Two rectangular blocks A and B of masses 2 kg and 3 kg respectively are connected by a spring of spring constant `10.8 Nm^(-1)` and are placed on a frictionless horizontal surface. The block A was given an initial velocity of `0.15ms^(-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_Au = (m_A + m_B)v`

`implies v=(m_Au)/(m_A+m_B) = (2xx0.15)/(2+3) = 0.06 ms^(-1)`
According to the conservation of energy, we get
`1/2 m_Au^2=1/2(m_A+m_B) v^2+1/2kx^2`
`1/2 m_Au^2-1/2(m_A+m_B) v^2=1/2kx^2`
`1/2 xx2 xx(0.15)^2-1/2(2+3)(0.06)^2 =1/2 kx^2`
` 0.0225 - 0.009=1/2 kx^2 implies 0.0135 =1/2 kx^2`
`implies x sqrt((0.027)/k)=sqrt((0.027)/(10.8))=0.05 m`