Bullet is Independent of it?

My insight was that if the mass would be increased twofold, while keeping the material the same, then that is equivalent to two bullets hitting the wall, each of which have the same rise in temperature of course ;).

The question is unclear. Are you asking if the wall heats up or the bullet? Assume instead of a bullet, an immovable wall is struck by a cannon ball, then the temperature change of the wall will be significantly greater than if struck by a .22 round.

The problem is that this is not physical. There will be heating of the wall, not just the bullet. As the bullet gets more massive it will require more friction work from the wall to stop it and the wall will heat up more. This does not scale with the mass. Your analysis requires the assumption that friction against the wall material (compressive and shear will take place) will not heat the wall. A greater fraction of the kinetic energy will turn to wall thermal energy as the mass increases.

So you assumed the wall temprature didnt change? We need friction to stop bullet, when this force wasting all the kinetics, some of the heat transfers to the concrete wall by conduction. Conduction is a form of heat transfer that depend on the surface and in comparison between 2 bullets, the ratio of volumes is different from the ratio of surfaces. While mass is a volumetric parameter, (the stored energy / temperature difference) in a bullet with same density is a volumateric parameter, but the friction loss has two side one side depends on volume and other depends on surface... Temperature will not be the same

velocity depends on time right?then why not the temperature?

Twice the mass with the same velocity = twice the energy at impact. Twice the energy = twice the heat imparted to the wall.

Heat conductivity takes time to transfer.. surely the heat would take longer to transfer throughout the larger bullet therefore meaning the large bullet heats slower than the small bullet?

The surface area is more for larger bullet so it must heat slowly

Fired with the same gun some residual heat from the last shot increases the temperature

I think the issue I had with this question is that it wasn't immediately clear that it's asking about the bullet heating up. If you double the number of the bullets then the KE doubles. But each bullet heats up the same amount even though the increase of heat applied to the wall has doubled.

The larger bullet has a larger area and bigger effect of the friction as an energi drain, the initial kinetic energy of the bullet at the instant before hitting the wall is dispersed in heat, sound the work of the friction over the area of contact between surfices and the work for making spliting the molecules apart while the bullet goes trough. So as a the friction is bigger, the bullet slows down faster, given a fixed radio, and takes less heating time. The temperature is not the same

This is a poor question and the answer is unsatisfactory.

By 'larger bullet', do they mean more mass or more volume? If it means more mass, then:

Let mass of bullet be $m$ and mass of the block be $M$ . If the bullet is fired with a velocity $v$ : By momentum conservation, $mv= (M+m) v' \leadsto v' = \frac{mv}{M+m}$ -- ( ) Now, change in energy would be $\frac{1}{2}(M+m){v'}^{2} - \frac{1}{2} m v^{2}$ Substituting the value of $v'$ from ( ), we get: $\Delta{E} = \frac{{m}^{2}v^{2} - m(M+m)v^{2}}{2(M+m)} \leadsto \Delta{E} = \boxed{\frac{-mMv^{2}}{2(M+m)}}$ This clearly depends on the mass of the bullet!

I agree with James Sifert. The question is ambiguous and non ideal conditions are added in the answer as though the non ideal conditions would render an exact result. The answer could still be that their delta T's would be different.

I also thought this depended on surface area... if it is going the same velocity but the wall contacts more surface are, that means the bullet would contract more friction than the first bullet if it was larger and had the same mass. If it had an equally proportionate mass, then no. The energy transfer into heat is the same... so I ended up answering yes because I overthink everything.

Temperature of the wall after impact is higher with the higher mass bullet. More energy is required to propel a higher mass to the same velocity that a lower mass requires. As this energy converts to heat upon impact with the wall, temperature in the wall increases. Given the same velocity, a larger mass add more energy, hence increasing the local temperature of the wall as this energy is transferred.

What about the air resistance!! I think with bigger mass then more air resistance to the object then less velocity then less temperature, on the other hand if we take the object as combination of many atoms, then the temperature in the

Anyway how to discuss solutions to these sort of weekly problems?can u pls someone tell me

The assumption that the mass at the two sides of the equation is the same is not necessarily correct, the mass displaced in the wall might not be equal to the mass of the bullet.

I was thinking about the friction between the wall and the bullet.

This answer ignores the ambient temp of the wall.

Ah, more heat, but not a higher temperature. Thanks!

the bullet mass is negligible to the\at of the wall and if the assumption that the contact with the wall is ideal its mass is irrelevant the problem was not phrased accurately enough...

Not sure I agree with the answer; in practice a heavier mass bullet to be able run at the same velocity means it would have to have more energy and so more impact/friction at the target. Wording of the question leaves the outcome a bit open.

Point to be noted if we assume the distance between bullet a and the wall as x When the bullet a is fired it will hit wall and penetrate it the reducing the distance. When bullet b is fired it will travel a larger distance. Further points to be looked upon if both the bullets will hit exact same point bullet b will hit the surface of bullet a and not the wall thus as on collision surface temperature will be different

Another consideration is the heat from the first bullet upon raising it's temperature will also raise the temperature of the surrounding wall, as the bullet is not in a vaccum; therefore, transferring some of the heat its surroundings. While that may be similar to the "earth moving when I jump (same Newton law as in your answer), it still has an impact regardless of how small that may be.

