Momentum and Energy

Momentum and Energy

Used together, the laws of conservation of momentum and energy allow us to solve otherwise impossible problems. We will often adopt the technique of setting up a systems of equations to do this.

1. Suppose a neutron in a reactor collides elastically with a carbon-12 nucleus at rest. What fraction the kinetic energy of the neutron is transferred to the carbon nucleus? If the initial kinetic energy of the neutron is $6.27 x 10^{-13}$ J then what is its final energy?

$E_{kinetic_nuetron} = \frac{1}{2}m_{neutron}v_{initial}^2$

$E_{kinetic_nuetron} = \frac{1}{2}m_{neutron}v_{final_neutron}^2$

$E_{kinetic_carbon} = \frac{1}{2}m_{carbon}v_{final_carbon}^2$

We will first invoke conservation of momentum. This yields the following:

$\rho_{initial} = \rho_{final}$

$m_{neutron}v_{initial} = m_{neutron}v_{final_neutron} + m_{carbon}v_{final_carbon}$

$m_{neutron}v_{initial} – m_{neutron}v_{final_neutron} = m_{carbon}v_{final_carbon}$

$v_{initial} – v_{final_neutron} = \frac{m_{carbon}}{m_{neutron}}v_{final_carbon}$

$v_{initial} – v_{final_neutron} = 12v_{final_carbon}$

Next, we employ the conservation of energy to obtain:

$E_{initial} = E_{final}$

$\frac{1}{2}m_{neutron}v_{initial}^2 = \frac{1}{2}m_{neutron}v_{final_neutron}^2 + \frac{1}{2}m_{carbon}v_{final_carbon}^2$

$m_{neutron}v_{initial}^2 – m_{neutron}v_{final_neutron}^2 = m_{carbon}v_{final_carbon}^2$

$v_{initial}^2 – v_{final_neutron}^2 = \frac{m_{carbon}}{m_{neutron}}v_{final_carbon}^2$

$v_{initial}^2 – v_{final_neutron}^2 = 12v_{final_carbon}^2$

At this point, the ratio of the ratios to obtain:

$\frac{v_{initial}^2 – v_{final_neutron}^2}{v_{initial} – v_{final_neutron}} = \frac{12v_{final_carbon}^2}{12v_{final_carbon}}$

$v_{initial} + v_{final_neutron} = v_{final_carbon}$

Combining this result with our result, we have:

$v_{initial} + v_{final_neutron} = v_{final_carbon}$

$v_{initial} – v_{final_neutron} = 12v_{final_carbon}$

$2v_{initial} = 13v_{final_carbon} \rightarrow v_{initial} = \frac{13}{2} v_{final_carbon}$

So the fraction we are looking for can be represented as

$\frac{\frac{1}{2}m_{carbon}v_{final_carbon}^2}{\frac{1}{2}m_{neutron}v_{initial}^2} = 12 \left(\frac{2}{13}\right)^2 = \frac{48}{169} = .2840$

Now, we use this result in conjunction with the law of conservation of energy to calculate the kinetic energy transferred to the carbon nucleus to obtain:

$E_{kinetic_neutron} – \frac{48}{169}E_{kinetic_neutron} = E_{final_kinetic_neutron}$

$6.27 x 10^{-13} – \frac{48}{169}6.27 x 10^{-13} = 4.4892 x 10^{-13} J$