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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 collisionbetween 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.
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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:
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{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}\\
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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}
Cancelling the masses and comparing to the equation of kinetic energy conservation will immediately yield the result that
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{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.
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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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{table:align=right|cellspacing=0|cellpadding=1|border=1|frame=box}{tr}{td:align=center|bgcolor=#F2F2F2}*[Examples from Momentum]* {td}{tr}{tr}{td}
{pagetree:root=Examples from Momentum}
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Deck of Cards
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Part A
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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
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Solution
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System:*
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Cue
ball
and
five
ball
as
[
point
particles
|point particle].
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{
.
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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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{
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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
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Image Added
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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.
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=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}\\
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{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}
}\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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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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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?
