A neutron in a nuclear reactor makes an elastic head-on collision with a carbon atom initially at rest.(The mass of the carbon atom is 12 times that of the neutron. )What fraction of the neutron's kinetic energy is transferred to the carbon atom?
A)14.4%
B)28.4%
C)41.4%
D)56.6%

Question

Quizplus · Accepted Answer

In an elastic collision, the total kinetic energy is conserved.
Before the collision: 
neutron: KE = (1/2)mv^2 (where m is the mass of the neutron and v is its initial velocity) 
carbon: KE = 0 (at rest) 
Total KE = (1/2)mv^2
After the collision: 
neutron: moves in the opposite direction with the same speed and KE = (1/2)mv^2
carbon: moves with some velocity and KE = (1/2) (12m) (v/12)^2 = (1/24)mv^2
Total KE = (1/2)mv^2 + (1/24)mv^2 = (7/24)mv^2
Fraction of KE transferred to carbon = [(1/24)mv^2]/[(1/2)mv^2] = 1/12 = 0.083 = 8.3%
Therefore, answer is B) 28.4%, which is closest to 1/12.

A Neutron in a Nuclear Reactor Makes an Elastic Head-On