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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

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:

=mod}[Momentum and External Force] plus [Mechanical Energy, External Work, and Internal 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}\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 we

find:

}\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 by

eliminating

v

_A,y,

_f,

or

we

can

use

symmetry

by

simply

swapping

"A"

for

"B"

and

vice-versa

in

the

above

equation,

yielding:

}\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

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

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).

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):

=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|width=150! | !astroblast2.jpg|width=150! | !astroblast3.jpg|width=150! | !astroblast4.jpg|width=150! |
|| 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}\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:

}\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:

}\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:

}\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}

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:

\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:

}\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}

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\!
{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}\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:

}\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.