Two blocks A and B of masses m and 2 m are connected by a massless spring of force constant k and are placed on a smooth horizontal plane. The spring is stretched by an amount x and then released. The relative velocity of the blocks when the spring comes to its natural length is

Step-by-step solution

Let us assume the velocity of block A and B are v 1 and v 2 respectively. Relative velocity will be v 1 + v 2 Using conservation of linear momentum m v 1 = 2 m v 2 ⇒       v 1 = 2 v 2 Using conservation of energy 1 2 k x 2 = 1 2 m v 1 2 + 1 2 2 m v 2 2 1 2 k x 2 = 1 2 m 2 v 2 2 + m v 2 2 1 2 k x 2 = 2 m v 2 2 + m v 2 2 3 m v 2 2 = k x 2 2 v 2 2 = k x 2 6 m v 2   k 6 m . x Relative velocity = 3 v 2 = 3 k 6 m . x =   3 k 2 m . x