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NBPhO 2024, Теория и експеримент — Задача 8. Airplanes

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Всички материали от NBPhO 2024

Nordic-Baltic Physics Olympiad 2024 · 7 т.

Условие

— Janis Huns, Jaan Kalda.

Two airplanes pass each other while flying at a constant altitudes. While they have identical airspeeds, their ground speeds are $v_1$ and $v_2$ , respectively, and the angle between the velocity vectors is $\alpha$ . Based on this knowledge, what is the minimal possible value

i) of the airspeed of the planes; [1 т.]

ii) of the wind speed at the altitudes of the planes? [3 т.]

iii) Answer the question ii) if now $v_1 = v_2 = v$ , but it is known that the airspeed of one of the planes is two times bigger than that of the other. [3 т.]

Решение

— Solution by Tudor Plopeanu.

i) (1 т.)

Vector diagram for parts i) and ii): from the origin O, vectors v1 = w + u1 to point A and v2 = w + u2 to point B; the wind vector w from O to O′ on the dashed perpendicular bisector l of AB; dash-dotted segments from O′ to A and to B; a dashed segment w_min from O perpendicular to l.

Let $\vec{w}$ be the speed of the wind at the airplanes’ altitude, and let $\vec{u}_1$ and $\vec{u}_2$ be the planes’ respective speeds in absence of wind. $\vec{v}_1 = \vec{w} + \vec{u}_1$ , $\vec{v}_2 = \vec{w} + \vec{u}_2$ . Let $O'$ be the point such that $\vec{OO'} = \vec{w}$ , $A$ the point such that $\vec{OA} = \vec{v}_1$ , and $B$ the point such that $\vec{OB} = \vec{v}_2$ . As $|\vec{w} - \vec{v}_1| = |\vec{w} - \vec{v}_2|$ , we know that $O'$ lies on the perpendicular bisector of $(AB)$ , which we shall denote as $l$ . A quick check also yields that all points on this line are valid selections for $\vec{w}$ . $|\vec{u}_1| = |AO'|$ . We can find the minimal airspeed as the length of the perpendicular from $A$ to $l$ , of length $|AB|/2$ . For a vector result in terms of $\vec{v}_1$ and $\vec{v}_2$ , we have

|\vec{u}_1|_{\min} = |\vec{v}_1 - \vec{v}_2|/2

= \frac{1}{2}\sqrt{v_1^2 + v_2^2 - 2v_1v_2\cos\alpha}.

Grading:

This is relevant for all three subproblems. The solutions are mainly geometric, however several students have tried solving it in a more algebraic manner. If it yielded the correct answer or near correct answer, points were given, but otherwise partial points were not given.

Representing the problem with vectors and adding the velocities correctly (0.2 pts)
Wrong order or sign (-0.1 pts)
The wind velocity vector $w$ on perpendicular bisector of $(AB)$ (0.4 pts)
The minimum airspeed needs $w$ to lie on the intersection of $AB$ and the perpendicular bisector (0.1 pts)
calculating correct answer (0.3 pts)
Small error but reasonable answer with correct units, or not expanded answer that is simple to expand (-0.1 pts)

ii) (3 т.)

The minimal wind speed is given by the length of the perpendicular from $O$ to $l$ . This is the length of $(\vec{v}_1 + \vec{v}_2)/2$ projected on $\vec{v}_1 - \vec{v}_2$ , and can be found as

|\vec{w}|_{\min} = \frac{|(\vec{v}_1 + \vec{v}_2)\cdot(\vec{v}_1 - \vec{v}_2)|}{2|\vec{v}_1 - \vec{v}_2|}

= \frac{\left||\vec{v}_1|^2 - |\vec{v}_2|^2\right|}{2|\vec{v}_1 - \vec{v}_2|}

= \frac{|v_1^2 - v_2^2|}{2\sqrt{v_1^2 + v_2^2 - 2v_1v_2\cos\alpha}}.

Grading:

Representing the problem with vectors and adding the velocities correctly (0.5 pts)
The wind velocity vector $w$ on perpendicular bisector of $(AB)$ (0.5 pts)
The wind velocity is smallest when it is perpendicular to $l$ (0.5 pts)
Calculating the correct answer (1.5 pts)
Small error but reasonable answer with correct units, or not expanded answer that is simple to expand (-0.5 pts)

iii) (3 т.)

Vector diagram for part iii): from the origin O, vectors v1 = w + u1 to A and v2 = w + u2 to B; the wind vector w from O to O′, which lies on a dashed circle (the Apollonius circle) with centre O″; a dotted line through A, B and O″; dashed segments from O′ to A and B.

In this situation, $|\vec{w} - \vec{v}_1| = 2|\vec{w} - \vec{v}_2|$ , so $O'$ is on the Apollonius’s circle with respect to the points $A$ and $B$ , such that the points on the circle are 2 times closer to $B$ than to $A$ . Let its center be $O''$ . Its radius is $R_A = 2|\vec{v}_1 - \vec{v}_2|/3$ (this can be found by considering the intersection points of the Apollonius circle with the line $AB$ ). We are tasked to find the shortest distance from $O$ to this circle, which is found as $|OO''| - R_A$ . We note that $\vec{OO''} = (4/3)\vec{v}_2 - (1/3)\vec{v}_1$ . The minimal wind speed is thus

|\vec{w}|_{\min} = \left|\frac{4}{3}\vec{v}_2 - \frac{1}{3}\vec{v}_1\right| - \frac{2}{3}|\vec{v}_2 - \vec{v}_1|

= \frac{\sqrt{17 - 8\cos\alpha} - 4\sin\frac{\alpha}{2}}{3}v.

Grading:

Representing the problem with vectors and adding the velocities correctly (0.2 pts)
The wind velocity vector $w$ needs to lie on the Apollonius circle (doesn’t need the name) (1.3 pts)
if one realises it is a circle, but not the correct one (0.5 pts) is given
Calculating the $R_A$ (0.5 pts)
Calculating the correct answer (1 pts)
Small error but reasonable answer with correct units, or not expanded answer that is simple to expand (-0.5 pts)

Оригинал в Архива: NBPhO_2024_problems_v1.0.pdf · официални решения: NBPhO_2024_solutions_v0.98.pdf