Elastic Collision problem with atoms.

AI Thread Summary
In an elastic collision involving a neutron and a carbon nucleus, the conservation of momentum and kinetic energy equations are applied to determine the final velocities of both particles. The initial velocity of the carbon nucleus is zero, simplifying the equations. The derived formulas for the final velocities are V1F = ((M1 - M2)/(M1 + M2))V1i and V2F = (2M1/(M1 + M2))V1i, where M1 is the mass of the neutron and M2 is the mass of the carbon nucleus. The solution involves rearranging the momentum and energy equations, allowing for the calculation of the fraction of kinetic energy transferred to the carbon nucleus. This approach confirms that the user is on the right track to solving the problem.
hellomister
Messages
29
Reaction score
0

Homework Statement



A neutron in a nuclear reactor makes an elastic head-on collision with the nucleus of a carbon atom initially at rest.

(a) What fraction of the neutron's kinetic energy is transferred to the carbon nucleus? (The mass of the carbon nucleus is about 12 times the mass of the neutron.)




Homework Equations



Conservation of momentum: M1V1i+M2V2i=M1V1f+M2V2F
Conservation of Kinetic Energy: 1/2M1V1i^2+1/2M2V2i^2=1/2M1V1F^2+1/2M2V2F^2



The Attempt at a Solution




Since V2i=0
M1V1i=M1V1F+M2V2F

and

1/2M1V1i^2=1/2M1V1F^2+1/2M2V2F^2 <--- 3 unknowns V1i,V1F,V2F so i need 3 equations to solve for them.

(1)(V1i)=(1)(V1F)+(12)(V2F)
V2F=V1i-V1F/12

umm so i was wondering if I am on the right track to solving this problem?
 
Physics news on Phys.org
You are on the right track. Since V2i = 0 and using the conservation of momentum and kinetic energy equations one gets

V1F = ((M1 - M2)/(M1 + M2))V1i

V2F = (2M1/(M1 + M2))V1i

You arrive at these equations by rearranging the momentum and energy equations by grouping the initial and final velocities of the first mass on one side of the equation and doing the same on the other side for the second mass. Then divide the rearrange energy equation by the rearranged momentum equation using the difference of two perfect squares factoring i.e. a2 - b2 = (a + b)(a - b). This gives

V1i - V2i = V2F - V1F

Using the above equation and the rearranged momentum equation the final velocities can be obtained.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top