Consider the collision depicted in figure to be between two billiard balls are equal masses m_(1)=m_(2) . The first ball iscalledthe 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 thatfriction and rotational motion are not important . Obtain theta_(1) .

Consider the collision depicted in figure to be between two billiard b

1.	Consider the collision depicted in figure to be between two billiard balls are equal masses m_(1)=m_(2) . The first ball iscalledthe 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 thatfriction and rotational motion are not important . Obtain theta_(1) .
Answer» Solution :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.