What is the temperature change of a bullet upon impact?

[AleksW]

Summary:: What is the temperature change of a bullet upon impact.

I have this problem to solve but I'm kinda stuck, would apricate any feedback.

We fire a silver bullet with a muzzle speed of 200 ms−1 into a sack of sand. What is the temperature change of the bullet, if 40 % of its kinetic energy is lost during the impact? Specific heat of the silver is 230 J(kg K) −1 .

 

[Helios]

Summary:: What is the temperature change of a bullet upon impact.

if 40 % of its kinetic energy

In order for the bullet to come to rest, it must lose all it's kinetic energy.

 

[AleksW]

In order for the bullet to come to rest, it must lose all it's kinetic energy.

Yeah It doesn't seem logical that's why I'm having a hard time with it.

 

[PeroK]

Summary:: What is the temperature change of a bullet upon impact.

I have this problem to solve but I'm kinda stuck, would apricate any feedback.

We fire a silver bullet with a muzzle speed of 200 ms−1 into a sack of sand. What is the temperature change of the bullet, if 40 % of its kinetic energy is lost during the impact? Specific heat of the silver is 230 J(kg K) −1 .

I've asked for this to be moved to the homework section. See you guidelines for homework:

https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/

As above, you need to ensure you copy the problem statement fully. It can't be right as it is. And, you need to make your best effort to solve the problem.

 

[AleksW]

I've asked for this to be moved to the homework section. See you guidelines for homework:

https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/

As above, you need to ensure you copy the problem statement fully. It can't be right as it is. And, you need to make your best effort to solve the problem.

Sorry I am new to this forum,, for the problem that is the full problem. There is no more added information its just that :/

 

[PeroK]

I'd assume that 40% of its KE is lost to the sand.

 

[anorlunda]

Perhaps you mean that the bullet passes through the sack of sand and comes out the other side at a slower speed. Is that the scenario?

No matter how much kinetic energy is lost, the result is to warm up both the sand and the bullet. Without knowing how much the sand warms, the temperature of the bullet can not be calculated.

 

[PeroK]

Perhaps you mean that the bullet passes through the sack of sand and comes out the other side at a slower speed. Is that the scenario?

No matter how much kinetic energy is lost, the result is to warm up both the sand and the bullet. Without knowing how much the sand warms, the temperature of the bullet can not be calculated.

I'd assume 40% goes to the sand and 60% to the bullet. That seems the most likely thing to me.

 

[gmax137]

@AleksW can you break the problem down and at least make a start?
1- determine the initial kinetic energy. You have said the velocity is 200 m/sec. What else do you know (or need to know) to calculate the associated kinetic energy?

2- determine the relationship between energy "lost" and the temperature change. What if ALL of the initial kinetic energy went into raising the bullet temp? By how many degrees would the temperature increase?

 

[AleksW]

@AleksW can you break the problem down and at least make a start?
1- determine the initial kinetic energy. You have said the velocity is 200 m/sec. What else do you know (or need to know) to calculate the associated kinetic energy?

2- determine the relationship between energy "lost" and the temperature change. What if ALL of the initial kinetic energy went into raising the bullet temp? By how many degrees would the temperature increase?

I have taken that the density of silver is 10.5 kg/m^3 and from the formula KE= (rho*V^2)/2 to get 210kJ, But the problem I am facing is that I need to find the Delta T of the bullet but I have no starting Temp.

 

[AleksW]

I have taken that the density of silver is 10.5 kg/m^3 and from the formula KE= (rho*V^2)/2 to get 210kJ, But the problem I am facing is that I need to find the Delta T of the bullet but I have no starting Temp.

I can take the starting temp to be 25 C room temp. But still I am confused as to how to link the KE to the change in Temp

 

[gmax137]

I have taken that the density of silver is 10.5 kg/m^3 and from the formula KE= (rho*V^2)/2 to get 210kJ, But the problem I am facing is that I need to find the Delta T of the bullet but I have no starting Temp.

show us your calculation of the KE along with the units.

 

[vanhees71]

I'd assume 40% goes to the sand and 60% to the bullet. That seems the most likely thing to me.

Based on what?

 

[AleksW]

show us your calculation of the KE along with the units.

I hope the image quality is ok.

 

[PeroK]

Based on what?

Fingerspitzengefühl!

 

[gmax137]

density of silver is 10.5 kg/m^3

This is off by a factor of 1000.

But the density is not what you need. What are the units of $\rho v^2$?

 

[PeroK]

View attachment 278999
I hope the image quality is ok.

Let me get this right. You don't know the mass of the bullet, but you can look up the density of silver, so you'll have to use that? That's not physics!

 

[AleksW]

This is off by a factor of 1000.

But the density is not what you need. What are the units of $\rho v^2$?

210,000 J (jules) = 210 kJ (kiloJules)
density is in kg/m^3
Velocity is in ms^-1

 

[AleksW]

Let me get this right. You don't know the mass of the bullet, but you can look up the density of silver, so you'll have to use that? That's not physics!

I really don't know what to tell you :D, except that my physics prof likes to make life hard.

 

[PeroK]

I really don't know what to tell you :D, except that my physics prof likes to make life hard.

Don't blame the prof! You use $m$ for the mass of the bullet and put your faith in algebra.

Life and university physics are supposed to be hard!

 

[gmax137]

210,000 J (jules) = 210 kJ (kiloJules)
density is in kg/m^3
Velocity is in ms^-1

No!

$\frac {kg} {m^3} \frac {m^2} {{sec}^2} = \frac{kg} {m \text{ } {sec}^2}$

That is not joules.

 

[AleksW]

No!

$\frac {kg} {m^3} \frac {m^2} {{sec}^2} = \frac{kg} {m \text{ } {sec}^2}$

That is not joules.

Yeah your right It's Newton so its going to be 210kN

 

[gmax137]

It's Newton

No it is not. Stop guessing. What is the formula for Kinetic Energy (hint: don't use density)

 

[hilbert2]

I'd assume the "lost" 40% of kinetic energy is what becomes heat and sound waves on impact and the rest 60% becomes the kinetic energy of bullet and sandbag combined after impact. In this kind of elementary problem it is most likely assumed that all lost kinetic energy goes to the heating of the bullet, not to anything external.

 

[AleksW]

No it is not. Stop guessing. What is the formula for Kinetic Energy (hint: don't use density)

I wasn't guessing I just mistook it for kg*m/s^2, and for the equation I know mass should be used but I have no idea of how to get it

 

[PeroK]

I wasn't guessing I just mistook it for kg*m/s^2, and for the equation I know mass should be used but I have no idea of how to get it

You don't need the mass of the bullet. The mass of the bullet is $m$, where $m$ could be anything. Deal with it!

 

[gmax137]

Maybe better to use M for the mass, and keep m only for unit of meters.

 

[vanhees71]

Typography helps: All units have to set upright (roman) and quantities in italics, but I think this is not the real problem here...

 

[gmax137]

@AleksW , write out the formula for kinetic energy, then write out the formula for temperature increase vs energy added. Both formulas will have a "m" for the bullet mass (which has a numerical value which is unknown). Then combine the two formulas to get the temperature rise given the entire KE. You will see, the value of the mass divides out, you don't need it to get the answer.

The "trick" here is to write out the equations using variable names ("m" for mass, "v" for velocity, etc) and then simplify the expression algebraically before plugging in the given values to get the specific answer.

Then the only thing left will be to decipher the "40%" part.