A New Look at Momentum — Question 7

Question 7

The center-of-momentum frame is not obtained by averaging velocities

Two equal-mass particles move at $u_1=0.60c$ and $u_2=0$ in a laboratory. Seek a frame moving at speed V in which their total momentum vanishes. Use $p_i\prime=\gamma_V\gamma_i m(u_i-V)$.

1. Derive V from the zero-total-momentum condition.

2. Transform both velocities and verify that they become equal and opposite.

Question 7 - Solution

The center-of-momentum frame is not obtained by averaging velocities

(a) Method and calculation

$$0=\sum_i\gamma_i m(u_i-V)\quad\Rightarrow\quad\boxed{V=\frac{\gamma_1u_1+\gamma_2u_2}{\gamma_1+\gamma_2}}.$$$$\boxed{V/c=\frac{1.25(.60)}{1.25+1}=1/3}.$$

(b) Interpretation and further calculation

$$\boxed{u_1\prime/c=\frac{.60-1/3}{1-.60/3}=1/3},\qquad\boxed{u_2\prime/c=-1/3}.$$$$\gamma_1\prime=\gamma_2\prime\quad\Rightarrow\quad p_1\prime+p_2\prime=0.$$

Physical check

The simple velocity average is 0.30c. Momentum weights each particle by its Lorentz factor as well as its mass.