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 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 $\theta_{2}=37^{\circ} .$ Assume that the collision is elastic and that friction and rotational motion are not important. Obtain $\theta_{1}$
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 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 $\theta_{2}=37^{\circ} .$ Assume that the collision is elastic and that friction and rotational motion are not important. Obtain $\theta_{1}$
Answer From momentum conservation, since the masses are equal
$v _{11}= v _{1 f}+ v _{2 f }$
$v_{1 i}^{2}=\left( v _{1 f}+ v _{2 f}\right) \cdot\left( v _{1 f}+ v _{2 f}\right)$
$=v_{1 f}^{2}+v_{2 f}^{2}+2 v _{1 f} \cdot v _{2 f}$
$=\left\{v_{1 f}^{2}+v_{2 f}^{2}+2 v_{1 f} v_{2 f} \cos \left(\theta_{1}+37^{\circ}\right)\right\}$
since the collision is elastic and $m_{1}=m_{2}$ it follows from conservation of kinetic energy that $v_{1 i}^{2}=v_{1 f}^{2}+v_{2 f}^{2}$
Comparing above equations we get
$\cos \left(\theta_{1}+37^{\circ}\right)=0$
or $\quad \theta_{1}+37^{\circ}=90^{\circ}$
Thus, $\quad \theta_{1}=53^{\circ}$
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.
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