h2. 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?
h4. Solution
{toggle-cloak:id=sysa} *System:* {cloak:id=sysa}Cue ball and five ball as [point particles|point particle].{cloak}
{toggle-cloak:id=inta} *Interactions:* {cloak:id=inta}Impulse from external forces is ignored since the collision is assumed to be instantaneous.{cloak}
{toggle-cloak:id=moda} *Models:* {cloak:id=moda}[Momentum and External Force] plus [Mechanical Energy and Non-Conservative Work].
{note}Note that the rule we have derived above relies on the fact that the kinetic energy remains constant during the collision. For this reason, we have stated that we are using the mechanical energy model, though it will not be explicitly used in the solution.{note}
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{toggle-cloak:id=appa} *Approach:*
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{toggle-cloak:id=diaga} {color:red} *Diagrammatic Representation* {color}
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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.
||Before Collision||After Collision||
|!pool1.jpg!|!pool2.jpg!|
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{toggle-cloak:id=matha} {color:red} *Mathematical Representation* {color}
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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.
{note}Recall that we are assuming external impulses are negligible compared to the internal impulse of the collision.{note}
{latex}\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}{latex}
Using the fact that the cue ball and 5-ball have equal masses, and substituting any known zeros, this becomes:
{latex}\begin{large}\[ v^{cue}_{i} = v^{cue}_{x,f}+v^{five}_{x,f}\]
\[0 = v^{cue}_{y,f} + v^{five}_{y,f}\]\end{large}{latex}
We can rewrite these equations using geometry:
{latex}\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}{latex}
We can now solve for each final speed in turn, finding:
{latex}\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}{latex}
These expressions are already fairly simple, but they can be made even simpler by realizing that since:
{latex}\begin{large}\[ \theta^{five}+\theta^{cue} = 90^{\circ} \]\end{large}{latex}
we know:
{latex}\begin{large}\[ |\sin(\theta^{five})|=|\cos(\theta^{cue})| \] \[|\cos(\theta^{five})|=|\sin(\theta^{cue})|\]\end{large}{latex}
Using this plus the fact that sin{color:black}^2^{color} +cos{color:black}^2^{color} = 1, it is possible to simplify our answers to the form:
{latex}\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}{latex}
{tip}Can you see that this result satisfies the elastic collision requirement (remembering that the masses are equal)? Considering the assignment of the cosine and the sine, we see that as the cue ball's angle grows the speed it retains shrinks. Does this make sense?{tip}
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