Question

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 12.0 times the mass of the neutron.)

__________

b) The initial kinetic energy of the neutron is 2.70 10-13 J. Find its final kinetic energy and the kinetic energy of the carbon nucleus after the collision.

neutron __________J

carbon nucleus __________J

Answer 1

For an elastic collision both momentum and energy are conserved.

First define appropriate variables:
m = mass of neutron
M = mass of carbon nucleus = 12*m
v = initial velocity of the neutron
u = final velocity of neutron
w = final velocity of carbon nucleus

Now let us look at what we are trying to find. And that is the ratio of the final KE of the carbon nucleus to the initial kinetic energy of the neutron.

KE0 = initial kinetic energy of the neutron
KE2 = final kinetic energy of the carbon nucleus

Fraction = KE2/KE0
Since kinetic energy is (1/2)m*v^2 we can rewrite this as.
Fraction = ([1/2)12m*w^2]/[(1.2)m*v^2] = 12*(w/v)^2

So we need the ratio of the velocities and we can find that using conservation of energy and conservation of momentum.

Conservation of energy.
(1/2)m*v^2 = (1/2)m*u^2 + (1/2)12m*w^2
v^2 = u^2 + 12*w^2

Conservation of momentum.
m*v = m*u + 12m*w

We want to solve for w as a function of v so get a raleation for u from the momentum equation and put this into the energy equation and solve.
u = v - 12*w
v^2 = (v - 12*w)^2 + 12*w^2
v^2 = v^2 - 24*v*w + 144*w^2 + 12*w^2

24*v*w = 156*w^2
w = (24/156)*v = (2/13)*v
(w/v) = 2/13

So the fraction of the kinetic energy is 12*(2/13)^2 = 0.284