Question

A block of mass m moves with a speed v towards the right block in equilibrium with a spring. If the surface is frictionless and collistions are elastic, the frequency of collisions between the masses will be :

• A.
• B.
• C.
• D. $\frac{4}{[\frac{2L}{v}+\frac{1}{\pi }\sqrt{\frac{K}{m}}]}$

Answer 1

The correct option is C

Time taken to collide on left wall and get back to the mass attached with spring is $t_1=\frac{2L}{v}$

Time to get compressed once and back is, $t_2=\frac{T}{2}=\frac{2\pi }{2}\sqrt{\frac{m}{K}}=\pi \sqrt{\frac{m}{K}}$

Remember the colliding block will come to rest on exchanging momentum and starts back when the mass connected to spring hits it, on its way back.

So, $\nu =\frac{1}{t_1+t_2}$

And the frequency of collision will be double the frequency of oscillation.

So frequency of collision.$2v=\frac{2}{[\frac{2L}{v}+\frac{1}{\pi }\sqrt{\frac{K}{m}}]}$