A neutron in a reactor makes an elastic head-on collision with a nitrogen atom that is initially at rest. (The mass of the nitrogen nucleus is about 14 times that of the neutron.)
(a) What fraction of the neutron's kinetic energy is transferred to the nitrogen nucleus?
(b) If the neutron's initial kinetic energy is 2.44 10-13 J, find its final kinetic energy and the kinetic energy of the nitrogen nucleus after the collision.
To determine the fraction of the neutron's kinetic energy transferred to the nitrogen nucleus in an elastic head-on collision, we need to apply the principles of conservation of momentum and conservation of kinetic energy.
(a) Conservation of momentum:
In an elastic collision, the total momentum before the collision is equal to the total momentum after the collision. Since the nitrogen atom is initially at rest, the total momentum before the collision is just the momentum of the neutron, given by:
p_neutron = m_neutron * v_neutron,
where m_neutron is the mass of the neutron and v_neutron is its velocity.
After the collision, the neutron and nitrogen nucleus will move in opposite directions. Let's denote the velocity of the nitrogen nucleus as v_nitrogen, which is the value we want to find.
The total momentum after the collision is:
p_neutron + p_nitrogen = 0,
since the neutron and nitrogen nucleus move in opposite directions.
Combining these two equations, we have:
m_neutron * v_neutron + m_nitrogen * v_nitrogen = 0, (Equation 1)
where m_nitrogen is the mass of the nitrogen nucleus.
(b) Conservation of kinetic energy:
In an elastic collision, the total kinetic energy before the collision is equal to the total kinetic energy after the collision. The kinetic energy of the neutron is given by:
KE_neutron = (1/2) * m_neutron * v_neutron^2.
The total kinetic energy after the collision is:
KE_neutron + KE_nitrogen = (1/2) * m_neutron * v_neutron_f^2 + (1/2) * m_nitrogen * v_nitrogen^2,
where v_neutron_f is the velocity of the neutron after the collision, which is the value we want to find.
Combining these two equations, we have:
(1/2) * m_neutron * v_neutron^2 = (1/2) * m_neutron * v_neutron_f^2 + (1/2) * m_nitrogen * v_nitrogen^2. (Equation 2)
Now, we have two equations, Equations 1 and 2, with two unknowns, v_neutron_f and v_nitrogen. Solving these equations simultaneously will give us the values we need.
To solve these equations manually, you can follow these steps:
1. From Equation 1, isolate v_nitrogen:
v_nitrogen = -(m_neutron / m_nitrogen) * v_neutron.
2. Substitute this expression for v_nitrogen into Equation 2:
(1/2) * m_neutron * v_neutron^2 = (1/2) * m_neutron * v_neutron_f^2 + (1/2) * m_nitrogen * (-(m_neutron / m_nitrogen) * v_neutron)^2.
3. Simplify the equation and solve for v_neutron_f:
v_neutron_f^2 = v_neutron^2 * (1 - (m_neutron^2 / m_nitrogen^2)).
4. Take the square root of both sides to find v_neutron_f.
5. Once you have v_neutron_f, you can determine the final kinetic energy of the neutron using the expression:
KE_neutron_final = (1/2) * m_neutron * v_neutron_f^2.
To find the kinetic energy of the nitrogen nucleus after the collision, you can substitute the value of v_nitrogen obtained in step 1 into the equation for kinetic energy:
KE_nitrogen = (1/2) * m_nitrogen * (-(m_neutron / m_nitrogen) * v_neutron)^2.
Plugging in the known values will give you the final kinetic energy of the neutron (KE_neutron_final) and the kinetic energy of the nitrogen nucleus (KE_nitrogen).