Rise in temperature after a collision

[runningninja]

Homework Statement

"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.)"

Homework Equations

$$\text{Coefficient of restitution}\ =~e~= \frac{v_{2F} - v_{1F}}{v_{1I} - v_{2I}}$$
$$p_I = p_F$$
$$\Delta K_2 = Q_2$$

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.
[tex] \frac{1}{3}\ = \frac{v_{2f} - 2v_{2f} + v}{v} [/tex]

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.

 

[voko]

I started off by resolving my momentum equation into
$$mv = -mv_{1F} + 2mv_{2F}$$

Why the minus sign?

 

[runningninja]

Why the minus sign?

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.

 

[voko]

Because it is a one dimensional collision. The first object has to bounce backwards because there is no other place for it to go.

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.

Even if I omit the minus sign, I get 8/81 for the fraction.

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

 

[Nickie]

Your approach is correct, but there is a mistake in your substitution in equation (2). It should be:

$$v_{2F} = \frac{2v}{3} + v$$

This is because in equation (1), the final velocity of the 2m body is equal to its initial velocity plus the final velocity of the m body. Therefore, the correct equation for ΔT is:

$$\Delta T = \frac{2v^2}{9c+6c} = \frac{2v^2}{15c} = \frac{2}{15}\left(\frac{v^2}{c}\right)$$

This gives the correct answer of 2/27.