Two rectangular blocks A and B of masses 2 kg and 3 kg respectively are connected by a spring of constant 10.8 Nm-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
As the block A moves with velocity 0.15 ms-1 it 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, i.e., 0.15 ms-1.Let this velocity be v.
Now, spring is in a state of maximum compression. Let x be the maximum compression at this stage.
According to the law of conservation of linear momentum, we get
⇒ v =
v = 0.06 ms-1
According to the law of conservation of energy
0.0225 - 0.009 =
⇒ x =
x = 0.05 m