m₁ and m₃ is inelastic, m₁ and m₃ will move together towards right and m₂ does not move due to the absence of friction.The velocity of the combined mass

= $v^{'} = \frac{m_{1}v}{m_{1} + m_{3}}$

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

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

Hence, the correct choice is (3).