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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.


  1. 7.7 × 10-2 J

  2. 7.7 × 10-1 J          

  3. 7.7 × 10-3 J          

  4. 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 v2v1 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 × 103 J.

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