The right block in figure moces at a speed `V` towards the left block placed in equilibrium. All the surfaces are smooth and all the collisions are elastic. Find the time period motion. Neglect the width of the blocks.

Options

A. `pisqrt((m)/(2k)) + (2L)/(v)`
B. `pisqrt((m)/(2k)) + (L)/(v)`
C. `pisqrt((m)/(2k)) - (L)/(v)`
D. `pisqrt((m)/(k)) + (L)/(v)`

Answer

A

Explanation

since all collision are elestic and total energy is conserved
Time period of motion of right block is `t_(1) = (2L)/(v)`.
Time period of motion of left block is `t_(2) = (1)/(2)2pisqrt((m)/(2k))`
So. Total time period `= (2L)/(v) + (1)/(2) [2pisqrt((m)/(2k))]`