Question

Protons and neutrons (together called nucleons) are held together in the nucleus of an atom by a force called the strong force. At very small separations, the strong force between two nucleons is larger than the repulsive electrical force between two protons—hence its name. But the strong force quickly weakens as the distance between the protons increases. A well-established model for the potential energy of two nucleons interacting via the strong force is

$U=U_0\left[1-e^{-x/x_0}\right]$

where x is the distance between the centers of the two nucleons, x0 is a constant having the value $x_o=2.0\times {10}^{-15}m$, and $U_o=6.0\times {10}^{-11}J.$

Quantum effects are essential for a proper understanding of nucleons, but let us innocently consider two neutrons as if they were small, hard, electrically neutral spheres of mass $1.67\times {10}^{-27}kg$and diameter $1.0\times {10}^{-15}m$. Suppose you hold two neutrons $5.0\times {10}^{-15}m$apart, measured between their centers, then release them. What is the speed of each neutron as they crash together? Keep in mind that both neutrons are moving.

Short answer

The speed of each neutron as they crash together is$1.373\times {10}^{8}m/s$.

Step-by-step solution

Given information 

The potential energy of the two nucleons is$U=U_0\left(1-e^{-x/x_0}\right)$,

where $x_0=2.0\times {10}^{-15}\mathrm{m}$, and $U_0=6.0\times {10}^{-11}J$.

The mass of the neutron$=1.67\times {10}^{-27}$

The diameter of the neutron$=1.0\times {10}^{-15}$

Explanation

When the two neutrons collide, the potential energy will be the sum of the kinetic energy of each neutron. Let v be the speed of each neutron. Where$x_2=1.0\times {10}^{-15}m$and$x_1=5.0\times {10}^{-15}m$.

localid="1648561439852" $$\begin{gathered} U_1-U_2=\frac{1}{2}{m}_n{v}^2+\frac{1}{2}{m}_n{v}^2 \\ {U}_0\left(1-e^{-x_1/x_0}\right)-U_0\left(1-e^{-x_2/x_0}\right)=\frac{1}{2}{m}_n{v}^2+\frac{1}{2}{m}_n{v}^2 \\ {U}_0\left(1-e^{-x_2/x_0}-1+e^{-x_1/x_0}\right)=m_n{v}^2 \\ {v}^2=\frac{U_0\left(e^{-x_2/x_0}-e^{-x_1/x_0}\right)}{m_n} \\ v=\sqrt{\frac{6.0\times {10}^{-11}\mathrm{J}\left(e^{-1.0\times {10}^{-15}\mathrm{m}/2.0\times {10}^{-15}\mathrm{m}}{e}^{-5.0\times {10}^{-15}\mathrm{m}/2.0\times {10}^{-15}\mathrm{m}}\right)}{1.67\times {10}^{-27}\mathrm{kg}}} \\ \nu =\sqrt{\frac{6.0\times {10}^{-11}J\left(e^{-0.5}-e^{-2.5}\right)}{1.67\times {10}^{-27}kg}} \\ v=1.373\times {10}^8\mathrm{m}/\mathrm{s} \end{gathered}$$