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