A New Look at Momentum — Question 2

Question 2

A constant laboratory force gives linear momentum growth but nonlinear speed growth

A particle starts from rest under a constant positive one-dimensional laboratory force F. Use $\mathbf p=\gamma m\mathbf v$, $\gamma=(1-v^2/c^2)^{-1/2}$, and $\mathbf F=d\mathbf p/dt$. Here m is invariant mass.

1. Find p(t) and v(t).

2. Find the time needed to reach $v=0.80c$ and the acceleration at that instant.

Question 2 - Solution

A constant laboratory force gives linear momentum growth but nonlinear speed growth

(a) Method and calculation

$$\boxed{p(t)=Ft},\qquad\boxed{v(t)=\frac{cFt}{\sqrt{m^2c^2+F^2t^2}}}.$$$$\frac{dv}{dt}=\frac Fm\left[1+\left(\frac{Ft}{mc}\right)^2\right]^{-3/2}=\frac{F}{m\gamma^3}.$$

(b) Interpretation and further calculation

$$\beta=.80:\quad p=\frac43mc\quad\Rightarrow\quad\boxed{t=\frac{4mc}{3F}}.$$$$\boxed{a=\frac Fm\left(\frac35\right)^3=\frac{27F}{125m}}.$$

Physical check

Constant force means constant dp/dt, not constant dv/dt. The acceleration falls as the particle approaches c.