Two blocks A and B of equal mass `m=1.0kg` are lying on a smooth horizontal surface as shown in figure. A spring of force constant `k=200N//m` is fixed at one end of block A. Block B collides with block A with velocity `v_0=2.0m//s`. Find the maximum compression of the spring.

Answer

A, C

Explanation

At maximum compression `(X_m)` velocity of both the blocks is same, say it is v. Applying conservation of linear momentum, we have
`(m_A+m_B)v=m_Bv_0`
or `(1.0+1.0)v=(1.0)v_0`
or `v=v_0/2=2.0/2=1.0m//s`

Using conservation of mechanical energy, we have
`1/2m_Bv_0^2=1/2(m_A+m_B)v^2+1/2kx_m^2`
Substituting the values, we get
`1/2xx(1)xx(2.0)^2=1/2xx(1.0+1.0)xx(1.0)^2+1/2xx(200)xxx_m^2`
or `2=1.0+100x_m^2`
or `x_m=0.1m=10.0cm`