A block A of mass 2m is hanging from a vertical massless spring of spring constant k and is in equilibrium. Another block B of mass m strikes the block A with velocity u and sticks to it as shown in the figure. The magnitude of the acceleration of the combined system of the blocks just after the collision is

Options

A. `g//2`
B. `g//3`
C. g
D. zero

Answer

B

Explanation

For block A to be in equilibrium, the spring force acting on it should be equal to its weight. This means
`F _("spring") = 2 mg`
Just after the collision, the combined mass of the system is 3m and the spring force is the same. Hence, thee magnitude of the acceleration of the combined system of the blocks just after the collision,
`a = ( 3 mg - 2 mg)/(3 m ) = (g)/(3)`