# Gas station without pumps

## 2012 April 27

### Astro-blaster

Filed under: home school — gasstationwithoutpumps @ 20:49
Tags: , , , , ,

Astro-blaster sold by Arbor Scientific. The bottom three balls have smaller holes, so do not come off the common shaft. The top ball has a large hole and is free to fly off the shaft. Much of the energy of the 120g system is transferred to the 4 top ball, so the little ball goes flying up to the ceiling from even a modest 10" drop

I did have one other toy that we used in today’s class—one I should have had 2 weeks ago when we were doing collisions.  Arbor Scientific calls this toy an Astro-blaster. The idea is simple: you drop the whole unit from a foot or two above the floor, and the little ball goes flying up to the ceiling at high speed.

I had done a crude version of the demo last week using a basketball and a small rubber ball, but neither of them had much bounce to them, and it was hard to keep them vertically aligned, so the demo was not very impressive. This demo unit, with high coefficient-of-restitution balls and the shaft to keep them aligned makes for a much more impressive demo.  I’m not sure why they use 4 balls rather than 2.  Does it make for more efficient energy transfer? Or is it just to keep things aligned more easily?

The balls have diameters 4.8cm, 3.7cm, 2.6cm, and 2.1cm, so should have masses roughly in the ratios 11.9, 5.5, 1.9, 1 (not correcting for the holes drilled in the balls—the total mass is about 120g but the smallest ball is more like 4g than 6g).  We should be able to figure out what happens when the bottom ball hits the floor.  At that point all the balls have the same vertical velocity $-v$ and momentums $-11.9 v, -5.5v, -1.9v, -v$.  The big ball is much less massive than the floor, so (assuming perfect elasticity) rebounds with velocity $v$, immediately colliding with the ball above it.  The momentum of this pair is $(11.9 -5.5)v$ and the energy is $1/2 (11.9+5.5) v^2$.  If we again assume an elastic collision (energy conserved), we can figure out the velocities of both balls after the collision.  Perhaps the easiest way to do this is to change to an inertial reference frame in which the total momentum is 0 (moving upward at $\frac{11.9-5.5}{11.9+5.5}v = 0.368v$).  In this frame the balls are moving toward each other with velocities $0.632v$ and  $-1.368v$, and after the collision will be moving apart with velocities $-0.632v$ and $1.368v$. That means that the second ball will be moving upward at $1.736 v$ with respect to the floor.

Now we have another collision between balls 2 and 3, with the momentum of the pair being $(5.5 * 1.736 - 1.9)v$, so the inertial reference frame in which they have zero momentum moves upward at $\frac{5.5*1.736-1.9}{5.5+1.9}v = 1.034v$.  The balls come together with velocities $0.702v$ and $-2.034v$, so ball 3 ends up moving upward at $(2.034+1.034)v=3.067v$.

The final collision between balls 3 and 4 has combined momentum $(1.9*3.067-1)v$, so the inertial reference frame in which they have zero momentum moves upward at $\frac{1.9*3.067-1}{1.9+1}v=1.665v$.  The balls move together with $1.402v$ and $2.665v$, so the little bounces off with $4.329v$.  That means it leaves with 18.7 times the kinetic energy it came in with, and should bounce to that multiple of the initial height.

What would happen if we used a 19.3 to 1 ratio with just 2 balls?  We’d have an inertial frame moving upward at $\frac{19.3-1}{19.3+1}v=0.901v$ and the small ball would leave at $2.803v$, which is much less impressive than $4.329v$.  So there is good reason to stack many balls.  One could write the equation for the bounce of the final ball, and figure out the optimal ratios of balls to get the maximum bounce with a fixed weight for the final ball and for the total mass.  I tried doing this for 3 balls, with the smallest one being 1 and the total mass being 20.3 (as in this example), and got 15.8, 3.5, 1, which would give $2.275v$ for the middle ball and  $4.094 v$ for the top ball, slightly less than for 4 balls, but I could easily have goofed in doing the algebra, even with the aid of Wolfram Alpha.  I’d have to write a program to do the computation for more balls.  Clearly there is an optimal number of balls to use given the total mass relative to the mass of the smallest ball, since making a string of 20 identical balls would cause an effect like Newton’s cradle, with the top ball only leaving at $v$.  I wonder whether 4 is indeed the optimum for this ratio of masses.

Of course, all the analysis above assumes that the collisions are perfectly elastic, which is nonsense.  The material of the balls provides a high coefficient of restitution, but nowhere near 1 (the closest material I know to that property is “liquidmetal“, which would probably shatter if used in this demo).  I did not measure the coefficient of restitution of the balls (but from my son’s 5th grade science fair project I would estimate it as about 0.86, based on the measurement of similar bouncy balls).  I don’t feel like redoing the analysis with a more realistic coefficient of restitution.  I did determine that dropping the Astro-blaster from 11″ resulted in not hitting the 8′ ceiling, but dropping from 12″ did, so the ratio of heights is between 8 and 8.7, not 18.7.  I wonder whether the Astro-blaster was optimized assuming elastic collisions or the actual coefficient of restitution.