Two rectangular blocks A and B of masses 2 kg and 3 gk respectively are connected by a spring of spring constant `10.8Nm^(-1)` and are placed on a frictionless horizontal surface. The block A was given an initial velocity of `0.15 ms^(-1)` in the direction shown in the figure. The maximum compression of the spring during the motion is

Options

A. 0.01 m
B. 0.02 m
C. 0.05 m
D. 0.03 m

Answer

C

Explanation

Block A moves with velocity `0.15ms^(-1)` compresses the spring which pushes B towards right. A goes on compressing the spring till the velocity acquired by B becomes equal to the velocity of A. Let this velocity be v. This state occurs when the spring is in a state of maximum compression. Let x be the maximum compression in this stage.
According to the law of conservation of linear momentum we get
`m_(A)u=(m_(A)+m_(B))v`
or `v=(m_(A)v)/(m_(A)+m_(B))=(2xx0.15)/(2+3)=0.06ms^(-1)`
According to the conservation of energy, we get
`1/2m_(A)u^(2)=1/2(m_(A)+m_(B))v^(2)+1/2kx^(2)`
`1/2m_(A)u^(2)-1/2(m_(A)+m_(B))v^(2)=1/2kx^(2)`
`1/2xx2xx(0.15)^(2)-1/2(2+3)(0.06)^(2)=1/2kx^(2)`
`0.0225-0.009=1/2kx^(2)` or `0.0135=1/2kx^(2)`
or `x=sqrt(0.027/k)=sqrt(0.027/10.8)=0.05m`