The left block in figure moves at a speed v towards the right block placed in equilibrium. All collisions to take place are elastic and the surfaces are frictionless. Show that motion of the two blocks are periodic. Find the time period of these periodic motions. Neglect the widths of the blocks.

Answer

B

Explanation

When the block A moves with velocity V and collides with the block B, it transfers all energy to the block B. (because it is an elastic collision). The block A will move a distance x against the spring, again the spring, again the block b will return to the origiN/Al point and completes half of the oscillation.
So, the time period of b is
`(2pisqrt(m/k))/2=m/k`
The block B collides with the block A and comes o rest at that point. The block AS again moves a further distance L to return to its origiN/Al position. The time taken is
`L/V+L/V=2(L/V)`
So the period of the periodic motion is
`2L/V+pi sqrt((m/k))`