The Strong Force on a Neutron Captured by a Nucleus

[mcdowellmg]

Homework Statement

A nucleus that captures a stray neutron [in a nuclear reactor, for example] 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 1.5 x 10^7 m/s is just barely captured by a nucleus with a diameter of 0.9 x 10^-14 m. Assuming the strong force on the neutron is constant, find the magnitude of that force. (The neutron's mass is 1.67x10^-27 kg.)

[HINT: When the acceleration is constant, the velocity changes from vi to vf during some time interval. The average velocity over that interval is (vi + vf)/2.]

Homework Equations

f=ma
force = mass*acceleration

also, probably one of these constant acceleration problems, since I am looking for the acceleration of the neutron.
v=v0+at
x-x0=v0t+.5at^2
v^2=v0^2+2a(x-x0)
x-x0=.5(v0+v)t
x-x0=vt-.5at^2

In those equations, v = final velocity, v0 = initial velocity, a = acceleration, t = time, and x-x0 = position on x axis.

The Attempt at a Solution

I do not have the time it took the neutron to stop, and I am looking for the acceleration in order to determine the force.

Therefore, I used v^2=v0^2 + 2a(x-x0), which is 0=(1.5 x 10^7)^2 + 2a(0.9 x 10^-14). Solving for a, I got -1.25x10^28, and then I multiplied that by the mass 1.67x10^-27, and got -20.875.

For one, I cannot tell if I did this correctly, and two, I do not know if that should be a negative force. I do not think it is, because the neutron is not traveling left along the x axis. But then again, the neutron is traveling right on the x-axis and thus the force should be going left to bring it to a stop. I only have one shot at turning in this assignment, so I thought I would ask if someone could check over my work.

Thank you very much!

 

[Disconnected]

You did it correctly. The value is negative because it is in the opposite direction as the velocity, which you took to be positive. No need to think about left/right on the x axis.