Homework Help: Elastic Collisions (Kinetic energy)

1. Oct 11, 2009

Magma828

1. The problem statement, all variables and given/known data

A nuclear fusion reaction occurs when a deuterium nucleus, mass 2m, and a tritium nucleus, mass 3m, combine (each with velocity v in opposite directions). Most of the energy released in the fusion is carried away in the kinetic energy of the product neutron, mass m, and velocity 5v. The other product is a helium nucleus, mass 4m, and velocity v.

(a) Show in terms of m and v that momentum is conserved in the process.
I've done this part, I don't need help with it. I'm just posting it incase it's a sub-step for the next part. The answer is -mv=-mv

(b) Calculate the kinetic energy released in the fusion in terms of m and v.

I'm stuck with part b. For part a I had to use all the data in the question, I'm guessing that for part b I only have to use the masses and velocities before the collision.

2. Relevant equations

Ek = 0.5mv2

3. The attempt at a solution

This is what I've done so far:
Ek = 0.5mv2

v1 = 1v
m1 = 2m
v2 = 1v
m2 = 3m

Ek1 = 0.5m1v12
Ek1 = 0.5x2m1v2
Ek1 = mv2

Ek2 = 0.5m2v22
Ek1 = 0.5x3m1v2
Ek1 = 1.5mv2

EkT = Ek1+Ek2
EkT = (mv2)+(1.5mv2)

But the answer in the book is EkT = 12mv2...

I think I may have messed up in the algebra, or maybe I need to use the data after the collision too.

2. Oct 11, 2009

tiny-tim

Welcome to PF!

Hi Magma828! Welcome to PF!
No, the total KE after is greater than the total KE before …

the difference in energy must have come from somewhere, and it comes from the "nuclear energy" released.

Try again.

3. Oct 11, 2009

Magma828

Re: Welcome to PF!

Ahh of course. I'd just spent the previous hour doing part a using the principle of conservation of momentum and decided to invent the principle of conservation of kinetic energy :tongue:

So it's the exact same method but just with the after-collision velocities?

4. Oct 11, 2009

tiny-tim

It's the exact same method with all the velocities.