How Does a Bullet Affect the Temperature of Strawberry Jam?

[beatem]

Homework Statement

A bullet of mass m and speed v0 is shot at a block of mass M, that is completely strawberry jam. The bullet enters and comes through other side at vf. Initial temperature of block is T0. The initial temperature of the bullet is much larger than T0, so 90% of the heat in the collision goes into the jam, 10% to the bullet. The specific heat of the jam is C.

The block travels to the right on a frictionless surface and hits a patch of sandpaper of length L, and coefficient of kinetic friction, mu-kinetic. Only half of the heat generated from friction with sandpaper flows to jam. Ignore the heat coming out of jam and ignore any jelly coming out from collision with bullet.

Homework Equations

What is the approximate temperature of the jam after it has passed over the sandpaper?

The Attempt at a Solution

So, I think it's an elastic collision, since they don't stick, so linear momentum and energy are conserved:

m*v0 = m*vf + Mv2 (v2 being an unknown)
(1/2)m*v0^2 = (1/2)m*vf^2 + (1/2)M*v2^2 (+ heat energy?)

The knowns for heat are the temperature of the block, T0, and its specific heat, C.

I'm not sure, but I think I might have missed some energy in the above energy equation due to heat. The problem is I don't know what equation to use. I think it's from heat transfer (Q = mc*delta-T) * .9, but I'm not sure about this. What side would it go on? The right?

I know the friction does work on the block when it is on there, and since only half flows into the block, the energy is (1/2)*mu-kinetic*MgL.

So overall, I think the temperature of the block is affected by two things: the heat transfer from the collision and the work done by friction when it passes over the sandpaper. These both affect its internal energy, delta-U. Do I use delta-U here?

My problem is I'm not sure how this all factors in. Could someone verify if I am right and/or give me a hint in the right direction?

I think what I would do is:

Create a dummy variable for the new temperature from the first heat transfer (the collision), so: Q = MC*(T-dummy - T0)*.90 and then add this equation on the right side of energy and solve. Then, I would do W(friction) = (1/2)mu-kinetic*MgL = MC(Tfinal - T-dummy). (work done by friction being negative, since non-conservative force).

But am I right?

 

[anathema]

Thank you for your interesting problem. You are on the right track with considering both the heat transfer from the collision and the work done by friction on the block. Here are some steps you can take to solve this problem:

1. Start by considering the conservation of linear momentum in the collision between the bullet and the jam block. This will give you an equation for the final velocity of the block, vf, in terms of the initial velocity of the bullet, v0.

2. Next, use the conservation of energy to write an equation for the final kinetic energy of the block, Kf, in terms of the initial kinetic energy of the bullet, K0. Remember to include the heat energy that is transferred to the block during the collision, which is given by Q = 0.9*m*C*(Tdummy - T0), where Tdummy is the unknown temperature of the jam after the collision.

3. Now, consider the work done by friction on the block as it moves over the sandpaper. This work will cause the block to lose kinetic energy and gain thermal energy, which will increase its temperature. Write an equation for the work done by friction, W, in terms of the coefficient of kinetic friction, mu-kinetic, the length of the sandpaper, L, and the change in temperature of the block, Tfinal - Tdummy.

4. Finally, combine the equations from steps 2 and 3 and solve for Tfinal, the final temperature of the jam block after it has passed over the sandpaper.

I hope this helps you in solving the problem. Good luck!