Question

Two blocks A and B each having mass 5 kg and 10 kg respectively. Block B is attached with a spring of spring constant 25 N/m. The rear block A is moving with a speed of 8 m/s toward the front block ‘B’ which is kept at rest. The maximum compression in the spring is: 

Answer 1

We can solve it by taking the initial kinetic energy of the moving block (1/2 mv² ) and the stored potential energy in the spring (1/2 kx² ). and equate it and find the change in position of displacement.

Also, Conservation of momentum will do the same (the spring force for the second block for final momentum) 

For maximum compression equate the final velocity of both the block.

Answer 2

When the moving block A comes in contact with the spring fixed with the  block B at rest ,the velocity of A is retarded and that of B is accelerated. As a result compression of spring goes on until maximum compression occurs.At this stage both the blocks attain same velocity (v) following conservation of momentum.

So we can write

m_Av_A+m_Bv_B=(m_A+m_B)v

=> 5×8+10×0=(5+10)v

=v=8/3 m/s

Now net loss in KE during this interaction will go to increase the PE of the spring in its compressed state.. So we can write using conservation of mechanical energy 

1/2m_Av_A² -1/2(m_A+m_B)v²= 1/2 kx² , where k is the force constant of the spring and x is its maximum compression

=> 5× 8²-(5+10)(8/3)² = 25x²

=>x² =( 5x8²×2)/3×1/25

=>x²= 128/15

=> x= 2.9m