NBPhO 2025

Part i (2 points)

In the drone’s reference frame, the effective gravitational field is $(g,\, g)$, which has a magnitude of $g\sqrt{2}$ pointing at a 45° angle. The initial kinetic energy of the ball is $\frac{1}{2}mv_0^2$, and the potential energy change is $m(g\sqrt{2})\cdot(h\sqrt{2})$, where $h\sqrt{2}$ is the displacement along the direction of the effective field. From the energy conservation law we obtain $\frac{1}{2}mv_0^2 = m(g\sqrt{2})\cdot(h\sqrt{2})$, hence

$$v_0 = 2\sqrt{gh}.$$

Solution 2 by Anne-Sofie Mårtensson and Adam Warnerbring: In time $t$ the drone travels a distance $s = \frac{1}{2}gt^2$. For a collision to occur at time $t$, the ball must satisfy $h = vt\sin\alpha - \frac{1}{2}gt^2$ and $h + s = vt\cos\alpha$. From these equations we get $vt\sin\alpha = vt\cos\alpha$, which means $\alpha = 45°$. The initial $x$ and $y$-components of the velocity must be the same, so the initial speed is minimised when the highest point of the trajectory is at height $h$ from the throwing point. From energy conservation, $\frac{1}{2}\left(\frac{v_0}{\sqrt{2}}\right)^2 = gh$, giving $v_0 = 2\sqrt{gh}$.

Alternatively, from $\alpha = 45°$ the condition for the ball and drone to meet becomes $\frac{1}{2}gt^2 - \frac{vt}{\sqrt{2}} + h = 0$. This quadratic in $t$ has a solution when its discriminant $\Delta \geq 0$:

$$\Delta = \frac{v^2}{2} - 2gh \geq 0.$$

The minimal speed is when $\Delta = 0$, giving $v = 2\sqrt{gh}$.

Some students interpreted the task such that the drone has an acceleration $(g,\,-g)$. This interpretation makes the problem more difficult and will be accepted. If we move to the coaccelerating frame of the drone, the ball will have a net acceleration $(-g,\,0)$, and the drone will stay in place. Rotating the coordinate axes so the acceleration of the ball is $(0,\,-g)$ and setting the origin at the throwing point, the drone will be at $(-h,\, h)$. All points reachable by the ball with initial speed $v$ are given by the envelope curve $y \leq \frac{v_0^2}{2g} - \frac{gx^2}{2v_0^2}$. With the minimal possible speed the drone will be on the envelope curve, so we need $h = v_0^2/(2g) - gh^2/(2v_0^2)$. Rearranging gives $v_0^4 - 2ghv_0^2 - g^2h^2 = 0$, which has solutions

$$v_0^2 = \frac{2gh \pm \sqrt{4g^2h^2 + 4g^2h^2}}{2} = gh \pm gh\sqrt{2}.$$

This is only positive if $\pm$ is $+$, so $v_0^2 = (1+\sqrt{2})gh$ and thus $v_0 = \sqrt{(1+\sqrt{2})gh}$.

Grading (preliminary)

• Idea of switching to the coaccelerating frame: 1 pts
• Ball gains horizontal acceleration $g$: 0.3 pts
• Effective gravitational field $g\sqrt{2}$ pointing from drone to throwing point: 0.2 pts
• Energy conservation or the respective kinematical equation: 0.3 pts
• Correct answer: 0.2 pts

Grading for Solution 2:

• Correct kinematic equations: 0.6 pts (0.2 for each)
• Getting the condition $\sin\alpha = \cos\alpha$: 0.3 pts
• $\alpha = 45°$: 0.1 pts
• $v_{x,0} = v_{y,0}$ implies that the highest point of the trajectory has to be as low as possible: 0.5 pts
• Energy conservation or the respective kinematical equation: 0.3 pts
• Correct answer: 0.2 pts

Grading for alternative interpretation:

• Idea of switching to the coaccelerating frame: 1 pts
• Ball gains vertical and horizontal acceleration $g$: 0.3 pts
• Effective gravitational field $(0,\,-g)$ (in the rotated coordinate frame): 0.2 pts
• Drone must be on the envelope curve: 0.2 pts
• Correct form for the envelope curve: 0.1 pts
• Correct answer: 0.2 pts

NB! If a student has interpreted that the drone’s acceleration is along the $y$-axis or antiparallel to the positive $y$-axis, it is automatically 0 points.

