Three balls `A, B` and `C(m_(A)=m_(C)=4m_(B))` are placed onn a smooth horizontal surface. Ball `B` collides with ball `C` with an initial velocity `v` as shown in figure. Find the total number of collision betwenent the balls (all collisions are elastic).

Answer

2

Explanation

For the first collision `e=1, v=v_(1)+v_(2)`

`implies v_(2)=v-v_(1)`
By momentum conservation
`m_(B)v=-m_(B)v_(1)+m_(C) v_(2)`
`m_(B)v=-m_(B)v_(1)+4m_(B)v_(2)`
`v_(2)=(v_(1)+v)/4`
From eqn i and ii, `v_(1)=3/5 v` and `v_(2) 2/5v`
For the second collision `e=1`

`v_(1)=v_(1)^(')+v_(3)impliesv_(3)=v_(1)-v_(1)'(`)
By momentum conservation `-m_(B)v_(1)=m_(B)v_(1)^(')-m_(A)v_(3)`
or `-m_(B)v_(1)=m_(B)v_(1)^(')-4m_(B)v_(3)` (`:'m_(A)=4m_(B))`
`v_(3)=(v_(1)^(')+v_(1))/4`
from eqn iii and iv `v_(1)^(')=3/5v_(1)(3/5v)=9/25v`
Clearly `9/25vlt2/5v`
therefore `B` cannot collide with `C` for the second time.
Hence, the total number of collisions is `2`.