Thread Closed

Calculating how much kinetic energy is transfer?

 
Share Thread Thread Tools
Oct20-09, 02:42 PM   #1
 

Calculating how much kinetic energy is transfer?


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



2. Relevant 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"


3. 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 dont know the initial speed either can some show how to solve this problem? please
 
PhysOrg.com
PhysOrg
science news on PhysOrg.com

>> Hong Kong launches first electric taxis
>> Morocco to harness the wind in energy hunt
>> Galaxy's Ring of Fire
Oct20-09, 04:35 PM   #2
 
Recognitions:
Homework Helper Homework Help
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.
 
Oct21-09, 02:41 AM   #3
 
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 becuase 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 ?????
 
Oct21-09, 03:47 AM   #4
 
Recognitions:
Homework Helper Homework Help

Calculating how much kinetic energy is transfer?


KE of neutron = 1/2*m*U^2
KE of carbon nucleus = 1/2*12m*[2U/13]^2
Take the ratio.
 
Oct21-09, 01:40 PM   #5
 
I got it thanks guys
 
Thread Closed
Thread Tools


Similar Threads for: Calculating how much kinetic energy is transfer?
Thread Forum Replies
Kinetic energy transfer problem Introductory Physics Homework 4
transfer of kinetic energy Introductory Physics Homework 5
Transfer of kinetic energy Classical Physics 1
calculating max. kinetic energy (photoelectric) Introductory Physics Homework 6
Transfer and Conversion of pressure to Energy (Kinetic) General Physics 0