Q: A block of mass 1 kg moving with a speed of 4 m/s, 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

(a) 8 J

(b) 4 × 10^{-7} J

(c) 4 J

(d) 0 J

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Ans: (c)

Sol: Applying conservation of momentum ,

1 ×4 + 2 ×0 = 1 ×0 + 2× v

4 = 2 v

v = 2 m/s

$ \displaystyle K.E_i = \frac{1}{2} \times 1 \times 4^2 = 8 J$

$ \displaystyle K.E_f = \frac{1}{2} \times 2 \times 2^2 = 4 J$

Loss in K.E = 8 – 4 = 4 J