Learn a valuable shortcut for dealing with a specific kind of elastic collision. 


The game of pool is an excellent showcase of interesting physics. One of the most common occurences in a game of pool is an almost perfectly elastic collision between a moving cue ball and a stationary ball of equal mass. These collisions occur in two dimensions.



One very useful rule of thumb for pool players is the right angle rule: after the collision the cue ball's final velocity and the final velocity of the struck ball will make a right angle (assuming that both are moving). 


Note that this rule is only valid in the limit that ball spin can be ignored. If the cue ball has significant spin, then the rule will be violated. 



 Proof of the Rule




The rule is generically valid for [*elastic* collisions|elastic collision] between *equal mass* objects when *one is stationary* before the collision. A short proof is to square the magnitude of each side of the vector version of the equation of momentum conservation:
\\
{latex}\begin{large}\[(m\vec{v}_{\rm 1,i})\cdot(m\vec{v}_{\rm 1,i})=(m\vec{v}_{\rm 1,f}+m\vec{v}_{\rm2,t})\cdot(m\vec{v}_{\rm 1,f}+m\vec{v}_{\rm2,t})\]\end{large}{latex}
{latex}\begin{large} \[ m^{2}v_{1,i}^{2} = m^{2}(v_{1,f}^{2} + 2\vec{v}_{1,f}\cdot \vec{v}_{2,f} + v_{2,f}^{2}) \]\end{large}{latex}
Dividing each side by the mass and by 2 yields the equation:
{latex}\begin{large} \[ \frac{1}{2}mv_{\rm 1,i}^{2} = \frac{1}{2}mv_{\rm 1,f}^{2} + m\vec{v}_{\rm 1,f} \cdot \vec{v}_{\rm 2,f} + \frac{1}{2}mv_{\rm 2,f}^{2}\]\end{large}{latex}
Compare this with the equation for the conservation of kinetic energy:
{latex}\begin{large} \[ \frac{1}{2}mv_{\rm 1,i}^{2} = \frac{1}{2}mv_{\rm 1,f}^{2} + \frac{1}{2}mv_{\rm 2,f}^{2}\]\end{large}{latex}
 
and it's clear that the cross term in the center of the right hand side must be zero (you can subtract the corresponding sides of each equation to get this)
 
{latex}\begin{large}\[ \vec{v}_{1,f}\cdot\vec{v}_{2,f} = 0 \]\end{large}{latex}
which implies that (a.) one of the objects has zero final velocity or else (b.) the objects move at right angles to one another after the collision.



When this rule applies, it is very powerful. The following examples showcase the utility of the right angle rule.








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



 System: 
Cue ball and five ball as point particles.



 Interactions: 
Impulse from external forces is ignored since the collision is assumed to be instantaneous.



 Models: 
Momentum and External Force plus Mechanical Energy, External Work, and Internal Non-Conservative Work.



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. 



 Approach:




 Diagrammatic Representation




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 






  


  









 Mathematical Representation




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. 


Recall that we are assuming external impulses are negligible compared to the internal impulse of the collision. 



\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}
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? 









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.



 Approach:


The same solution method as in Part A will lead to the same relationships:
{latex}\begin{large}\[v^{cue}_{f} = v^{cue}_{i}\cos(\theta^{cue})\]
\[ v^{five}_{f} = v^{cue}_{i}\sin(\theta^{cue})\] \end{large}{latex}
This time, however, we solve for the angles:
{latex}\begin{large}\[ \theta^{cue} = \cos^{-1}\left(\frac{v^{cue}_{f}}{v^{cue}_{i}}\right) = 41^{\circ} \]
\[ \theta^{five} = 90^{\circ} - \theta^{cue} = 49^{\circ}\]\end{large}{latex}