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BPO 2026, Теория — Задача 2. Trajectory

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EIGHTH BALKAN PHYSICS OLYMPIAD (BPO-8) · 29 юни 2026 г. · 25 т.

Условие

There is a planet without an atmosphere. The acceleration of free fall on its surface is $a = 10{,}0\ \mathrm{m/s^2}$ . Two balls are thrown simultaneously from its surface from a point with Cartesian coordinates $(x, y) = (0{,}0)$ with one and the same speed $v_0$ ( $x$ is the horizontal axis, $y$ is the vertical axis). The angles of throw for the first and second balls are respectively $\alpha$ and $\beta$ . Two photos are made during their flight. The coordinates of the first ball on the first photo and on the second photo are respectively $(x_1, y_1)$ and $(x_2, y_2)$ .

A) Obtain a formula $y = f(x)$ for the trajectory of the first ball expressed in terms of the given quantities $x_1, y_1, x_2, y_2$ . [6 т.]

B) If $(x_1, y_1) = (5{,}0\ \mathrm{m}, 9{,}0\ \mathrm{m})$ and $(x_2, y_2) = (15{,}0\ \mathrm{m}, 21{,}0\ \mathrm{m})$ , find the values of $\alpha$ (or its trigonometric function) and $v_0$ . [3 т.]

C) Obtain the formula for the horizontal distance $L$ of the flight (expressed in terms of $a$ , $v_0$ and $\alpha$ ) of the first ball and calculate its value. [3 т.]

D) It turned out that the flight distance $L$ of the second ball is the same. Express $\beta$ through $\alpha$ . [2 т.]

E) Calculate the moments $t_1$ and $t_2$ (after the throw) when the first and second photo were taken. [3 т.]

F) Calculate the coordinates $(x'_1, y'_1)$ and $(x'_2, y'_2)$ of the second ball on the first and second photo, respectively. [5 т.]

G) Calculate how long it took the first ball to fall after the second (the time delay $T$ ). [3 т.]

Отговори

Решение

A) The motion of the first ball along the $x$ -axis is uniform: $= v_0\cos\alpha\, t$ . (1) (0.5 p) The motion of the first ball along the $y$ -axis is with constant acceleration $a$ : $y = v_0\sin\alpha\, t - \frac{1}{2}at^2$ . (2) (0.5 p) Expressing $t$ from Eq. (1) and substituting the result in (2): $= \tan\alpha\, x - \frac{ax^2}{2{v_0}^2\cos^2\alpha}$ . (1.0 p) (3) Lets $A = \tan\alpha$ and $= \frac{a}{2{v_0}^2\cos^2\alpha}$ . Then (3) is transformed to $y = Ax - Bx^2$ . (4) Substituting in (4) the two positions of the first ball on the photos: $y_1 = Ax_1 - B{x_1}^2$ (5) and $y_2 = Ax_2 - B{x_2}^2$ (6). Multiplying Eq. (5) by ${x_2}^2$ and Eq. (6) by ${x_1}^2$ and subtracting them, we obtain $y_1{x_2}^2 - y_2{x_1}^2 = Ax_1x_2(x_2 - x_1)$ . Then $A = \frac{y_1{x_2}^2 - y_2{x_1}^2}{x_1x_2(x_2 - x_1)}$ (7). (1.5 p) Similarly, multiplying Eq. (5) by $x_2$ and Eq. (6) by $x_1$ and subtracting them, we obtain $y_1x_2 - y_2x_1 = Bx_1x_2(x_2 - x_1)$ . Then $B = \frac{y_1x_2 - y_2x_1}{x_1x_2(x_2 - x_1)}$ . (1.5 p) (8) Finally, the formula for the trajectory is $y = \frac{y_1{x_2}^2 - y_2{x_1}^2}{x_1x_2(x_2 - x_1)}x - \frac{y_1x_2 - y_2x_1}{x_1x_2(x_2 - x_1)}x^2$ . (9) (1.0 p)

B) $\tan\alpha = A = \frac{y_1{x_2}^2 - y_2{x_1}^2}{x_1x_2(x_2 - x_1)} = \frac{9.15^2 - 21.5^2}{5.15(15 - 5)} = 2$ . $\alpha \approx 63.4^{\circ}$ . (1.0 p) As $\frac{1}{\cos^2\alpha} \equiv \tan^2\alpha + 1$ , then $B = \frac{a}{2{v_0}^2}(A^2 + 1)$ and $v_0 = \sqrt{\frac{a(A^2+1)}{2B}}$ . (10) $B = \frac{y_1x_2 - y_2x_1}{x_1x_2(x_2 - x_1)} = \frac{9.15 - 21.5}{5.15(15 - 5)} = 0.0040\ \mathrm{m^{-1}}$ . Then using (10), $v_0 = \sqrt{\frac{10(2^2+1)}{2.0{,}04}} = 25{,}0\ \mathrm{m/s}$ . (2.0 p)

