A New Look at Momentum — Question 9

Question 9

Deflecting momentum through an angle requires a vector impulse

A particle moves at $0.80c$ before and after a scattering event. Its velocity is initially along +x and finally at $60^\circ$ above +x. 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 x and y components of impulse.

2. Find the impulse magnitude and derive its dependence on a general deflection angle $\chi$.

Question 9 - Solution

Deflecting momentum through an angle requires a vector impulse

(a) Method and calculation

$$p=\gamma mv=\frac43mc,\quad J_x=p(\cos60^\circ-1)=\boxed{-\frac23mc}.$$$$J_y=p\sin60^\circ=\boxed{\frac{2\sqrt3}3mc}.$$

(b) Interpretation and further calculation

$$|\mathbf J|^2=p^2[(\cos\chi-1)^2+\sin^2\chi]=4p^2\sin^2(\chi/2).$$$$\boxed{|\mathbf J|=2p|\sin(\chi/2)|=\frac43mc\quad(\chi=60^\circ)}.$$

Physical check

Unchanged momentum magnitude does not imply unchanged momentum vector. The net impulse is the chord between the initial and final vectors.