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A block of mass 1 kg moving with a speed of $4 {ms}^{- 1}$ , collides with another block of mass 2 kg which is at rest. The lighter block comes to rest after collision. The loss in KE of the system is

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8 J

$4 \times 10^{- 7} J$

4 J

0 J

answer is C.

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Detailed Solution

$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2} 1 \times 4 = 2 \times v_{2} v_{2} = 2 m / s$

$∆ KE = \frac{1}{2} m_{1} {u_{1}}^{2} - \frac{1}{2} m_{2} {v_{2}}^{2} = \frac{1}{2} (1 \times 4^{2} - 2 \times 2^{2}) = \frac{1}{2} 8 = 4 J$

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