C) From (3) it follows $0 = \tan\alpha\, L - \frac{aL^2}{2{v_0}^2\cos^2\alpha} = L\left(\tan\alpha - \frac{aL}{2{v_0}^2\cos^2\alpha}\right)$ . (0.5 p) Then $L = \frac{2\tan\alpha\cos^2\alpha\, {v_0}^2}{a} = \frac{2\sin\alpha\cos\alpha\, {v_0}^2}{a}$ . (11) (0.5 p) As $\tan\alpha = 2$ , then $\cos\alpha = \frac{1}{\sqrt{5}}$ and $\sin\alpha = \frac{2}{\sqrt{5}}$ . Substituting in (11), $L = \frac{2\frac{2}{\sqrt{5}}\frac{1}{\sqrt{5}}25^2}{10} = 50.0\ \mathrm{m}$ . (2.0 p)

D) From Eq. (11) it follows $= \frac{2\sin\alpha\cos\alpha\, {v_0}^2}{a} = \frac{\sin(2\alpha){v_0}^2}{a}$ . (0.5 p) As $L(\alpha) = L(\beta)$ , then $\sin(2\alpha) = \sin(2\beta)$ , $2\beta = \pi - 2\alpha$ , $\beta = \frac{\pi}{2} - \alpha$ . (12) (1.5 p)

E) From (1) $t_1 = \frac{x_1}{v_0\cos\alpha} = \frac{5}{25.\frac{1}{\sqrt{5}}} = \frac{1}{\sqrt{5}}\ \mathrm{s} \approx 0.45\ \mathrm{s}$ , (1.5 p) $t_2 = \frac{x_2}{v_0\cos\alpha} = \frac{15}{25.\frac{1}{\sqrt{5}}} = \frac{3}{\sqrt{5}}\ \mathrm{s} \approx 1{,}34\ \mathrm{s}$ . (1.5 p)

F) Calculate the coordinates $(x'_1, y'_1)$ and $(x'_2, y'_2)$ of the second ball on the first and second photo, respectively.

From (1) and (12) $x'_1 = v_0\cos\beta\, t_1 = v_0\sin\alpha\, t_1 = \frac{v_0\sin\alpha\, x_1}{v_0\cos\alpha} = \tan\alpha\, x_1 = 2.5 = 10{,}0\ \mathrm{m}$ . (1.0 p)

From (2) and (12) $y'_1 = \tan\beta\, {x_1}' - \frac{a{x_1'}^2}{2{v_0}^2\cos^2\beta} = \frac{{x_1}'}{\tan\alpha} - \frac{a{x_1'}^2}{2{v_0}^2\sin^2\alpha} = \frac{10}{2} - \frac{10.10^2}{2.25^2\left(\frac{2}{\sqrt{5}}\right)^2} = 4{,}0\ \mathrm{m}$ . (1.5 p)

From (1) and (12) $x'_2 = v_0\cos\beta\, t_2 = v_0\sin\alpha\, t_2 = \frac{v_0\sin\alpha\, x_2}{v_0\cos\alpha} = \tan\alpha\, x_2 = 2.15 = 30{,}0\ \mathrm{m}$ . (1.0 p)

From (2) and (12) $y'_2 = \tan\beta\, {x_2}' - \frac{a{x_2'}^2}{2{v_0}^2\cos^2\beta} = \frac{{x_2}'}{\tan\alpha} - \frac{a{x_2'}^2}{2{v_0}^2\sin^2\alpha} = \frac{30}{2} - \frac{10.30^2}{2.25^2\left(\frac{2}{\sqrt{5}}\right)^2} = 6{,}0\ \mathrm{m}$ . (1.5 p)

g) (3 points) Calculate how long it took for the first ball to fall after the second (the time delay $T$ ).

From (1) $T = \frac{L}{v_0\cos\alpha} - \frac{L}{v_0\cos\beta}$ $\mathbf{(0.5\ p)}$ $= \frac{L}{v_0\cos\alpha} - \frac{L}{v_0\sin\alpha} = \frac{L}{v_0}\left(\frac{1}{\cos\alpha} - \frac{1}{\sin\alpha}\right)$ $\mathbf{(1.0\ p)}$ $= \frac{50}{25}\left(\sqrt{5} - \frac{\sqrt{5}}{2}\right) \approx 2{,}2\ \mathrm{s}$ . (1.5 p)

Оригинал в Архива: problems.pdf · официални решения: solutions.pdf