Yes but sharpshooter that I am, my second bullet did not hit the wall rather it embedded in the first bullet. Since concrete and lead have a different add specific heat capacity then clearly even if the velocities are the same delta T will be different. I strongly suggest you rethink your answer or be more specific about the trajectory of bullet number two.

I confused energy and temperature since

1/2mv2>1/2Mv2

Ok sorry I'm new the sight.

isnt the second bullet hotter from barrel heat

Not understanding

Im only 11 soo...

I thought that because of the mass of the bullet, the bullet will slightly go slower and therefore will have a slighty decrease in temperature, but you don't mention the distance and you say that the bullet melted beacause of the heat. In that ecuation can be added and the distance?

what are the units of "S"?

This is a very awkwardly-constructed puzzle. It is quite unclear to me, whether the second bullet strikes at another point along the wall (which should be at the same temperature and affected by the same humidity as the area receiving bullet #1), or fired so as to strike the base of the first bullet (i.e., "skewering"). The heat from both bullets will add warmth to the surrounding aggregate the same, only if fired at two separate points along the wall. If however, the second bullet strikes in the immediate vicinity of #1, and before the dispersion of that heat to the surrounding aggregate material has completed, there will be an additive effect from the friction of both strikes in close sequence. The mathematics appear solid, but the parameters of the question are somewhat less than 100% clear (IMHO).

In both cases every thing is constant, only the mass difference which does not contribute any effect as long as the velocity and material are the same in both conditions!

A "larger" lead bullet would have more surface area to rub against the material of the wall causing greater friction and increasing the amount of heat generated.

Omg ppl stop adding parameters. Its the same wall, same temp, same speed, different slug. The mass is greater. The answer is wrong. But thats my opinion. It just more energy to disipate. What is heat? A form of energy! What would cause this heat? Resistance! More mass more resistance. No, I didnt do any math. But do you really need to?

I agree with your equation. But is the heat energy the same as the rise in temperature? As heat energy depends on surface (friction, deformation), rise in temperature on the volume of the bullet, and bullets of the same material with different masses can not have the same volume (nor surface), even if the heat Energy is the same, the rise in temperature should not.

why no more problems like this

The answer is wrong, and is easy to tell why. The question is 'Is the increase in temperature different this time' and the answer is 'Yes'.

You have two bullets - Lead. Lead is relatively soft. It is hitting brick.

How much can the smaller bullet deform? How much can the bigger bullet deform?

That deformation will take up some of the energy.

Here is an experiment: Take a really big pillow. Just a bunch of them. Thrown them on the floor and jump off your couch to belly flop on it. Does it hurt?

Great. Now, take away all those pillows except one - A flat one, very small. Do the same thing.

How much deformation took place in the pillows? Did you feel it absorb more or less kinetic energy?

Do you want to do this experiment? If not, why not?

Now, anyone who knows basic physics knows that Force = Mass x Velocity.

Which will have the greater Force? Which will be deformed more because it has the greater Force?

If a greater mass has a greater velocity, it will have a greater force. If it has a greater mass and a greater surface area, there will be more Frictional Forces applied to it(Adjusted for having higher deformation).

There are so many variables that are different in this equation that the answer will have differences in both how quickly the object is heated, how quickly it dissipates and how and where the heat increase will take place and how the heat increase radiates out.

The temperature increase will be 'different' in many qualitative ways.

I object to the solution, or rather the statement of the problem. What is "it"? The bullet or the wall?

"This is regarding the comments below. Many of you are getting confused that as the mass was doubled the change in temperature gets doubled and so do the friction force and other resistance forces. These forces do exist and will effect the the rising temperature to some small extent. But another main thing to note is that although these forces gets doubled they are acting on twice the mass so the raise in temperature remains the same in both the cases independent of the mass."

That's... Not true. I discussed some of why below, but I want you to consider:

Mass vs surface area. This is not linear: For instance, an object with many tiny spikes will have a very high surface area-to-mass comparison to other objects of like substance. The larger object will strike with greater force, meaning greater deformation. The impact will not be a straight 1-to-1. The drag will not be the same. The deformation will not create an exactly similar shape.

Greater object. Greater deformation. Greater force. Greater area. Too many variables for it to ever work out exactly the same. What you are doing is a univariate analysis on something that, by its nature, has a greater complexity attached to it than your mathematical analysis allows. I invite you to do this study yourself and come back to tell us if the temperature increase was -exactly- the same for two bullets of differing masses. I would like to see the methodology used to achieve that, too.

I thought you are asking about the heating of wall

This question is unclear.

You all making it so confusing, actually both bullets have equal Momentum. Now consider the following statements- --force = change in momentum/time --so both bullet exert same force on wall. --heat is generated due to contact friction. --friction depends on the amount of external force and not on contact surface area.. Thus it can be concluded that both have same value of friction,
So it helps in generating same amount of heat. Need clarification then comment below. Physics is all about fun.....