The Man with the Golden Gun

Gold has a significantly greater density than lead (by about 50%). A golden bullet has therefore more kinetic energy than a lead bullet moving at the same speed.

(The reason why lead is often used is that lead is denser than most metals, yet also affordable.)

Relevant wiki: Momentum - Problem Solving 1D

The impact of a projectile on a target can ideally be considered as an elastic collision of two bodies in which momentum and energy conservation are fulfilled. Of course, in reality, deformation of the bullet and the target material will result in energy being lost, resulting in an inelastic collision, but the argument remains the same.

We assume that a cylindrical bullet with length $l$ and diameter $d$ is fired at a wall of thickness $t$ . If the bullet breaks through the wall, a cylindrical part of the wall with diameter $d$ must be knocked out. If the mass $m_2$ of the wall piece just corresponds to the mass $m_1$ of the bullet, the bullet just barely stuck in the wall ( $v_1' = 0$ ), while the wall piece continues to fly with the initial speed of the bullet ( $v_2' = v_1$ ), since the momentum $p = m_1 v_1 = m_2 v_2'$ has to be conserved. The wall thickness $t$ , which is just penetrated by the bullet, can be determined by the condition $m_1 = m_2$ : $\begin{aligned} m_1 = \frac{1}{4} \pi d^2 l \rho_1 &= m_2 = \frac{1}{4} \pi d^2 t \rho_2 \\ \Rightarrow \quad t &= \frac{\rho_1}{\rho_2} l \end{aligned}$ with the mass densities $\rho_1$ and $\rho_2$ of the bullet and the wall,respectivily. The bullet with the greater density $\rho_1$ can thus penetrate a wall with a greater thickness $t$ . Since gold ( $\rho \approx 19.3 \,\text{g}/\text{cm}^3$ ) is heavier than lead ( $\rho \approx 11 \,\text{g}/\text{cm}^3$ ), the gold bullet also has a higher penetrating power. If gold was not expensive, it would make sense to use gold ammunition. However, another material with similar density to gold is uranium, which is used in armor piercing ammunition. The main reason why uranium is used is its high density $\rho \approx 19.1 \,\text{g}/\text{cm}^3$ . (In addition, in the production of uranium ammunition low-level radioactive waste arising from uranium enrichment can be disposed of inexpensively.)

Since the two bullets are told to have the same size and shape, a golden bullet will be heavier than the lead one. I believe it is safer to assume that the bullets will be loaded with the same amount of energy when they are fired from the gun, than to assume that they somehow have the same initial speed.

Golden bullet will be slower, therefore its impact will be poorer. Not so! Since it is more massive, it would have the same momentum, therefore the same impact. The difference could come from the materials' properties itself, however, which is why I gave the answer as the golden bullet.

Lead is an easily deformable material, and therefore the bullet would lose its shape very quickly throughout the impact. Without its piercing shape, it wouldn't travel as much as the golden bullet that wouldn't deform so much.

1) Gold is much denser than lead.

2) The problem states that the bullets are the same size (or volume), so the gold bullet has more mass (mass = density x volume)

3) The problem states that the bullets leave with the same velocity, and thus the same acceleration

4) The gold bullet will hit with more force (Force = mass x acceleration)

5) The bullet that hits with more force will take longer to slow down, so the gold bullet will have deeper penetration

Because the problem states the bullets won’t deform, you can assume all of the energy spent in this example is kinetic and not lost to heat/deformation

Gold has a larger density than lead, and therefore a gold bullet of the same size has larger mass. This makes it a more effective bullet, but not in a simple way.

The energy of the bullet is is equal to the work done by the expanding gases created by the gunpowder, $W=\int p d V$ . If we treat the gases as an ideal gas, created instantaneously, it does not matter what the mass of the bullet is, the work is the same. For larger mass that means the gold bullet will leave the gun with a smaller velocity. While they start with the same energy, the gold bullet will have more energy when it reaches the target, because it looses less energy to air resistance.

There is another effect: The gunpowder gases cannot be treated as an ideal gas. The burning process is incomplete when the bullet starts to move, and it is ongoing even after the bullet leaves the gun (hence the flash of the gunfire). For a larger mass it takes a longer time to move through the barrel, and therefore there is more time to complete the burning. As a result, the gold bullet will experience higher pressure (at each position, or volume $V$ ) relative to the lead bullet. As a result its energy will be, in fact, larger.

The relative weight of these two advantages depends on the range to the target.

Note: a gunpowder burning too fast may create so much pressure that the barrel explodes. The "good" gunpowder is tuned so that it maintains a high pressure even as the bullet moves and the volume increases.

It's easier to deform lead than gold. Deformation costs energy. The less energy the bullet has the less the impact it's going to make. Since the lead bullet will deform more it will have less impact. Furthermore, the golden bullet will not only lose less kinetic energy at impact, it will have more kinetic energy to begin with.

Gold is denser than lead (MOHS hardness scale).

Actually the gold bullet has greater density as compared to the lead one thus gold one posses more (k.E) as compared to lead thus have a greater impact!