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{excerpt:hidden=true}A series of elastic collisions.{excerpt}

!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).

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|external force] 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:

{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:

{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~ 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:

{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}

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).  

{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:

{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}

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 overabout 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:

{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}