Relative Motion in Two Dimensions — Question 8

Question 8

Translate the frame and rotate its axes consistently

A particle’s ground velocity is $(5,1)$ m/s. Observer B moves with ground velocity $(1,-2)$ m/s and uses axes turned $45^\circ$ counterclockwise from the ground axes. For a vector $(w_x,w_y)$, the rotated components are $((w_x+w_y)/\sqrt2,(-w_x+w_y)/\sqrt2)$.

1. Find particle velocity relative to B, first in ground-oriented axes.

2. Express that relative velocity in B’s rotated axes and check its magnitude.

3. Explain why rotating the particle velocity and then subtracting the unrotated pair $(1,-2)$ is invalid. How could rotating first be done correctly?

Question 8 - Solution

Translate the frame and rotate its axes consistently

(a) Subtract velocities in one common basis.

$$\boxed{\vec v_{A/B}=(5,1)-(1,-2)=(4,3)\,\mathrm{m/s}}$$ $\vec v_{A/B}=(5,1)-(1,-2)=(4,3)$ m/s in ground-oriented axes. This relative physical vector is defined before choosing how to describe its components.

(b) Project the relative vector onto the rotated basis.

$$\vec v'_{A/B}=\left(\frac{4+3}{\sqrt2},\frac{-4+3}{\sqrt2}\right)=\left(\frac7{\sqrt2},-\frac1{\sqrt2}\right)\,\mathrm{m/s},\quad v'=\sqrt{49/2+1/2}=5\,\mathrm{m/s}$$ The new pair is $(7/\sqrt2,-1/\sqrt2)$ m/s. Its magnitude is $\sqrt{49/2+1/2}=5$ m/s, matching $\sqrt{4^2+3^2}$. The negative second component reflects the new axis orientation, not a different physical relative motion.

(c) Avoid mixing bases in one subtraction.

$$Q(\vec v_A-\vec v_B)=Q\vec v_A-Q\vec v_B=\frac{(6,-4)-(-1,-3)}{\sqrt2}=\frac{(7,-1)}{\sqrt2}\quad\mathrm{m/s}$$ The proposed shortcut subtracts components referenced to different directions. If rotating first, transform both input velocities: A becomes $(6/\sqrt2,-4/\sqrt2)$ and B becomes $(-1/\sqrt2,-3/\sqrt2)$ m/s. Their difference again gives $(7/\sqrt2,-1/\sqrt2)$ m/s because rotation is linear.