Calculating how much kinetic energy is transfer?

In summary: I forgot to square the 12 in the denominator so it should have been KE of neutron = 1/2*m*U^2KE of carbon nucleus = 1/2*12m*[(2U/13)^2]Take the ratio.In summary, the question is asking for the fraction of kinetic energy transferred from a neutron to a carbon nucleus in a head on collision in a nuclear reactor. The solution involves solving for the velocities using momentum and energy equations, and then taking the ratio of their kinetic energies.
  • #1
Jac8897
25
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
 
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  • #2
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.
 
  • #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 ?
 
  • #4
KE of neutron = 1/2*m*U^2
KE of carbon nucleus = 1/2*12m*[2U/13]^2
Take the ratio.
 
  • #5
I got it thanks guys
 

1. How do you calculate kinetic energy?

Kinetic energy is calculated by multiplying the mass of an object by its velocity squared and then dividing by 2. The formula is KE = 1/2 * m * v^2, where KE is kinetic energy, m is mass, and v is velocity.

2. What units are used to measure kinetic energy?

Kinetic energy is typically measured in joules (J) in the International System of Units (SI). However, other units such as foot-pounds (ft-lb) or calories (cal) may also be used depending on the context.

3. How is kinetic energy transferred?

Kinetic energy is transferred when a force is applied to an object, causing it to accelerate. The energy is then transferred from the object's motion to the surrounding environment.

4. Can kinetic energy be converted into other forms of energy?

Yes, kinetic energy can be converted into other forms of energy such as potential energy, heat, or light. For example, when a moving object collides with another object, some of its kinetic energy may be transferred as heat due to friction.

5. Why is it important to calculate kinetic energy transfer?

Calculating kinetic energy transfer is important in understanding the impact and potential dangers of moving objects. It is also crucial in engineering and designing systems to efficiently transfer and utilize kinetic energy, such as in transportation or energy production.

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