Versions Compared

Old Version 32

changes.mady.by.user Andrew E Pawl

Saved on

compared with

New Version Current

changes.mady.by.user Andrew E Pawl

Saved on

Key

  • This line was added.
  • This line was removed.
  • Formatting was changed.
Comment: Migration of unmigrated content due to installation of a new plugin

Excerpt
hiddentrue

A series of elastic collisions.

Composition Setup
 HTML Table
 Wiki Markup
{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}
 Table Row (tr)
valigntop
 Table Cell (td)
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
 Toggle Cloak
idsys
 Systems: 
 Cloak
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. 
 Toggle Cloak
idint
 Interactions: 
 Cloak
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. 
 Toggle Cloak
idmod
 Models: 
 Cloak
idmod
 plus .
 Toggle Cloak
idapp
 Approach: 
 Cloak
idapp
 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}
{
 Toggle Cloak
:
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}
diag
 Diagrammatic Representation
 Cloak
iddiag
With the preliminary math out of the way, we sketch the situation in more detail: 
 Image Added 
 Image Added 
 Image Added 
 Image Added 
 Labels 
 1st Collision (Schematic) 
 2nd Collision (Schematic) 
 3rd Collision (Schematic) 
 Cloak
diag
diag
 Toggle Cloak
idmath
 Mathematical Representation
 Cloak
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?
 Cloak
math
math
 Toggle Cloak
idke
 Finding the Kinetic Energy
 Cloak
idke
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.
 Cloak
ke
ke
 Cloak
app
app
 Table Cell (td)
width250
Image Added