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

Explanation

From momentum conservation, since the masses are equal
`v_(1i)= v_(1f)+v_(2f)`
or, `v_(1i)""^(2)=(v_(1f)+v_(2f))*(v_(1f)+v_(2f))`
`=v_(1i)""^(2)+v_(2f)""^(2)+2v_(1f)*v_(2f)`
`={v_(1f)""^(2)+v_(2f)""^(2)+2v_(1f)v_(2f) cos(theta_1 +37^(@))}" "(6.32)`
Since the collision is elastic and `m_(1)= m_(2)` it follows from conservation of kinetic energy that
`v_(1f)""^(2)= v_(1f)""^(2)+v_(2f)""^(2)` (6.33)
Comparing Eqs. (6.32) and (6.33), we get
`cos (theta_(1)+37^(@))=0`
or, `theta_(1)+ 37^(@)= 90^(@)`
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 to each other.