Two blocks with masses m 1 = 1 k g a n d m 2 = 2 kg are connected by a spring constant k = 24 N m - 1 and placed on a frictionless horizontal surface. The block m 1 is imparted an initial velocity v 0 = 12 c m s - 1 to the right, the value of amplitude of oscillation (in cm ) of the block with respect to other is

Step-by-step solution

The amplitude of oscillations will be the maximum compression in the spring. At the time of maximum compression velocities of both the blocks are equal say v , then using law of conservation of momentum, m 1 v 0 = m 1 + m 2 v or 1 × 12 = 1 + 2 v o r v = 4 c m s - 1 Using law of conservation of energy, we have 1 2 m 1 v 0 2 = 1 2 k x 2 + 1 2 m 1 + m 2 v 2 Putting the value and solving we get x = 2 cm