  Question

# A block of mass 1.2 kg moving at a speed of 20 cm/s collides head-on with a similar block kept at rest. The coefficient of restitution is 35. Find the loss of kinetic energy during the collision.7.7 × 10-2 J 7.7 × 10-1 J           7.7 × 10-3 J           7.7 × 10-4 J

Solution

## The correct option is C 7.7 × 10-3 J           Suppose the first block moves at a speed v1 and the second at v2 after the collision. Since the collision is  head-on, the two blocks move along the original direction of motion of the first block. By conservation of linear momentum, (1.2 kg)(20 cm/s) = (1.2 kg)v1 + (1.2 kg)v2 or,                       v1 + v2 = 20 cm/s              ...........(i) The velocity of separation is v2−v1 and the velocity of approach is 20 cm/s. as the coefficient of restitution is 35, we have,  v2 − v1 = 35 × 20 cm/s = 12 cm/s              ...........(ii) by (i) and (ii) v1 = 4 cm/s and v2 = 16 cm/s the loss in kinetic energy is 12(1.2kg)[(20cm/s)2−(4cm/s)2−(16cm/s)2] = (0.6 kg)[0.04 m2/s2 - 0.0016 m2/s2 - 0.0256 m2/s2] = (0.6 kg)(0.0128 m2/s2) = 7.7 × 10−3 J.  Suggest corrections   