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Excerpt
hiddentrue

A series of elastic collisions.

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{table:align=right|cellspacing=0|cellpadding=1|border=1|frame=box|width=30%} {tr} {td:align=center|bgcolor=#F2F2F2}*[Examples from Momentum]* {td} {tr} {tr} {td} {pagetree:root=Examples from Momentum} {search-box} {td} {tr} {table}{excerpt:hidden=true}A series of elastic collisions.{excerpt} {composition-setup}{composition-setup} !astroblaster.JPG! Shown above is an Astro-Blaster toy. This toy consists of a set of 4 rubber balls that can be stacked on a plastic rod. The balls are stacked in order of size with the largest at the bottom. The masses of the balls decrease with their size, so that for the example shown the balls have masses of about 68 g, 28 g, 10 g and 4 g respectively. When the stack is dropped to the ground, the balls undergo a series of collisions which causes the top ball (the small red ball) to launch upward to a height considerably larger than the original drop height. Assuming that all the collisions are elastic and that the assembly hits the ground moving at a speed _v_, find the speed of the red ball as it launches up from the top in terms of _v_ and find the fraction of the initial kinetic energy deposited in each ball (ignoring subsequent collisions and friction due to the rod). h4. Solution h6. {toggle-cloak:id=sys} {color:maroon}*Systems:*{color} {cloak:id=sys} We will consider three separate collisions: the bottom ball with the second ball, the second with the third, and the third with the top. We will assume that all the balls are slightly separated in flight, so that each collision becomes a separate interaction. The largest ball (Ball 1) strikes the ground first, and undergoes a perfectly elastic collision with the ground, so that the magnitude of its velocity is unchanged, but it completely reverses direction, as shown in Schematic for the First Collision  under "Diagrammatic Representation" below. \\ {cloak} {toggle-cloak:id=int} {color:maroon}*Interactions:*{color} {cloak:id=int}For each collision, we will consider the system to be made up of the two balls that are colliding. We will assume that each collision is instantaneous, so that [external forces|external force] will provide negligible [impulse]. We are told to assume that the collisions are elastic, so we will assume there is no [non-conservative work|work#nonconservative] done on any of the systems. \\ {cloak} {toggle-cloak:id=mod} {color:maroon}*Models:*{color} {cloak:id=mod}[Momentum and External Force] plus [Mechanical Energy and Non-Conservative Work]. {cloak} {toggle-cloak:id=app} *Approach:* \\ {cloak:id=app} {color:red}{*}Preliminaries for a Complicated Problem{*}{color} With three collisions to evaluate, it will save time to derive general formulas for the results of a one-dimensional elastic collision. Taking upward to be the positive _y_ direction, we can write the equations of momentum conservation and energy conservation (the collision is assumed elastic) for a two-object collision: {latex}
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Shown at right is an Astro-Blaster toy. This toy consists of a set of 4 rubber balls that can be stacked on a plastic rod. The balls are stacked in order of size with the largest at the bottom. The masses of the balls decrease with their size, so that for the example shown the balls have masses of about 68 g, 28 g, 10 g and 4 g respectively. When the stack is dropped to the ground, the balls undergo a series of collisions which causes the top ball (the small red ball) to launch upward to a height considerably larger than the original drop height. Assuming that all the collisions are elastic and that the assembly hits the ground moving at a speed v, find the speed of the red ball as it launches up from the top in terms of v and find the fraction of the initial kinetic energy deposited in each ball (ignoring subsequent collisions and friction due to the rod).

Solution

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Systems:
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idsys

We will consider three separate collisions: the bottom ball with the second ball, the second with the third, and the third with the top. We will assume that all the balls are slightly separated in flight, so that each collision becomes a separate interaction. The largest ball (Ball 1) strikes the ground first, and undergoes a perfectly elastic collision with the ground, so that the magnitude of its velocity is unchanged, but it completely reverses direction, as shown in Schematic for the First Collision  under "Diagrammatic Representation" below.

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idint
Interactions:
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idint

For each collision, we will consider the system to be made up of the two balls that are colliding. We will assume that each collision is instantaneous, so that external forces will provide negligible . We are told to assume that the collisions are elastic, so we will assume there is no non-conservative work done on any of the systems.

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Models:
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plus .

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Approach:

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Preliminaries for a Complicated Problem
With three collisions to evaluate, it will save time to derive general formulas for the results of a one-dimensional elastic collision.

