Linear Momentum — Question 8

Question 8

Rotating the momentum axes preserves its length

A particle has momentum $(6,8)\,\mathrm{kg\,m/s}$ in Cartesian axes. New unit axes are $\hat e_1=(3/5,4/5)$ and $\hat e_2=(-4/5,3/5)$.

1. Find the new components and check the magnitude.

2. Does the zero second component mean the particle has no momentum? Explain.

Question 8 - Solution

Rotating the momentum axes preserves its length

(a) Derivation and calculation

Project onto the new orthonormal axes: $$\boxed{p_1=\vec p\cdot\hat e_1=6\frac35+8\frac45=10\ \mathrm{kg\,m/s}},$$ $$\boxed{p_2=\vec p\cdot\hat e_2=6\left(-\frac45\right)+8\frac35=0}.$$

(b) Result and interpretation

The invariant magnitude is $$\boxed{|\vec p|=\sqrt{6^2+8^2}=\sqrt{10^2+0^2}=10\ \mathrm{kg\,m/s}}.$$ The new first axis aligns with momentum. One zero component does not imply a zero vector; all components would have to vanish.

Physical check

The basis vectors have unit magnitude and zero mutual dot product, so the transformation preserves length.