Problem Set

NBPhO 2025

5. Throwing 6 pts

Mechanics · Kinematics, Projectile motion, Non-inertial frames

Minimum-speed projectile to intercept an accelerating drone, and graphical determination of speeds in a stone–ball intercept problem.

High-level summary by Claude.

Ingredients relative motioneffective gravity in accelerating frameperpendicularity conditiongraphical figure measurement
Tags mechanicskinematicsprojectile-motionoptimisationrelative-motionnon-inertial-frameminimum-speedperpendicularity-conditiongraphical-measurement

Difficulty medium

Prerequisites

  • Projectile kinematics, r⃗(t)=r⃗0+v⃗0t−12gy^t2\vec{r}(t) = \vec{r}_0 + \vec{v}_0 t - \tfrac{1}{2} g\hat{y} t^2
  • Vector decomposition and dot products
  • Pseudo-forces in a uniformly accelerating reference frame
  • Energy conservation for vertical throws, vmin⁡=2gHv_{\min} = \sqrt{2 g H}
  • Reading distances from a scale bar in a figure

Learning objectives

  • Eliminate gravity by transforming to a co-falling or co-accelerating frame, replacing g⃗\vec{g} with an effective g⃗eff=g⃗−a⃗frame\vec{g}_\text{eff} = \vec{g} - \vec{a}_\text{frame}
  • Recognise that two free-fall projectiles share gravitational acceleration, so their relative motion is uniform and gravity drops out of r⃗b(t)−r⃗s(t)\vec{r}_b(t) - \vec{r}_s(t)
  • Derive the geometric minimum-speed condition u⃗min⁡⊥D⃗\vec{u}_{\min} \perp \vec{D} for intercepting a co-falling target, where D⃗\vec{D} is the line from launcher to target's release point
  • Reduce a moving-target intercept to a stationary-target problem by choosing coordinates aligned with the effective gravity
  • Extract physical lengths from a figure using a scale bar and propagate measurement uncertainty into the final result

Watch out for

  • In part (i), the symmetric setup forces vx=vyv_x = v_y for every solution time tt, not only the optimum. Mistaking this for an extra constraint (rather than a symmetry of the problem) leads to a wrong number of unknowns when minimising over tt.
  • The minimum-speed condition u⃗⊥D⃗\vec{u} \perp \vec{D} in part (ii) refers to the line BSBS — from the boy to the stone's release point — not to the line BCBC from the boy to the collision point. The trajectory chord BCBC is parabolic, not straight; the launch direction is steeper than the chord.
  • τ2=2[b(cx−b)−hcy]/(gh)\tau^2 = 2[b(c_x-b) - h c_y]/(gh) is a small difference of large quantities (here ≈22.8−13.2\approx 22.8 - 13.2). Figure-reading errors of ∼0.2 m\sim 0.2\,\text{m} propagate to ∼5%\sim 5\% in τ\tau and in the speeds u,vu, v. Quoting answers to three significant figures over-states the precision of a graphical reading.
  • The familiar fixed-target minimum-speed result u2=g(L+Δy)u^2 = g(L + \Delta y) does not apply here, because the target moves with the stone. Using it gives a numerically different (and larger) ball speed than the correct moving-target answer.
  • Treating the drone in part (i) as a stationary target at its eventual intercept position misses the entire point of the problem. The drone's acceleration gx^g\hat{x} is what creates the symmetry vx=vyv_x = v_y and the effective 45∘45^\circ launch angle.