Consider the collision depicted in Fig. 6.10. to be between two billiard balls with 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 an angle theta_(2)= 37^(@). Assume that the collision is elastic and that friction and rotational motion are not important. Obtain theta_(1).

Consider the collision depicted in Fig. 6.10. to be between two billia

1.	Consider the collision depicted in Fig. 6.10. to be between two billiard balls with 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 an angle theta_(2)= 37^(@). Assume that the collision is elastic and that friction 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.