To solve this problem, we can use the laws of conservation of momentum and conservation of kinetic energy.
1. Conservation of momentum in the x-direction:
Initial momentum = Final momentum
m1*v1 = m2*v2x + m3*v3x
Where m1 = mass of neutron, v1 = velocity of neutron before collision, m2 = mass of lithium nucleus, v2x = velocity of lithium nucleus after collision in x-direction, m3 = mass of neutron after collision, v3x = velocity of neutron after collision in x-direction.
(1.7 * 10^27 kg) * (2.7 km/s) = (1.2 * 10^26 kg) * (0.40 km/s * cos(54°)) + (1.7 * 10^27 kg) * (2.5 km/s * cos(angle))
2. Conservation of kinetic energy:
Initial kinetic energy = Final kinetic energy
(1/2)*m1*v1^2 = (1/2)*m2*v2^2 + (1/2)*m3*v3^2
(1/2)*(1.7 * 10^27 kg)*(2.7 km/s)^2 = (1/2)*(1.2 * 10^26 kg)*(0.40 km/s)^2 + (1/2)*(1.7 * 10^27 kg)*(2.5 km/s)^2
Now solve the two equations simultaneously to find the angle at which the neutron is now traveling.
2. A neutron of mass 1.7 1027 kg, travelling at 2.7 km/s,
hits a stationary lithium nucleus of mass 1.2 1026 kg.
After the collision, the velocity of the lithium nucleus is
0.40 km/s at 54° to the original direction of motion of the
neutron. If the speed of the neutron after the collision is
2.5 km/s, in what direction is the neutron now travelling?
1 answer