Taking upward to be the positive y direction, we can write the equations of momentum conservation and energy conservation (the collision is assumed elastic) for a two-object collision:

Latex
\begin{large}\[ m_{A}v_{A,y,i} + m_{B}v_{B,y,i} = m_{A}v_{A,y,f} + m_{B}v_{B,y,f}\]
\[ \frac{1}{2}m_{A}v_{A,y,i}^{2} + \frac{1}{2}m_{B}v_{B,y,i}^{2} = \frac{1}{2}m_{A}v_{A,y,f}^{2} + \frac{1}{2}m_{B}v_{B,y,f}^{2} \]\end{large
}{latex
}

By

algebraically

eliminating

_

v

{_}{~}

B,y,

f~

f we

find:

{

Latex
}
\begin{large}\[ v_{A,y,f} = \frac{2m_{B}}{m_{A}+m_{B}}v_{B,y,i} + \frac{m_{A}-m_{B}}{m_{A}+m_{B}}v_{A,y,i}\]\end{large}
{latex}

We

can

also

solve

for

_

v

{_}{~}

B,y,

f~

f by

eliminating

_

v

{_}{~}

A,y,

f~

f,

or

we

can

use

symmetry

by

simply

swapping

"A"

for

"B"

and

vice-versa

in

the

above

equation,

yielding:

{

Latex
}
\begin{large}\[ v_{B,y,f} = \frac{2m_{A}}{m_{A}+m_{B}}v_{A,y,i} + \frac{m_{B}-m_{A}}{m_{A}+m_{B}} v_{B,y,i}\]\end{large}
{latex} {

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:

id

=diag} {color:red}{*}Diagrammatic Representation{*}{color}\\ {cloak:id=diag} With the preliminary math out of the way, we sketch the situation in more detail: \\ | !astroblast1.jpg! | !astroblast2.jpg! | !astroblast3.jpg! | !astroblast4.jpg! | || Labels || 1st Collision (Schematic) || 2nd Collision (Schematic) || 3rd Collision (Schematic) || {cloak:diag} {toggle-cloak:id=math} {color:red}{*}Mathematical Representation{*}{color}\\ {cloak:id=math} The task of finding the final velocity of ball 4 is accomplished by repeated application of the formula derived above. When ball 1 bounces off the ground, it rebounds with the same speed _v_ that it had on impact (though now directed upward). \\ {note}If this is not obvious, simply apply our collision formula to the collision between ball 1 and the earth, treating the earth as a ball of infinite mass. {note} Using that fact, we can find the speed of ball 2 after its collision with ball 1, remembering that ball 2 is still moving downard with speed _v_ (it hasn't collided with anything yet): {latex}

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Diagrammatic Representation

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With the preliminary math out of the way, we sketch the situation in more detail:

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Labels

1st Collision (Schematic)

2nd Collision (Schematic)

3rd Collision (Schematic)

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idmath
Mathematical Representation

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idmath

The task of finding the final velocity of ball 4 is accomplished by repeated application of the formula derived above. When ball 1 bounces off the ground, it rebounds with the same speed v that it had on impact (though now directed upward).

Note

If this is not obvious, simply apply our collision formula to the collision between ball 1 and the earth, treating the earth as a ball of infinite mass.

Using that fact, we can find the speed of ball 2 after its collision with ball 1, remembering that ball 2 is still moving downard with speed v (it hasn't collided with anything yet):

Latex
\begin{large}\[ v_{2,1} = \frac{2m_{1}}{m_{2}+m_{1}}v + \frac{m_{2}-m_{1}}{m_{1}+m_{2}}(-v) = \frac{3m_{1}-m_{2}}{m_{1}+m_{2}}v\]\end{large}
{latex}

This

becomes

the

initial

velocity

for

ball

2

in

its

subsequent

collision

with

ball

3

(which

is

still

moving

downward

with

speed

_

v

_

)

giving

a

speed

for

ball

3

after

that

collision

of:

{

Latex
}
\begin{large}\[ v_{3,2} = \frac{2m_{2}}{m_{3}+m_{2}}\left(\frac{3m_{1}-m_{2}}{m_{1}+m_{2}} v\right) + \frac{m_{3}-m_{2}}{m_{3}+m_{2}}(-v) = \frac{7m_{2}m_{1}-m_{2}^{2}-m_{1}m_{3}-m_{2}m_{3}}{(m_{1}+m_{2})(m_{2}+m_{3})} \:v\]\end{large}
{latex}

Repeating

the

process

for

ball

4

gives:

