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| <h2>Solutions of Savchenko Problems in Physics</h2> | | <h2>Solutions of Savchenko Problems in Physics</h2> |
| <p class="author"> | | <p class="author"> |
| Aliaksandr Melnichenka <br/> | | Aliaksandr Melnichenka <br/> |
| October 2023 | | October 2023 |
| </p> | | </p> |
| </header> | | </header> |
| | | |
| <h3 id="back-link"><a href="../../#1.3">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#1.3">$\leftarrow$Back</a></h3> |
| | | |
| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $1.3.17.$ In a rectangular box, elastically hitting the bottom and the right wall, a ball jumps back and forth along one trajectory. The time interval between hitting the bottom and the wall is $\Delta t$. The bottom of the box forms an angle $\alpha$ with the horizon. Find the speed of the ball immediately after hitting. | | $1.3.17.$ In a rectangular box, elastically hitting the bottom and the right wall, a ball jumps back and forth along one trajectory. The time interval between hitting the bottom and the wall is $\Delta t$. The bottom of the box forms an angle $\alpha$ with the horizon. Find the speed of the ball immediately after hitting. |
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| For problem $1.3.17$ | | For problem $1.3.17$ |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| </p> | | </p> |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| <p>The most important thing in this problem is that the ball bounces along the same trajectory. Indeed, in an elastic collision, the angle of incidence is equal to the angle of rebound. If the ball falls on the wall of the box so that its speed is not perpendicular to the wall, then when it bounces, its trajectory would differ from the trajectory of its approach - I showed this with an additional small drawing. Therefore, $v_2$ is perpendicular to the wall. And this means that the velocity vectors $v_1$ and $v_2$ are perpendicular to each other. | | <p>The most important thing in this problem is that the ball bounces along the same trajectory. Indeed, in an elastic collision, the angle of incidence is equal to the angle of rebound. If the ball falls on the wall of the box so that its speed is not perpendicular to the wall, then when it bounces, its trajectory would differ from the trajectory of its approach - I showed this with an additional small drawing. Therefore, $v_2$ is perpendicular to the wall. And this means that the velocity vectors $v_1$ and $v_2$ are perpendicular to each other. |
| </p> | | </p> |
| <p> | | <p> |
| Let's draw a triangle of velocities. | | Let's draw a triangle of velocities. |
| </p> | | </p> |
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| <figure> | | <figure> |
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| loading="lazy" width="350" /> | | loading="lazy" width="350" /> |
| <figcaption> | | <figcaption> |
| Velocity triangle | | Velocity triangle |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| </p>It follows that</p> | | </p>It follows that</p> |
| $$v_1 = g \tau \cos \alpha$$ | | $$v_1 = g \tau \cos \alpha$$ |
| $$v_2 = g \tau \sin \alpha$$ | | $$v_2 = g \tau \sin \alpha$$ |
| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$v_{1}=g\Delta t\sin\alpha\text{, }v_{2}=g\Delta t\cos\alpha$$ | | $$v_{1}=g\Delta t\sin\alpha\text{, }v_{2}=g\Delta t\cos\alpha$$ |
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