Question
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?
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?
Answers
GPT 3.5
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.
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.