Question

The left block in figure (12-E13) 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 the motions of the two blocks are periodic. Find the time period of these periodic motions. Neglect the widths of the blocks.

Answer 1

When the block A mvoes 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 block b will return to the original point and completes half of the oscillation. So, the time period of b is,

$\frac{2\pi \sqrt{\left(\frac{m}{k}\right)}}{2}=\frac{m}{k}$

The block B collides with the block A and comes to rest at that point. The block A again moves a further distance 'L' to return to its original position. The time taken is,

$\frac{L}{V}+\frac{L}{V}=2\left(\frac{L}{V}\right)$

So, time period of the periodic motion is,

$2\frac{L}{V}+\pi \sqrt{\left(\frac{m}{k}\right)}$