Question

In this section, all collisions will be completely elastic

Answer 1

Given that,

m1 = 1 kg ; u1 = 5.8 m/s

m2 = 6 kg ; u2 = 0(as it is at rest intially)

The collision is completely elastic.

(1) let the speed of m1 after collision be v1. It will be given by:

v1 = u1 x (m1 - m2 ) / (m1 + m2) = 5.8 x [( 1 - 6) / ( 1 + 6)] = 4.143 m/s

hence v1 = 4.143 m/s

(2) Let the speed of m2 be v2 . So

v2 = u1 (2m1 / (m1+m2) = 5.8 x ( 2 x 1 / ( 1 + 6) = 1.66 m/s

hence v2 = 1.66 m/s

(3)Let the velocity after the second collision be V1'

As the energy will be conserved. So intial KE of block 1 before and after will be same, so the velocity remains same too after it has collided with the wall.

We already have determind the velocity of block two after first collision. So. Velocity of the first block after second collsion will be given by:

V1 = v1 (m1 - m2) + 2 m2 v2 / ( m1 + m2) = 4.143 ( 1 - 6 ) + 2 x 6 x 1.66 / ( 1 + 6 ) = 0.113 m/s

hence the velocity of block 1 (m1) after second collsion is = V1 = 0.113 m/s