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BPO 2023, Теория — Задача 3. Tennis ball

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5th Balkan Physics Olympiad, BPO 5 - Albania, 24 June 2023, Durrës, ALBANIA — Problems · 24 юни 2023 г.

Условие

A. Warming up.

During warm up, a tennis ball (A) is dropped from Novak Djokovic’s pocket at a height H. At the same time, a grasshopper (B) just underneath Novak’s pocket, jumps from the ground, vertically towards the ball with initial velocity $v_0$ . When they collide, the ball has twice the speed of the grasshopper. The collision occurs at height h. [15 т.]

A1. Write down the equation of motion for the ball, $y_A$ as a function of time, in terms of H, $v_0$ and gravitational acceleration g. [1 т.]

A2. Write down the equation of motion for the grasshopper, $y_B$ as a function of time, in terms of $v_0$ and g. [1 т.]

A3. Derive the expression for the velocity of the ball $v_A(t)$ in terms of H, g and $y_A(t)$ . [3 т.]

A4. Derive the expression for the velocity of the grasshopper $v_B(t)$ in terms of $v_0$ , g and $y_B(t)$ . [1 т.]

A5. Derive the expression for the initial velocity $v_0$ of the grasshopper in terms of h, H and g. [3 т.]

A6. Derive the expression for the moment of the collision $t_c$ in terms of $v_0$ and g. [3 т.]

A7. Calculate the numerical value of the ratio $h/H$ . [3 т.]

B. Match

During match, at serve, Novak aims to hit tennis ball horizontally. [10 т.]

B1. Write down the expression for the equation of motion of the ball in the horizontal direction $x(t)$ . [1 т.]

B2. Write down the expression for the equation of motion of the ball in the vertical direction $y(t)$ . [1 т.]

B3. Calculate the numerical value of the minimal initial velocity $v_i$ required for the ball to pass just above the 0.9 m high net, 15 m in front of Novak, if the ball is launched (horizontally) from a height of 2.5 m. [3 т.]

B4. Where will the ball land in the case given under B3? [4 т.]

B5. How long will the ball be in the air before it lands in the case given under B3? [1 т.]

The gravitational acceleration is $g = 9.8\ m/s^2$ .

Отговори

Решение

A. Warming up

During warm up, a tennis ball (A) is dropped from rest from Novak Djokovic pocket of height H. At the same instant of time grasshopper (B) stands on the ground below Novak’s pocket and starts to move vertically towards the ball with initial velocity $v_0$ . When they collide, the ball has twice the speed of the grasshopper. The collision occurs at height h.

A1.

y_{\mathrm{A}}(t) = H - \frac{1}{2}gt^2.

A2.

y_{\mathrm{B}}(t) = v_0 t - \frac{1}{2}gt^2.

A3.

v_A(t) = -gt \;\Rightarrow\; t = \left|-\frac{v_A(t)}{g}\right| = \frac{v_A(t)}{g},

\Rightarrow y_{\mathrm{A}}(t) = H - \frac{1}{2}g\left(\frac{v_A(t)}{g}\right)^2,

\Rightarrow v_A^2(\mathrm{t}) = 2g(H - y_{\mathrm{A}}(t)) \;\Rightarrow\; v_B(t) = \sqrt{2g(H - y_{\mathrm{A}}(t))}.

A4.

v_B^2(\mathrm{t}) = v_0^2 - 2gy_B(t) \;\Rightarrow\; v_B(t) = \sqrt{v_0^2 - 2gy_B(t)}.

A5.

Using condition $v_A(t_C) = 2v_B(t_C)$ at the moment of collision $t_C$ and $y_{\mathrm{A}}(t_C) = y_{\mathrm{B}}(t_C) = h$ we can write

2g(H - h) = 4v_0^2 - 8gh,

\Rightarrow v_0^2 = \frac{1}{2}g(H + 3h).

A6.

Using again condition $|v_A(t_C)| = 2v_B(t_C)$ we can write

gt_C = 2v_0 - 2gt_C,

\Rightarrow t_C = \frac{2v_0}{3g}.

A7.

H = v_0 t_C = v_0\frac{2v_0}{3g} = \frac{2}{3g}v_0^2 = \frac{2}{3g}\frac{1}{2}g(H + 3h),

\Rightarrow \frac{h}{H} = \frac{2}{3}.

B. Match

During match, at serve, Novak aims to hit tennis ball horizontally.

B1.

x(t) = v_i t.

B2.

y(t) = \mathrm{y}_0 - \frac{1}{2}gt^2.

B3.

For $x_1(t_1) = 15\ m$ , $\mathrm{y}_1(t_1) = 0.9\ m$ and $\mathrm{y}_0 = 2.5\ m$ , where $t_1$ is instant of time when the ball passed just above the net, we can write

x_1(t_1) = v_i t_1 \;\Rightarrow\; v_i = \frac{x_1(t_1)}{t_1},

\mathrm{y}_1(t_1) = \mathrm{y}_0 - \frac{1}{2}gt_1^2 \;\Rightarrow\; t_1 = \sqrt{\frac{2}{g}(\mathrm{y}_0 - \mathrm{y}_1(t_1))},

\Rightarrow v_i = x_1(t_1)\sqrt{\frac{g}{2(\mathrm{y}_0 - \mathrm{y}_1(t_1))}} = 26.3\ \frac{m}{s}.

B4.

For $x_2(t_2) = v_i t_2$ and $\mathrm{y}_2(t_2) = \mathrm{y}_0 - \frac{1}{2}gt_2^2 = 0$ , where $t_2$ is instant of time when the ball lands, we can write

t_2 = \sqrt{\frac{2}{g}\mathrm{y}_0},

\Rightarrow x_2(t_2) = v_i\sqrt{\frac{2}{g}\mathrm{y}_0} = x_1(t_1)\sqrt{\frac{g}{2(\mathrm{y}_0 - \mathrm{y}_1(t_1))}}\sqrt{\frac{2}{g}\mathrm{y}_0},

\Rightarrow x_2(t_2) = x_1(t_1)\sqrt{\frac{\mathrm{y}_0}{(\mathrm{y}_0 - \mathrm{y}_1(t_1))}} = 18.8\ m.

B5.

Gravitational acceleration is $g = 9.8\ \mathrm{m/s^2}$ .

t_2 = \sqrt{\frac{2}{g}\mathrm{y}_0} = 0.71\ m.

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