A ball of mass m is moving normally towards a wall of mass M(>> m) with a velocity 4 m/s.The wall is also moving. After striking elastically with the wall, velocity of ball becomes 8 m/s. Find the speed of the ball:

6 m / s

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2 m / s

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2 \sqrt{2} m / s

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4 \sqrt{2} m / s

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Solution

The correct option is A $6 m / s$
$\begin{matrix} T h e m o m e n t u m a n d c o n s e r v a t i o n o f e n e r g y, t h e e q u a t i o n s f o r t h e v e l o c i t i e s o f t h e s p h e r e s a f t e r c o l l i s i o n a r e a . V_{1} = (\frac{m_{1} - m_{2}}{m_{1} + m_{2}}) U_{1} + (\frac{2 m_{2}}{m_{1} + m_{2}}) U_{2} a n d V_{2} = (\frac{m_{2} - m_{1}}{m_{1} + m_{2}}) U_{2} + (\frac{2 m_{1}}{m_{1} + m_{2}}) U_{1} h e r e, t h e m a s s o f t h e w a l l i s l e c o m p a r e d t o t h e v e l o c i t y o f t h e b a l l . s o, w e c a n c o n s i d e r t h e m a s s o f t h e b a l l (m_{1} = 0) . T h e r e f o r e, V_{1} = (\frac{0 - m_{2}}{0 + m_{2}}) U_{1} + (\frac{2 m_{2}}{0 + m_{2}}) U_{2} V_{1} = - U_{1} + 2 U_{2} a n d V_{1} = (\frac{m_{2} - 0}{0 + m_{2}}) U_{2} + (\frac{2 \times 0}{m_{1} + m_{2}}) U_{1} V_{2} = U_{2} s o, t h e v e l o c i t y o f t h e b a l l a f t e r t h e c o l l i s i o n i s, V_{1} = 2 U_{2} - U_{1} V_{1} = 2 \times 1 - (- 4) V_{1} = 2 + 4 = 6 m / s \end{matrix}$

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