Question

It has been said that sometimes lead bullets melt upon impact. Assume that a bullet receives \(75.0 \%\) of the work done on it by a wall on impact as an increase in internal energy. a) What is the minimum speed with which a 15.0 -g lead bullet would have to hit a surface (assuming that the bullet stops completely and all the kinetic energy is absorbed by it) in order to begin melting? b) What is the minimum impact speed required for the bullet to melt completely?

Short answer

The minimum speed for the bullet to hit a surface in order to begin melting is approximately 324.50 m/s. What is the minimum impact speed for the bullet to melt completely? The minimum impact speed required for the bullet to melt completely is approximately 413.09 m/s.

Step-by-step solution

Define the known variables.

We are given the following information: - Bullet mass (m): 15.0 g = 0.015 kg - Work done on the bullet: 75% (\(0.75\)) - Latent heat of fusion of lead (L): 24.5 kJ/kg (\({2.45 * 10^4} J/kg\)) - Specific heat capacity of lead (c): 128 J/kg°C - Melting point of lead (T): 327°C - Initial temperature of the bullet (T₀): 20°C (room temperature)

Calculate the energy required to begin melting the bullet. (part a)

First, we need to calculate the energy required to raise the temperature of the bullet from room temperature to its melting point. Energy required to raise temperature (Q) is given by the formula: \(Q = mcΔT\) Where: m = mass of the bullet c = specific heat capacity of lead ΔT = change in temperature ΔT = T - T₀ = 327°C - 20°C = 307°C Now, we can calculate the energy required to raise the temperature: \(Q = (0.015 kg)(128\frac{J}{kg°C})(307°C) = 592.32 J\)

Calculate the work done on the bullet.

Since only \(75\%\) of the work is being used as an increase in internal energy, we need to find the total work done on the bullet (W) which is given by the formula: \(W = \frac{Q}{0.75}\) Now, we can calculate the work done on the bullet: \(W = \frac{592.32 J}{0.75} = 789.76 J\)

Calculate the minimum speed using the work-energy theorem. (part a)

According to the work-energy theorem, the total work done on the bullet is equal to its kinetic energy (K). \(K = \frac{1}{2}mv^2\) Where: m = mass of the bullet v = bullet's speed Now, we can solve for the bullet's minimum speed (v): \(789.76 J = \frac{1}{2}(0.015 kg)v^2\) \(v^2 = \frac{789.76 J * 2}{0.015 kg}\) \(v^2 = 105302.4\) \(v = \sqrt{105302.4}\) \(v \approx 324.50 \frac{m}{s}\) So, the minimum speed for the bullet to hit a surface in order to begin melting is approximately 324.50 m/s.

Calculate the energy required to melt the bullet completely. (part b)

Now, we need to calculate the energy required to change the bullet's state from solid to liquid (melting) using the latent heat of fusion formula: \(Q_m = mL\) Where: m = mass of the bullet L = latent heat of fusion of lead Now, we can calculate the energy required for melting: \(Q_m = (0.015 kg)(2.45 * 10^4 \frac{J}{kg}) = 367.5 J\)

Calculate the total energy required for the complete bullet melting. (part b)

To find the total energy, we need to add the energy required to raise its temperature to the melting point calculated previously (Q) with the energy required to melt it completely (Q_m): Total energy (E_total) = Q + Q_m \(E_{total} = 592.32 J + 367.5 J = 959.82 J\)

Calculate the minimum impact speed for the bullet to melt completely. (part _Сb)

Now, we need to find the work done on the bullet to achieve this total energy: \(W = \frac{E_{total}}{0.75}\) \(W = \frac{959.82 J}{0.75} = 1279.76 J\) Using the work-energy theorem and the same process as in step 4, we can find the minimum impact speed required for the bullet to melt completely: \(1279.76 J = \frac{1}{2}(0.015 kg)v^2\) \(v^2 = \frac{1279.76 J * 2}{0.015 kg}\) \(v^2 = 170635.2\) \(v = \sqrt{170635.2}\) \(v \approx 413.09 \frac{m}{s}\) So, the minimum impact speed required for the bullet to melt completely is approximately 413.09 m/s.