How Fast Must a Bullet Travel to Melt on Impact?

Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving the calculation of the minimum muzzle velocity of a lead bullet that melts upon impact. The context is set in a forensic investigation scenario, focusing on the thermal properties of lead and energy transformations during the impact.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the application of heat transfer equations and kinetic energy formulas to determine the bullet's velocity. There are discussions about the correct values for latent heat of fusion and specific heat, as well as unit conversions between grams and kilograms.

Discussion Status

The conversation has progressed with some participants providing clarifications on unit conversions and the importance of using consistent units in calculations. There is acknowledgment of a mistake regarding the conversion of heat values, leading to a resolution for one participant, though questions about unit consistency remain.

Contextual Notes

There are constraints regarding the specific values used for thermal properties and the necessity to convert units appropriately for the calculations. The original poster's calculations were based on the assumption that all values were correctly applied, which was later questioned by others.

yossup
Messages
28
Reaction score
0

Homework Statement


At a crime scene, the forensic investigator notes that the 7.2- lead bullet that was stopped in a doorframe apparently melted completely on impact. Assuming the bullet was shot at room temperature 30, what does the investigator calculate as the minimum muzzle velocity of the gun?

Homework Equations


Q = mc(change in T)
Q = mL
K = .5mv^2

The Attempt at a Solution


So this is really ticking me off because I was so certain that what I was doing was correct. So the melting point of lead is 327C. Heat of fusion (c) is .025J/g. Specific heat is .130J/g.

Basically I plugged these values into the corresponding equations, and

Q(heat of fusion) + Q(specific heat) = .5mv^2

and tried to solve for V but it won't give me the correct answer :(
 
Physics news on Phys.org


Are you sure your value for latent heat of fusion is correct?
 


yup...it's 25 kJ/kg but since I'm using grams since the mass of the bullet is in grams, i converted it to .025 J/g
 


What about the kilojoules to joules conversion though?
 


wait...so 25 kJ/kg doesn't equal .025 J/g?? O_O
 


haha wow...sorry...i am so dumb...sorry. :(
 


No. It equals 0.025 KJ[/color]/kg.
 


so Q(heat of fusion) = (7.2g) (25J/g) = 180

and Q(specific heat) = (7.2g) (.130J/g) (327-30C) = 277.992

277.992 + 180 = 457.992 = .5mv^2 = .5(7.2)(v^2)

the problem is when i solve for v, it turns out too small and the wrong answer.
 


The mass on the right hand side should be converted to kilograms.
 
  • #10


thanks, i got the right answer. but would you care to explain why i need to use kilgrams for the .5mv^2 equation when i used grams in all my other calculations? thanks.
 
  • #11


Well the constants in your other equations were given in terms of grams for convenience because the bullet weight was only a few grams. You gained your answer in Joules which is the SI unit of energy, and thus you must use SI units in the kinetic energy equation. The SI unit of mass being the kilogram. Just try and keep the units in mind as you're working through any problem. As you can see they are very important.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 21 ·
Replies
21
Views
2K
Replies
1
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 13 ·
Replies
13
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K