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Part A

The cue ball is moving at 4.0 m/s when it impacts the five ball, which is at rest prior to the collision. The cue ball exits the (perfectly elastic) collision at an angle of 30 degrees from its original direction of motion. What are the final speeds of each ball?

Solution

We begin with a picture. Since the initial direction of motion of the cue ball is not specified, we arbitrarily assign it to move along the x-axis prior to the collision. We also arbitrarily assign it positive x- and y-velocity components after the collision.

Rather than implementing the elastic collision constraint of constant mechanical energy in the usual way, we will use the right angle rule described above. Using that rule plus conservation of y-momentum, we know that the five ball will have a final velocity directed at 60° below the positive x-axis in our coordinates. With this information in hand, we can simply write the equations of momentum conservation for our system.

\begin{large}\[ p^{cue}_{x,i} + p^{five}_{x,i} = p^{cue}_{x,f} + p^{five}_{x,f}\]
\[ p^{cue}_{y,f} + p^{five}_{y,f} = p^{cue}_{y,f} + p^{five}_{y,f}\]\end{large}

Using the fact that the cue ball and 5-ball have equal masses, and substituting any known zeros, this becomes:

\begin{large}\[ v^{cue}_{i} = v^{cue}_{x,f}+v^{five}_{x,f}\]
\[0 = v^{cue}_{y,f} + v^{five}_{y,f}\]\end{large}

We can rewrite these equations using geometry:

\begin{large}\[ v^{cue}_{i} = v^{cue}_{f}\cos(\theta^{cue})+v^{five}_{f}\cos(\theta^{five})\]
\[v^{five}_{f}\sin(\theta^{five}) = v^{cue}_{f}\sin(\theta^{cue}) \]\end{large}

We can now solve for each final speed in turn, finding:

\begin{large}\[v^{cue}_{f} = \frac{v^{cue}_{i}}{\cos(\theta^{cue}) + \cos(\theta^{five})\sin(\theta^{cue})/\sin(\theta^{five})}\]
\[ v^{five}_{f} = \frac{v^{cue}_{i}}{\cos(\theta^{five}) + \cos(\theta^{cue})\sin(\theta^{five})/\sin(\theta^{cue})}\]\end{large}

These expressions are already fairly simple, but they can be made even simpler by realizing that since:

\begin{large}\[ \theta^{five}+\theta^{cue} = 90^{\circ} \]\end{large}

we know:

\begin{large}\[ |\sin(\theta^{five})|=|\cos(\theta^{cue})| \] \[|\cos(\theta^{five})|=|\sin(\theta^{cue})|\]\end{large}

Using this plus the fact that sin² +cos² = 1, it is possible to simplify our answers to the form:

\begin{large}\[v^{cue}_{f} = v^{cue}_{i}\cos(\theta^{cue}) = \mbox{3.5 m/s}\]
\[ v^{five}_{f} = v^{cue}_{i}\sin(\theta^{cue})= \mbox{2.0 m/s}\] \end{large}

Part B

The cue ball is moving at 2.0 m/s when it impacts the five ball, which is at rest prior to the collision. The cue ball exits the (perfectly elastic) collision with a speed of 1.5 m/s. What angle with respect to the original direction of travel of the cue ball is made by each ball after the collision?

Solution

System, Interactions and Models: As in Part A.