|Nov16-10, 08:02 PM||#1|
Collision and Heat of Block
1. The problem statement, all variables and given/known data
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.
2. Relevant equations
What is the approximate temperature of the jam after it has passed over the sandpaper?
3. 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?
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