K_(2f)=K_(C)
=(1.20 imes 10^(-13) J)
=1.20 imes 10^(-14) J
The collision is elastic, so the neutron retains the rest of the energy, and we have, for its final kinetic energy,
K_(1f)=K_(n)
=(1.20 imes 10^(-13) J)
= Your response is off by a multiple of ten. 1.20 imes 10^(-13) J.
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 1.20 imes 10^(-13) J. Find its final kinetic energy and the kinetic energy of the carbon nucleus after the collision.
Part 1 of 4 - Conceptualize
Visualize a light projectile such as a rubber ball striking a bowling ball. The rubber ball will bounce back at a less than its original speed, giving perhaps ten percent of its energy to the bowling ball, keeping the other ninety percent.
Part 2 of 4 - Categorize
Momentum and energy are both conserved for the neutron-nucleus system. This is a head-on collision, so we can use the relative velocity equation.
Part 3 of 4 - Analyze
For the relative velocity equation, we have the following.
v_(11)-v_(21)=-(v_(1f)-v_(21))
(a) Let object 1 be the neutron and object 2 be the carbon nucleus of mass m_(2)=12m_(1). Since v_(21)=0, we have v_(2f)=v_(11)+v_(1f). By conservation of momentum for the system of colliding particles, we have the following equation.