A nucleus that captures a stray neutron must bring the neutron to a stop within the diameter of the nucleus by means of the strong force.That force, which “glues” the nucleus together, is approximately zero outside the nucleus. Suppose that a stray neutron with an initial speed of$\mathbf{1}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathbf{m}\mathbf{/}\mathbf{s}$is just barely captured by a nucleus with diameter$\mathbf{d}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{14}}\mathbf{m}$. Assuming the strong force on the neutron is constant, find the magnitude of that force. The neutron’s mass is$\mathbf{1}\mathbf{.}\mathbf{67}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{27}}\mathbf{kg}$.

1)$m=1.67\times {10}^{-27}\mathrm{kg}$

2)$v_i=1.4\times {10}^7\mathrm{m}/\mathrm{s}$

3)$d=1\times {10}^{-14}\mathrm{m}$

Newton’s law states that the product of mass and its acceleration is equal to the net force acting on the object.

Using the kinematic equations, we can find the acceleration of the object, as we know the initial velocity and stopping distance. From Newton’s law of motion, we can find the force acting on it.

Formula

${\mathbf{F}}_{\mathbf{net}}\mathbf{=}\mathbf{m}\mathbf{\times }\mathbf{a}$

Here,${\mathbf{F}}_{\mathbf{net}}$ is the force, is the mass of the object, and is the acceleration of the object.

Use the kinematic equation to find the acceleration.

$$\begin{gathered} V_f^2=v_i^2+2ad \\ 0={\left(1.4\times {10}^7\mathrm{m}/\mathrm{s}\right)}^2+2\mathrm{a}\left(1\times {10}^{-14}\mathrm{m}\right) \\ \mathrm{a}=-\frac{{\left(1.4\times {10}^7\mathrm{m}/\mathrm{s}\right)}^2}{2\left(1\times {10}^{-14}\mathrm{m}\right)} \\ =-9.8\times {10}^{27}\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the acceleration is $-9.8\times {10}^{27}\mathrm{m}/{\mathrm{s}}^2$.

Use the formula from Newtons’ second law to calculate the net force,

$$\begin{gathered} F_{net}=ma \\ =1.67\times {10}^{-27}\mathrm{kg}\times \left(9.8\times {10}^{27}\mathrm{m}/{\mathrm{s}}^2\right) \\ =-16.366\mathrm{N} \end{gathered}$$

Therefore, the magnitude of the net force is 16 N.