Calculating how much kinetic energy is transfer?

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 10K views
Jac8897
Messages
25
Reaction score
0

Homework Statement


A neutron in a nuclear reactor makes an elastic head on collision with with the nucleus of a carbon atom initially at rest. (the mass of the carbon nucleus is 12 time the mass of the neutron)

a) what fraction of the neutron's kinetic energy is transferred to the carbon nucleus?



Homework Equations



momentum P(after)=P(before)
M(1a)V(1a)+M(2a)V(2a)=M(1b)V(1b)

Kinetic Energy
M(1a)V(1a)+M(2a)V(2a)=M(1b)V(1b) "I cancel the 1/2 already"


The Attempt at a Solution


I know that I can make the mass any number I want but for simplicity it would be
M(1)=1
M(2)=12

I though I could calculate the speed of the second mass after the collision and then subtracted from the speed of the first mass and calculate the difference but I don't know the initial speed either can some show how to solve this problem? please
 
Physics news on Phys.org
You must pretend that you know the initial velocity. You could just use any number for that and you'll get the same answer regardless of its value, but it will be much more satisfying for you to use a variable. I used u, v and w for the three velocities to save writing. I solved the momentum equation for v, subbed into the energy equation (note that you need to square your velocities!) and got a quadratic in w that I could solve for the non-trivial case and get the ratio of w to u. That is all you need to find the relative kinetic energies you need.
 
OK I did something like this.
c= carbom
n=nucleus

Mn=1kg
Mc=12kg
Vin=U
Vfn=V
Vfc=W

so
MnV+McW=MnU "now solve for V"

V=(MnU-McW)/Mn or V=U-12W "now subbed into the energy equation"

MnV^2+McW^2=MnU^2 "going to make it easier by diving everything by Mn"
V^2+12W^2=U^2
(U-12W)^2+12W^2=U^2

U^2-24WU+144W^2+12W^2=U^2 "I subtract U^2 from both sides "

-24WU+156W^2=0

12W(-2U+13W)=0 "divide by 12W"

-2U+13W=0 "solve for W"

13W=2U >>>>>> W=(2U)/13

now I am stuck here because if I choose an arbitrary number for U let say 3

W=6/13 >>>> W=.4615 but the answer says .284 ??

is something I did wrong or is there more steps I need to do ?