{

Latex
}
\begin{large}\[ v_{4,3} = \frac{2m_{3}}{m_{4}+m_{3}}\frac{7m_{2}m_{1}-m_{2}^{2}-m_{1}m_{3}-m_{2}m_{3}}{(m_{1}+m_{2})(m_{2}+m_{3})} \:v + \frac{m_{4}-m_{3}}{m_{4}+m_{3}}(-v) \]
\[= \frac{15m_{1}m_{2}m_{3} -m_{2}^{2}m_{3}-m_{1}m_{3}^{2}-m_{2}m_{3}^{2} - m_{1}m_{3}m_{4}-m_{1}m_{2}m_{4}-m_{2}^{2}m_{4}-m_{2}m_{3}m_{4}}{(m_{1}+m_{2})(m_{2}+m_{3})(m_{3}+m_{4})}\:\:v\]\end{large}
{latex}

Substituting

in

the

masses

of

the

balls

gives:

{

Latex
}
\begin{large}\[ v_{4,3} = 5.0 v\]\end{large}
{latex} {tip}The manufacturer claims that the ball will rise to over 5 times its initial height. What does this claim imply about the accuracy of our assumptions? {tip} {cloak:math} {toggle-cloak:id=ke} {color:red}{*}Finding the Kinetic Energy{*}{color}\\ {cloak:id=ke} We now find the fraction of the initial kinetic energy of the assembly that is carried by each ball after its collision with the one above. The initial kinetic energy is: \\ {latex}
Tip

The manufacturer claims that the ball will rise to over 5 times its initial height. What does this claim imply about the accuracy of our assumptions?

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math

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idke
Finding the Kinetic Energy

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We now find the fraction of the initial kinetic energy of the assembly that is carried by each ball after its collision with the one above. The initial kinetic energy is:

Latex
\begin{large}\[ K_{i} = \frac{1}{2}(m_{1}+m_{2}+m_{3}+m_{4})v^{2} \]\end{large}
{latex}

Since

we

already

have

the

final

velocity

of

ball

4,

we

can

calculate

the

fraction

it

has

after

it

has

been

launched:

{

Latex
}
\begin{large}\[ \frac{K_{f,4}}{K_{i}} = \frac{\frac{1}{2}m_{4}(5v)^{2}}{\frac{1}{2}(m_{1}+m_{2}+m_{3}+m_{4})v^{2}} = 0.91\]\end{large}
{latex} {note}The red ball carries away about 90% of the initial kinetic energy of the system\! {note} To find the final velocity of the third ball after its collision with the fourth, we use our result for the velocity after its collision with ball 2, and the fact that the fourth ball was moving downward with speed _v_ prior to the collision: {latex}
Note

The red ball carries away about 90% of the initial kinetic energy of the system!

To find the final velocity of the third ball after its collision with the fourth, we use our result for the velocity after its collision with ball 2, and the fact that the fourth ball was moving downward with speed v prior to the collision:

Latex
\begin{large}\[ v_{3,4} = \frac{2m_{4}}{m_{4}+m_{3}}(-v) + \frac{m_{3}-m_{4}}{m_{4}+m_{3}}\left(\frac{7m_{2}m_{1}-m_{2}^{2}-m_{1}m_{3}-m_{2}m_{3}}{(m_{1}+m_{2})(m_{2}+m_{3})}\right) \:v
= 0.79v\]\end{large}
{latex}

which

gives

a

kinetic

energy

ratio

of:

{

Latex
}
\begin{large}\[ \frac{K_{f,3}}{K_{i}} = \frac{\frac{1}{2}m_{3}(0.79v)^
{2}}{\frac{1}{2}(m_{1}+m_{2}+m_{3}+m_{4})v^
{2}}
= 0.057 \]\end{large}{latex} Similar calculations will show that the 2nd ball and 1st ball end up with 3% and about 2%, respectively. {note}Compare these percentages to the percent of the kinetic energy carried by each ball before contacting the ground. {note} {cloak:ke} {cloak:app} | !copyright and waiver^copyrightnotice.png! | RELATE wiki by David E. Pritchard is [licensed|copyright and waiver] under a [Creative Commons Attribution-Noncommercial-Share Alike 3.0 United States License|http://creativecommons.org/licenses/by-nc-sa/3.0/us/]. |
{\frac{1}{2}(m_{1}+m_{2}+m_{3}+m_{4})v^{2}} = 0.057 \]\end{large}

Similar calculations will show that the 2nd ball and 1st ball end up with 3% and about 2%, respectively.

Note

Compare these percentages to the percent of the kinetic energy carried by each ball before contacting the ground.

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