Comment:
Migration of unmigrated content due to installation of a new plugin
Composition Setup
Excerpt
hidden
true
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
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.
Toggle Cloak
id
proof
Proof of the Rule
Cloak
id
proof
When this rule applies, it is very powerful. The following examples showcase the utility of the right angle rule.
Deck of Cards
id
bigdeck
Wiki Markup
{composition-setup}{composition-setup}
{table:cellpadding=8|cellspacing=0|border=1|frame=void|rules=cols}
{tr:valign=top}
{td:width=325|bgcolor=#F2F2F2}
{live-template:Left Column}
{td}
{td}
{excerpt:hidden=true}This problem presents a valuable shortcut for dealing with a specific kind of elastic collision. {excerpt}
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}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. {note}{info}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}\[(m\vec{v}_{\rm 1,i})\cdot(m\vec{v}_{\rm 1,i})=(m\vec{v}_{\rm 1,f}+m\vec{v}_{\rm2,t})\cdot(m\vec{v}_{\rm 1,f}+m\vec{v}_{\rm2,t})\]\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.{info}
When this rule applies, it is very powerful. The following examples showcase the utility of the _right angle rule_.
{deck:id=bigdeck}
{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
{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}
{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° 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}
Card
label
Part A
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
Toggle Cloak
id
sysa
System:
Cloak
id
sysa
Cue ball and five ball as point particles.
Toggle Cloak
id
inta
Interactions:
Cloak
id
inta
Impulse from external forces is ignored since the collision is assumed to be instantaneous.
Toggle Cloak
id
moda
Models:
Cloak
id
moda
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.
Toggle Cloak
id
appa
Approach:
Cloak
id
appa
Toggle Cloak
id
diaga
Diagrammatic Representation
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
Image Added
Image Added
Cloak
diaga
diaga
Toggle Cloak
id
matha
Mathematical Representation
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° 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}{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}
{cloak:appa}
{card}
{card:label=Part B}
h2. 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?
h4. Solution
*System, Interactions and Models:* As in Part A.
{toggle-cloak:id=appb} *Approach:*
{cloak:id=appb}
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}
{cloak}
{card}
{deck}
{td}
{tr}
{table}
{live-template:RELATE license}
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?
Cloak
matha
matha
Cloak
appa
appa
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?
