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Bouncing Cannon Balls Puzzle

As if we didn’t need one, a new type of weapon is being designed which transfers the energy from a large moving mass to a much smaller and faster projectile.

The large mass is fired at the smaller projectile at rest in a tube which is closed at the far end. The projectile can bounce off the far end and come back to hit the input mass again, or the far end can be opened letting the projectile out. All collisions are elastic, there is no energy loss.

For each number of hits between the input mass and the projectile, there is a size for the large input mass such that all its energy is transferred to the projectile, which is finally fired at a higher speed.

 
v1: v2: Hits:
m1
m2
 

The first few optimal input masses and resultant output velocities are as follows:

N m1/m2 v2/v1
111
23+2√21+√2
37+4√32+√3
NNumber of hits between the input mass and the projectile
m1Mass fired in
m2Mass of the projectile
v1Initial velocity of input mass
v2Final velocity of projectile

A good guess from a quick inspection might be that the velocity ratio v2/v1 goes up as (N−1)+√N. However it isn't that easy I’m afraid!

So the problem is get a simple formula for the Nth velocity ratio v2/v1.
The Nth optimal mass ratio m1/m2 is this squared as the energy ∑½mv2 is conserved.

Solution

Firstly we can consider a bounce of the projectile off the end and then a hit with the input mass as a unit as they always come in pairs, we can treat that the iniial state as having its zero velocity reversed!

To cut through the straightforward dynamics here is how the velocities of the input mass and projectile are changed when the projectle bounces off the end and then collides with the input mass (i.e. v2 → −v2 first):

v1′ = ((m1−m2)v1 − 2m2v2) / (m1+m2)
v2′ = (2m1v1 + (m1−m2)v2) / (m1+m2)

These are got from the conservation of momentum ∑mv and the conservation of energy ∑½mv2.

Thus the velocities are changed by the matrix:

 
m1−m2
m1+m2
−2m2
m1+m2
2m1
m1+m2
m1−m2
m1+m2
 

We want the result of applying the matrix N times to [v1, 0] to result in [0, v2] i.e. v1′ is zero.

In fact this determinant of this matrix is 1 (It had to be because of conservation of energy but it's good to see that it actually is!). Also it can be written as:

 
d-10
0d
 
 
c-s
sc
 
 
d0
0d-1
 

where:

d =
4
m1
m2
    (Yes that’s right, the fourth root)
c =
m1−m2
m1+m2
s =
−2√(m1m2)
m1+m2

So the matrix can be decomposed as:

D-1RD

Multiplying the matrix N times gives D-1RND.

R is a rotation matrix as c2+s2 is 1. So N of them are equivalent to rotating by N times the amount. To get the final v1′ equal to zero we need the cosine of the rotation to be zero.

The rotation is by:

φ = cos-1 (
m1−m2
m1+m2
)

So we need cos(Nφ) = 0. To do this we need (higher values of φ won't occur in the problem):

φ = π / (2N)

(m1+m2) cos φ = m1−m2

m1(1−cos φ) = m2(1+cos φ)

m1 / m2 = (1+cos φ) / (1−cos φ)

Multiply top and bottom by (1+cos φ) and you’ll see that the square root of that last expression, which has to be the required result v2 / v1 because of energy conservation, is:

v2 / v1 = (1+cos φ) / sin φ

where φ = π / (2N)

The limit as N→∞ is 4N/π.