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Part ii (4 points)

Solution 1. Let us use the free-falling frame of reference wherein the ball and the stone move along straight lines. That frame fell together with points $S$ and $B$ during the flight time of duration $t$ by $h = \frac{1}{2}gt^2$, so in that frame the position of the collision point $C'$ is obtained by shifting $C$ relative to $S$ and $B$ by distance $h$ upwards. In that frame, $|SC'| = vt$ and $|BC'| = ut$, hence $|SC'|/|BC'| = v/u$.

We have $v$ fixed and want $u$ as small as possible, so $|SC'|/|BC'|$ needs to be maximal. From the sine theorem, $|SC'|/|BC'| = \sin\angle SBC'/\sin\angle C'SB$. As $\angle C'SB$ is defined by the stone-throwing angle and is therefore fixed, the boy needs to maximize $\sin\angle C'BS$. The maximum of 1 is reached for $\angle C'BS = 90°$. Therefore, we draw a perpendicular to $SB$ at $B$, and find $C'$ as its intersection with the vertical line drawn through $C$. Then $|CC'| = \frac{1}{2}gt^2$ from where we obtain

$$t = \sqrt{\frac{2|CC'|}{g}}, \quad v = \frac{\sqrt{g\,|SC'|}}{\sqrt{2|CC'|}} \approx 12.3\,\text{m\,s}^{-1}, \quad u = \frac{\sqrt{g\,|BC'|}}{\sqrt{2|CC'|}} \approx 10.9\,\text{m\,s}^{-1}.$$

Solution 2. After going to the free fall frame, go to the frame moving with velocity $\vec{v}$. For the ball to hit the stone, its velocity in this frame $\vec{u}'$ must point at the stone: $\vec{u}' = k\overrightarrow{BS}$ where $k > 0$. On the other hand $\vec{u}' = \vec{u} - \vec{v}$. This means that $\vec{v} - \vec{u}$ must end up on the line $SB$. The possible ending points of $\vec{v} - \vec{u}$ are achieved by drawing a circle of radius $u$ around the ending point of $\vec{v}$. With the minimal $|\vec{u}|$, the circle will be tangent to $SB$, which means $\vec{u} \perp \overrightarrow{BS}$.

Solution 3. Without loss of generality, put $S$ at $(0,0)$ and $B$ at $(q,0)$. The $SC'$ line is $y = kx$, so a general point $C'$ is $(x,\,kx)$. The ratio of lengths is

$$R = \frac{\sqrt{(x-q)^2 + k^2x^2}}{\sqrt{(k^2+1)x^2}}.$$

Differentiating (preferably logarithmically) and setting the derivative to zero gives $x = q$, i.e. $C' = (q,\,kq)$ and thus $C'B \perp BS$.

Grading (preliminary)

• The idea of switching to the free-falling frame: 0.5 pts
• Stating (explicitly or implicitly) that the stone and the ball travel in straight lines: 0.5 pts
• $SC' = vt$ and $BC' = ut$: 0.2 pts (0.1 for each)
• Noticing that we need to minimise $|SC'|/|BC'|$ (or maximise the reciprocal): 0.3 pts
• Sine theorem (to minimise): 0.5 pts
• $\angle C'BS = 90°$: 0.5 pts
• In the free falling frame $C'$ is shifted upwards with respect to $S$, $B$, $C$: 0.5 pts
• $|CC'| = \frac{1}{2}gt^2$: 0.1 pts
• Well drawn geometrical construction (that is correct): 0.5 pts
• $v = \sqrt{g|SC'|}/\sqrt{2|CC'|}$ and $u = \sqrt{g|BC'|}/\sqrt{2|CC'|}$: 0.2 pts (0.1 for each)
• $v \in [11.8,\,12.7]\,\text{m\,s}^{-1}$, $u \in [10.5,\,11.4]\,\text{m\,s}^{-1}$: 0.2 pts (0.1 for each, only if correct method)

Note. Well drawn means that straight lines are straight lines and the 90° angles look like 90° angles. Points will only be given if the construction is relevant for a correct solution.

Grading for Solution 2:

• Noticing that $\vec{v} - \vec{u}$ is on $SB$: 0.3 pts
• A valid argument that $u$ is minimal when $\vec{u} \perp \overrightarrow{SB}$: 0.5 pts

Grading for Solution 3:

• Correct $R$ function to maximise/minimise: 0.2 pts
• Derivative $= 0$: 0.1 pts
• Performing the rest of the calculation fully correct: 0.2 pts