One billiard ball is shot east at $2.0$ m/s. A second, identical billiard ball is shot west at $1.0$ m/s. The balls have a glancing collision, not a head-on collision, deflecting the second ball by 90° and sending it north at $1.41$ m/s. What are the speed and direction of the first ball after the collision? Give the direction as an angle south of east.

At a speed of $2.0$ m/s, one billiard ball is fired east. At $1.0$ m/s, a second identical billiard ball is fired west. The balls collide glancingly rather than head-on, diverting the second ball 90 degrees and propelling it north at 1.41 m/s.

Ball 1 represents a ball moving east (positive x-axis direction) while ball 2 represents a ball moving west (negative x-axis direction). Then there's the fact that the north is up, or that the y axis is moving in the right direction. We know that momentum can only be conserved if all of its components are conserved. As a result, for the x component, we have:

$$\begin{gathered} m_1{v}_{1,x_i}+m_2{v}_{2,x_i}=m_1{v}_{1,x_f}+m_2{v}_{2,x_f} \\ {m}_0\cdot 2m/s+m_0\cdot \left(-1m/s\right)=m_0.v_{1,x_f}+0 \\ {v}_{1,x_f}=1m/s \end{gathered}$$

For y-axis:

$\begin{array}{l}{m}_1{v}_{1,y_i}+m_2{v}_{2,y_i}=m_1{v}_{1,y_f}+m_2{v}_{2,y_f}\\ m_0\cdot 0m/s+m_0\cdot 0m/s=m_0{v}_{1,x_f}+m_0\cdot 1.41m/s\\ v_{1,y_f}=-1.41m/s\end{array}$

The final speed, $v_{1,f}=\sqrt{v_{1,x_f}^2+v_{1,y_f}^2}=1.73m/s$

The direction of the first ball,

$\begin{array}{rr}\text{tan}\theta =\frac{v_{1,y_f}}{v_{1,x_f}}& =\frac{1.41m/s}{1m/s}\\ \Rightarrow \theta & ={54.65}^{\circ }\end{array}$