Consider the collision depicted in the 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 $\theta_{1}$.

$(53^{\circ})$ From the law of conservation of linear momentum, since the masses are equal $(m_{1} = m_{2})$, we have:
$\vec{v}_{1i} = \vec{v}_{1f} + \vec{v}_{2f}$
Taking the dot product of the velocity vector with itself:
$v_{1i}^{2} = (\vec{v}_{1f} + \vec{v}_{2f}) \cdot (\vec{v}_{1f} + \vec{v}_{2f})$
$v_{1i}^{2} = v_{1f}^{2} + v_{2f}^{2} + 2\vec{v}_{1f} \cdot \vec{v}_{2f}$
$v_{1i}^{2} = v_{1f}^{2} + v_{2f}^{2} + 2v_{1f}v_{2f} \cos(\theta_{1} + \theta_{2})$
Since the collision is elastic and the masses are equal, the conservation of kinetic energy implies:
$\frac{1}{2}m v_{1i}^{2} = \frac{1}{2}m v_{1f}^{2} + \frac{1}{2}m v_{2f}^{2}$
$v_{1i}^{2} = v_{1f}^{2} + v_{2f}^{2}$
Comparing the two expressions for $v_{1i}^{2}$, we get:
$2v_{1f}v_{2f} \cos(\theta_{1} + \theta_{2}) = 0$
Since $v_{1f} \neq 0$ and $v_{2f} \neq 0$, we must have $\cos(\theta_{1} + \theta_{2}) = 0$, which implies $\theta_{1} + \theta_{2} = 90^{\circ}$.
Given $\theta_{2} = 37^{\circ}$, we find:
$\theta_{1} = 90^{\circ} - 37^{\circ} = 53^{\circ}$.
Thus, the balls move at right angles to each other after the collision.