Question

The fission of uranium nuclei in a nuclear reactor produces high speed neutrons. Before a neutron can trigger additional fissions, it has to be slowed down by collisions with carbon nuclei in the moderator of the reactor. Suppose a neutron with mass of 1.0u travelling at 2.0x107m/s undergoes a head-on elastic collision with a carbon nucleus with mass of 12u initially at rest. Calculate the velocities of the neutron and carbon after the collision.

Answer 1

Let us write the conservation of momentum law:

$m_n v_{n,0} + m_cv_{c,0} = m_n v_{n,1} + m_cv_{c,1}$ where $v_{c,0}=0.$

$m_n v_{n,0} = m_n v_{n,1} + m_cv_{c,1}$

Due to elastic collision, the energy is conserved, so

$\dfrac{m_n v_{n,0}^2}{2} + \dfrac{m_cv_{c,0} ^2}{2}= \dfrac{m_n v_{n,1}^2}{2} + \dfrac{m_cv_{c,1}^2}{2}$ or $m_n v_{n,0}^2 = m_n v_{n,1}^2 + m_cv_{c,1}^2$.

So we have a system of equations

$\begin{cases} m_n v_{n,0} = m_n v_{n,1} + m_cv_{c,1},\\ m_n v_{n,0}^2 = m_n v_{n,1}^2 + m_cv_{c,1}^2 \end{cases},$

$\begin{cases} 1 v_{n,0} = 1 v_{n,1} + 12v_{c,1},\\ 1 v_{n,0}^2 = 1 v_{n,1}^2 + 12v_{c,1}^2 \end{cases}$

We solve these equation by substituting $v_{n,1} =v_{n,0} - 12v_{c,1}$ into the second equation and get $v_{c,1} = \dfrac{2}{13}v_0 = \dfrac{2}{13}\cdot 2.0\cdot10^7\,\mathrm{m/s}= 3.1\cdot10^6\,\mathrm{m/s}$ .

$v_{n,1} = v_{n,0} - 12\cdot v_{c,1} = -1.7\cdot10^7\,\mathrm{m/s}.$