How Fast Must a Bullet Travel to Melt on Impact?

[riverjib]

Homework Statement

A forensic investigator notes a 7.2g lead bullet stopped in a door frame melted completely on impact. Assuming the bullet was shot at RT (20C) what would the investigator calculate as the minimum nozzle velocity of the gun to be? (C_lead = 130J/kg*C), (L_lead = 2.5*10^4J/kg) Lead melts at 327C

Homework Equations

Q = mc(delta)T
K = 1/2*m*v^2 (?)

The Attempt at a Solution

I know that the objective of this problem is to use the energy to determine the velocity. I found that Q = 287.4J. I'm not sure what to do from here, however. If I multiply 2.5*10^4J/kg by 0.0072kg, I get 180J. Do I subtract this from 287.4J, and use the result to calculate velocity? Is the difference kinetic energy? I'm not sure if I'm on the right track here. Any help is appreciated!

 

[Shooting Star]

What is the total energy needed to melt the bullet? Remember, it has to be first heated up to the melting point, and the total energy is equal to energy needed to heat it and then melt it. This energy has to come from the kinetic energy.

 

[nlightner]

I would approach this problem by first calculating the amount of heat energy required to melt the lead bullet. Using the equation Q = mc(delta)T, we can calculate that the bullet absorbed 287.4J of heat energy in order to melt completely.

Next, we can use the conservation of energy principle to equate the heat energy absorbed by the bullet to the kinetic energy it had before impact. This can be represented as:

Q = K = 1/2*m*v^2

Solving for v, we get v = √(2Q/m)

Substituting the values we have, we get v = √(2*287.4J/0.0072kg) = 400 m/s

Therefore, the minimum nozzle velocity of the gun would be 400 m/s in order for the bullet to melt completely on impact.

It's important to note that this is a simplified calculation and does not take into account factors such as air resistance or the specific gun used. However, it gives us a rough estimate of the minimum velocity required.