A New Look at Momentum — Question 6

Question 6

Zero total momentum does not require equal and opposite velocities

A particle of mass m moves at $+0.80c$. A second particle of mass 2m moves in the opposite direction. Their total momentum is zero. 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 the second particle speed.

2. Check both signed momenta and compare with a classical mass-weighted speed estimate.

Question 6 - Solution

Zero total momentum does not require equal and opposite velocities

(a) Method and calculation

$$p_1=\frac43mc,\quad2m\gamma_2v_2=-\frac43mc\quad\Rightarrow\quad\gamma_2\beta_2=-2/3.$$$$\boxed{\beta_2=-\frac2{\sqrt{13}}\simeq-0.554700}.$$

(b) Interpretation and further calculation

$$\gamma_2=\sqrt{13}/3,\quad\boxed{p_2=-4mc/3}.$$$$m(.80c)+2mv_{2,\mathrm{cl}}=0\quad\Rightarrow\quad v_{2,\mathrm{cl}}=-.40c.$$

Physical check

At unequal speeds the Lorentz factors differ, so balancing mv alone does not balance relativistic momentum.