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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).

The rule is generically valid for *elastic* collisions 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} \[ m^{2}v_{1,i}^{2} = m^{2}(v_{1,f}^{2} + 2\vec{v}_{1,f}\cdot \vec{v}_{2,f} + v_{2,f}^{2}) \]\end{large}{latex} \\ Cancelling the masses and comparing to the equation of kinetic energy conservation will immediately yield the result that \\ {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.

Note
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.
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Proof of the Rule

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

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Deck of Cards

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Card

label	Part A

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 {

Solution

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:

id

=

	sysa

} *

System:

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id	sysa

Cue ball and five ball as point particles.

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Interactions:

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

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id	moda

Models:

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plus .

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.

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Approach:

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Diagrammatic Representation

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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
Image Added	Image Added

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	diaga
	diaga

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Mathematical Representation

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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.

Latex

*  {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}
{cloak}

{toggle-cloak:id=appa} *Approach:*  

{cloak:id=appa}
{toggle-cloak:id=diaga} {color:red} *Diagrammatic Representation* {color}

{cloak:id=diaga}
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!|

{cloak:diaga}
{toggle-cloak:id=matha} {color:red} *Mathematical Representation* {color}

{cloak:id=matha}
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&deg; 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 sin² +cos² = 1, it is possible to simplify our answers to the form:

Latex

 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}

{cloak:matha}
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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?

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Card

label	Part B

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.

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id	appb

Approach:

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Part

A

Solution

Part B

Solution

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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? h4. Solution {

Solution

Part B

Solution

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