# Rise in temperature after a collision

1. Mar 24, 2013

### runningninja

1. The problem statement, all variables and given/known data
"A body of mass m, moving with velocity v, collides with a body of mass 2m at rest, in a head-on collision. The coefficient of restitution is 1/3. If the 2m body has a specific heat c, and if it is assumed that the two bodies share the heat generated in the collision equally (not a very reasonable assumption), and that no heat is lost (a ridiculous assumption), how much does the temperature of the 2m body rise? (Keep fractions throughout in solution.)"
2. Relevant equations
$$\text{Coefficient of restitution}\ =~e~= \frac{v_{2F} - v_{1F}}{v_{1I} - v_{2I}}$$
$$p_I = p_F$$
$$\Delta K_2 = Q_2$$
3. The attempt at a solution
Since it is head on, this is a one dimensional problem, with $v_{2I} = 0$ and $v_{1I} = v$. I started off by resolving my momentum equation into
$$mv = -mv_{1F} + 2mv_{2F}$$
(1) $v_{1F} = 2v_{2f} - v$
I then set my restitution equation equal to 1/3 and substituted (1) in.
$$\frac{1}{3}\ = \frac{v_{2f} - 2v_{2f} + v}{v}$$

Which simplifies to

(2) $$v_{2F} = \frac{2v}{3}$$

I then expand my Q equation to

(3) $$\frac{1}{2}\ 2 m (v_{2f})^2 = 2 m c \Delta T$$

I then plug (2) into (3) and solve for $\Delta T$ :

$$\Delta T\ = \frac{2v^2}{9c}$$

However, the answer has a fraction of 2/27 rather than 2/9. I'm off by a factor of a third, and I don't know why.

Last edited: Mar 24, 2013
2. Mar 25, 2013

### voko

Why the minus sign?

3. Mar 25, 2013

### runningninja

Because it is a one dimensional collision. The first object has to bounce backwards because there is no other place for it to go. Even if I omit the minus sign, I get 8/81 for the fraction.

4. Mar 25, 2013

### voko

It does not have to go backward. It can continue forward at a reduced speed. You do not have to guess what is going to happen. Use the equations.

Then you must have made another error somewhere. Show your new derivation.