Elastic Collisions (Kinetic energy)

AI Thread Summary
The discussion revolves around a nuclear fusion reaction involving deuterium and tritium nuclei, focusing on the conservation of momentum and the calculation of kinetic energy released. The user successfully demonstrated momentum conservation but struggled with calculating the total kinetic energy after the fusion. It was clarified that the total kinetic energy after the reaction is greater than before, indicating that the difference arises from nuclear energy released during the process. The user was advised to apply the same method used for momentum conservation, incorporating the velocities and masses of all products post-collision. The conversation emphasizes the importance of considering both pre- and post-collision data to accurately determine the kinetic energy released.
Magma828
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Homework Statement



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.

Homework Equations



Ek = 0.5mv2

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.
 
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Welcome to PF!

Hi Magma828! Welcome to PF! :wink:
Magma828 said:
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.

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. :smile:​
 


tiny-tim said:
Hi Magma828! Welcome to PF! :wink:


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. :smile:​

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 :-p

So it's the exact same method but just with the after-collision velocities?
 
It's the exact same method with all the velocities. :wink:
 
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