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).
Systems: We will consider three separate collisions: the bottom ball with the second ball, the second with the third, and the third with the top. 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 impulse.
Models: [Momentum and Impulse] plus [Mechanical Energy and Non-Conservative Work].
Approach: 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:
\begin
[ m_
v_
+ m_
v_
= m_
v_
+ m_
v_
]
[ \frac
m_
v_
^
+ \frac
m_
v_
^
= \frac
m_
v_
^
+ \frac
m_
v_
^
]\end
By algebraically eliminating vB,y,f we find:
\begin
[ v_
= \frac{2m_{B}}{m_
+m_{B}}v_
+ \frac{m_
-m_{B}}{m_
+m_{B}}v_
]\end
We can also solve for vB,y,f by eliminating vA,y,f, or we can use symmetry by simply swapping "A" for "B" and vice-versa in the above equation, yielding:
\begin
[ v_
= \frac{2m_{A}}{m_
+m_{B}}v_
+ \frac{m_
-m_{A}}{m_
+m_{B}} v_
]\end
With this 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).
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):
\begin
[ v_
= \frac{2m_{1}}{m_
+m_{1}}v + \frac{m_
-m_{1}}{m_
+m_{2}}(-v) = \frac{3m_
-m_{2}}{m_
+m_{2}}v]\end
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
[ v_
= \frac{2m_{2}}{m_
+m_{2}}\left(\frac{3m_
-m_{2}}{m_
+m_{2}} v\right) + \frac{m_
-m_{2}}{m_
+m_{2}}(-v) = \frac{7m_
m_
-m_
^
-m_
m_
-m_
m_{3}}{(m_
+m_
)(m_
+m_
)} \:v]\end
Repeating the process for ball 4 gives:
\begin
[ v_
= \frac{2m_{3}}{m_
+m_{3}}\frac{7m_
m_
-m_
^
-m_
m_
-m_
m_{3}}{(m_
+m_
)(m_
+m_
)} \:v + \frac{m_
-m_{3}}{m_
+m_{3}}(-v) ]
[= \frac{15m_
m_
m_
-m_
^
m_
-m_
m_
^
-m_
m_
^
- m_
m_
m_
-m_
m_
m_
-m_
^
m_
-m_
m_
m_{4}}{(m_
+m_
)(m_
+m_
)(m_
+m_
)}\:\:v]\end
Substituting in the masses of the balls gives: