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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.

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

$N$ $\raisebox{1ex}{${m}_{1}$}\!\left/ \!\raisebox{-1ex}{${m}_{2}$}\right.$ $\raisebox{1ex}{${v}_{2}$}\!\left/ \!\raisebox{-1ex}{${v}_{1}$}\right.$ $1$ $1$ $1$ $2$ $3+2\sqrt{2}$ $1+\sqrt{2}$ $3$ $7+4\sqrt{3}$ $2+\sqrt{3}$

$N$ | Number of hits between the input mass and the projectile |
---|---|

${m}_{1}$ | Mass fired in |

${m}_{2}$ | Mass of the projectile |

${v}_{1}$ | Initial velocity of input mass |

${v}_{2}$ | Final velocity of projectile |

A good guess from a quick inspection might be that the velocity ratio $\raisebox{1ex}{${v}_{2}$}\!\left/ \!\raisebox{-1ex}{${v}_{1}$}\right.$ goes up as $\left(N-1\right)+\sqrt{N}$ . However it isn't that easy I’m afraid!

So the problem is get a simple formula for the
${N}^{\mathrm{th}}$
velocity ratio
$\raisebox{1ex}{${v}_{2}$}\!\left/ \!\raisebox{-1ex}{${v}_{1}$}\right.$
.

The
${N}^{\mathrm{th}}$
optimal mass ratio
$\raisebox{1ex}{${m}_{1}$}\!\left/ \!\raisebox{-1ex}{${m}_{2}$}\right.$
is this squared as the energy
$\sum \frac{1}{2}m{v}^{2}$
is conserved.

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. ${v}_{2}\to -{v}_{2}$ first):

${v}_{1}^{\text{'}}=\frac{\left({m}_{1}-{m}_{2}\right){v}_{1}-2{m}_{2}{v}_{2}}{{m}_{1}+{m}_{2}}$

${v}_{2}^{\text{'}}=\frac{2{m}_{1}{v}_{1}+\left({m}_{1}-{m}_{2}\right){v}_{2}}{{m}_{1}+{m}_{2}}$

These are got from the conservation of momentum $\sum mv$ and the conservation of energy $\sum \frac{1}{2}m{v}^{2}$ .

Thus the velocities are changed by the matrix:

$\left[\begin{array}{cc}\frac{{m}_{1}-{m}_{2}}{{m}_{1}+{m}_{2}}& \frac{-2{m}_{2}}{{m}_{1}+{m}_{2}}\\ \frac{2{m}_{1}}{{m}_{1}+{m}_{2}}& \frac{{m}_{1}-{m}_{2}}{{m}_{1}+{m}_{2}}\end{array}\right]$

We want the result of applying the matrix $N$ times to $\left[{v}_{1},0\right]$ to result in $\left[0,{v}_{2}\right]$ i.e. ${v}_{1}^{\text{'}}$ 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:

$\left[\begin{array}{cc}{d}^{-1}& 0\\ 0& d\end{array}\right]\times \left[\begin{array}{cc}c& -s\\ s& c\end{array}\right]\times \left[\begin{array}{cc}d& 0\\ 0& {d}^{-1}\end{array}\right]$

where:

$d=\sqrt[4]{\frac{{m}_{1}}{{m}_{2}}}$ (Yes that’s right, the fourth root)

$c=\frac{{m}_{1}-{m}_{2}}{{m}_{1}+{m}_{2}}$

$s=\frac{-2\sqrt{{m}_{1}{m}_{2}}}{{m}_{1}+{m}_{2}}$

So the matrix can be decomposed as:

${D}^{-1}RD$

Multiplying the matrix $N$ times gives ${D}^{-1}{R}^{N}D$ .

$R$ is a rotation matrix as ${c}^{2}+{s}^{2}$ is $1$ . So $N$ of them are equivalent to rotating by $N$ times the amount. To get the final ${v}_{1}^{\text{'}}$ equal to zero we need the cosine of the rotation to be zero.

The rotation is by:

$\phi ={\mathrm{cos}}^{-1}\left(\frac{{m}_{1}-{m}_{2}}{{m}_{1}+{m}_{2}}\right)$

So we need $\mathrm{cos}\left(N\phi \right)=0$ . To do this we need (higher values of $\phi $ won't occur in the problem):

$\phi =\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$\left(2N\right)$}\right.$

$\left({m}_{1}+{m}_{2}\right)cos\phi ={m}_{1}-{m}_{2}$

${m}_{1}\left(1-cos\phi \right)={m}_{2}\left(1+cos\phi \right)$

$\raisebox{1ex}{${m}_{1}$}\!\left/ \!\raisebox{-1ex}{${m}_{2}$}\right.=\raisebox{1ex}{$\left(1+cos\phi \right)$}\!\left/ \!\raisebox{-1ex}{$\left(1-cos\phi \right)$}\right.$

Multiply top and bottom by $\left(1+cos\phi \right)$ and you’ll see that the square root of that last expression, which has to be the required result $\raisebox{1ex}{${v}_{2}$}\!\left/ \!\raisebox{-1ex}{${v}_{1}$}\right.$ because of energy conservation, is:

$\raisebox{1ex}{${v}_{2}$}\!\left/ \!\raisebox{-1ex}{${v}_{1}$}\right.=\raisebox{1ex}{$\left(1+cos\phi \right)$}\!\left/ \!\raisebox{-1ex}{$sin\phi $}\right.$

where $\phi =\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$\left(2N\right)$}\right.$

The limit as $N\to \mathrm{\infty}$ is $\raisebox{1ex}{$4N$}\!\left/ \!\raisebox{-1ex}{$\pi $}\right.$ .