Consider the collision depicted in Figure, to be between two billiard balls with equal masses `m_(1)=m_(2)`. The first ball is called the cue and the second ball is called the target. The billiard player wants to sink the target ball in a corner pocket, which is at an angle `theta_(2)=phi=37^(@)`. Assume that the collision is elastic and that friction and rotational motion are not important. Obtain `theta_(1)=theta` .

Explanation

From momentum conservation, since the masses are equal
`v_(li)= v_(lf) + v_(2f)`
or `v_(li)^(2) = (v-(1f)+v_(2f)).(v_(1f) + v_(2f))`
`=v_(1)f^(2)+v_(2)f^(2)+2v_(1)f. v_(2)f`
`={v_(1)f^(2)+v_(2)f^(2)+2v_(1)v_(2)fcos (theta_(1)+37^(@))`}............(i)
Since, the collision is elastic and `m_(1)=m_(2)` it follows from conservation of kinetic energy that
`v_(1t)^(2) = v_(1)f^(2)+v_(2)f^(2)`......................(ii)
Comparing Eqs. 1) and 2), we get
`cos(theta_(1)+37^(@)) = 0`
or `theta_(1)+37^(@)) =0`
Thus, `theta_(1)=53^(@)`.
This proves the following result, when two equal masses undergo a glancing elastic collision with one of them at rest, after the collision, they will move at right angles of each other.