Question

A ball of mass m moving with speed V collides with another ball of mass 2m (e= 1/2) in a horizontal smooth fixed circular tube of radius R (R is sufficiently large R>>>d). The time after which next collision will take place is:________

Answer 1

`$ (4\pi R)/(V)$`

Step-by-step explanation:

Given :

Mass of ball 1 = m

Mass of ball 2 = 2m

Since, R>>>d, the collision is head on.

Therefore, we get

`$ (v_1 -v_2)/(V)=(1)/(2)$`

`$ \therefore \frac{\text{velocity of seperation}}{\text{velocity of approach}}= v_1-v_2 = (V)/(2)$`

Relative velocity is given by V/2. So, we get the time when the masses will again collide as

`$ t = (2\pi R)/((V)/(2))=(4\pi R)/(V) $`