Relative Motion in Two Dimensions — Question 9

Question 9

A common gravitational acceleration cancels in relative motion

Projectile A starts at $(0,20)$ m with velocity $(4,6)$ m/s. Projectile B starts at $(12,20)$ m with velocity $(-2,6)$ m/s. Both accelerate at $(0,-10)$ m/s$^2$, with no air resistance, until first floor contact at y=0.

1. Find B’s relative position, velocity and acceleration with respect to A.

2. Find whether they collide before floor contact and the meeting point.

3. If only B’s initial vertical velocity changes to 8 m/s, can they still collide? Explain why shared acceleration alone is not sufficient.

Question 9 - Solution

A common gravitational acceleration cancels in relative motion

(a) Subtract both equal-acceleration histories.

$$\vec r_{B/A}=(12-6t,0),\quad\vec v_{B/A}=(-6,0)\,\mathrm{m/s},\quad\boxed{\vec a_{B/A}=\vec0}$$ $\vec r_{B/A}=(12-6t,0)$ m, $\vec v_{B/A}=(-6,0)$ m/s and $\vec a_{B/A}=\vec0$. Gravity changes both ground velocities identically, so it cancels in their difference.

(b) Solve the relative position and check the physical domain.

$$12-6t=0\ \Rightarrow\ t=2\,\mathrm s,\quad\vec r_{\mathrm{meet}}=(8,20+12-20)=(8,12)\,\mathrm m$$ They meet at t=2 s, when the horizontal difference vanishes. A is then at $(8,20+12-20)=(8,12)$ m, also B’s position. The height is positive, so the meeting occurs before either hits the floor.

(c) Retain the changed relative vertical velocity.

$$\vec r'_{B/A}=(12-6t,2t),\quad\vec r'_{B/A}(2)=(0,4)\,\mathrm m\ne\vec0$$ The modified relative position is $(12-6t,2t)$ m. Horizontal agreement still requires t=2 s, but the vertical separation then is 4 m. Zero vertical separation occurs only at t=0, when horizontal separation is 12 m. Equal accelerations preserve relative velocity; they do not force an encounter.