Two identical blocks, each of mass 0.5 kg , are connected by a light spring of spring constant 100 N m ⁻¹ and placed on a smooth horizontal surface. The spring is initially compressed by 0.1 m and the system is held at rest. If the system is then released, the maximum speed acquired by each block is

Step-by-step solution

Let the maximum speed of each block be v . By the principle of conservation of mechanical energy, the initial elastic potential energy of the spring is completely converted into the kinetic energy of the two blocks when the spring passes through its natural length. Initial energy of the system, E_ i = 1 2 kx₀² Final energy of the system, E_ f = 1 2 mv² + 1 2 mv² = mv² Equating E_ i and E_ f : 1 2 kx₀² = mv² 1 2 100 (0.1)² = 0.5 v² 50 0.01 = 0.5 v² 0.5 = 0.5 v² v² = 1 v = 1 m s ⁻¹ Answer: 1 m s ⁻¹