Animated Solution for Physics - Properties of Solids and Liquids: A lead bullet just melts when stopped by an obstacle. Assuming that $25$ per cent of the heat is absorbed by the obstacle, find the velocity of the bullet if its initial temperature is $27^{\circ }\text{C}$. (Melting point of lead $=327^{\circ }\text{C}$, specific heat of lead $=0.03{\text{ cal/g-}}^{\circ }\text{C}$, latent heat of fusion of lead $=6\text{ cal/g}$, $J=4.2\text{ J/cal}$).

Imagine a high-speed lead bullet tearing through the air and slamming into a solid steel obstacle. In a fraction of a second, the bullet goes from traveling at hundreds of meters per second to a complete standstill. But the story doesn't end there. The immense kinetic energy of the bullet cannot simply vanish; the universe demands that energy be conserved. Instead, it transforms violently into thermal energy, generating so much heat that the solid lead bullet literally melts into a puddle of liquid metal.

This problem is a beautiful synthesis of classical mechanics and thermodynamics. It bridges the gap between the macroscopic world of moving objects and the microscopic world of vibrating atoms and phase changes. Let's break down the physics of this fiery collision step by step.

When the bullet is in flight, it possesses kinetic energy. For a bullet of mass $m$ moving with velocity $v$, this energy is given by the familiar equation:

Upon impact, the bullet stops, meaning its final kinetic energy is zero. According to the law of conservation of energy, this lost kinetic energy must be converted into another form. In an inelastic collision like this, the vast majority of that energy becomes heat.

However, there is a crucial detail we must not overlook. The problem states that $25\%$ of the generated heat is absorbed by the obstacle. The obstacle acts as a heat sink, drawing away a quarter of the thermal energy. This means that only the remaining $75\%$ of the heat stays within the bullet.

Therefore, the actual thermal energy available to heat and melt the bullet is:

This is the energy budget we have to work with. If this energy is sufficient to raise the bullet's temperature to its melting point and then completely melt it, the condition given in the problem is satisfied.

Heating Up

Now, let's look at the thermal side of the equation. The bullet starts at an initial room temperature of $27^{\circ }\text{C}$. However, lead does not melt until it reaches a scorching $327^{\circ }\text{C}$.

Before the bullet can even begin to melt, it must first absorb enough heat to raise its temperature by $300^{\circ }\text{C}$. This type of heat, which causes a change in temperature, is called sensible heat. The formula for sensible heat is:

Here, $s$ is the specific heat capacity of lead, which is given as $0.03{\text{ cal/g-}}^{\circ }\text{C}$. Because the specific heat is given in calories per gram, we must be very careful with our units. Let's express the mass of the bullet in grams as $m\times 10^3$ (where $m$ is the mass in kilograms).

Substituting the values, we get:

But our kinetic energy is measured in Joules, the standard SI unit of energy. We cannot equate Joules directly to calories. We must use the mechanical equivalent of heat, $J=4.2\text{ J/cal}$, to convert our sensible heat into Joules:

This is the energy required just to get the bullet to the starting line of the melting process.

Melting the Lead

Once the bullet reaches $327^{\circ }\text{C}$, it is hot enough to melt, but it won't melt instantly. It requires an additional injection of energy to break the rigid crystalline bonds holding the solid lead atoms together. This energy, which changes the state of the material without changing its temperature, is called latent heat.

The formula for the latent heat of fusion is:

The latent heat of fusion for lead is given as $L_f=6\text{ cal/g}$. Again, using our mass in grams and converting to Joules, we calculate the required latent heat:

This is the energy required to completely transform the solid lead at $327^{\circ }\text{C}$ into liquid lead at $327^{\circ }\text{C}$.

We now know the total thermal energy required to achieve the melting of the bullet. It is simply the sum of the sensible heat and the latent heat:

This total required heat must be supplied by the $75\%$ of the kinetic energy that the bullet absorbed upon impact. We can now set up our master conservation of energy equation:

At this point, you might have noticed something mathematically beautiful. The mass of the bullet, $m$, appears on both sides of the equation.

On the left side, a heavier bullet has more kinetic energy. On the right side, a heavier bullet requires more thermal energy to heat up and melt. Because both the available energy and the required energy scale linearly with mass, the mass perfectly cancels out!

This is a profound physical insight. It tells us that the critical velocity required to melt a lead bullet upon impact is a fundamental property of the material itself (its specific heat, latent heat, and melting point), and is completely independent of the size or weight of the bullet. A tiny lead pellet and a massive lead cannonball will both just melt if they hit the wall at this exact same speed.

All that remains is to solve for the velocity $v$. We divide both sides by $0.375$:

Taking the square root of both sides gives us the final velocity:

The bullet must be traveling at approximately $409.8\text{ m/s}$—faster than the speed of sound—for its kinetic energy to generate enough heat to completely melt it upon a sudden stop. This problem elegantly demonstrates how macroscopic motion and microscopic thermal agitation are simply two sides of the same energetic coin.