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| <h3 id="back-link"><a href="../../#1.3">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#1.3">$\leftarrow$Back</a></h3> |
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| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $1.3.16^*.$ A ball flies into a tube of length $l$, inclined at an angle $\alpha$ to the horizon, with a horizontal velocity $v$. Determine the time of the ball's stay in the pipe, if the ball hits its walls elastic. | | $1.3.16^*.$ A ball flies into a tube of length $l$, inclined at an angle $\alpha$ to the horizon, with a horizontal velocity $v$. Determine the time of the ball's stay in the pipe, if the ball hits its walls elastic. |
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| For problem $1.3.16^*$ | | For problem $1.3.16^*$ |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| <p>a) Find the condition under which there will be no collisions$(x=0)$</p> | | <p>a) Find the condition under which there will be no collisions$(x=0)$</p> |
| <p>From the law of conservation of energy:</p> | | <p>From the law of conservation of energy:</p> |
| <p class="exp"> | | <p class="exp"> |
| $$ \frac{mv_0^2}{2} = mgl \cdot \sin \alpha $$ | | $$ \frac{mv_0^2}{2} = mgl \cdot \sin \alpha $$ |
| </p> | | </p> |
| <p class="exp"> | | <p class="exp"> |
| $$ v_{0}\leq\frac{\sqrt{2gl\sin \alpha}}{\cos\alpha} $$ | | $$ v_{0}\leq\frac{\sqrt{2gl\sin \alpha}}{\cos\alpha} $$ |
| </p> | | </p> |
| <p>In this case, time will pass</p> | | <p>In this case, time will pass</p> |
| <p class="exp"> | | <p class="exp"> |
| $$ t = \frac{2v_0}{g} \text{ctg} \alpha $$ | | $$ t = \frac{2v_0}{g} \text{ctg} \alpha $$ |
| </p> | | </p> |
| <p>b) Now let's look at those cases when touching occurs:</p> | | <p>b) Now let's look at those cases when touching occurs:</p> |
| <p class="exp"> | | <p class="exp"> |
| $$ v_{0}>\frac{\sqrt{2gl\sin \alpha}}{\cos\alpha} $$ | | $$ v_{0}>\frac{\sqrt{2gl\sin \alpha}}{\cos\alpha} $$ |
| </p> | | </p> |
| <p>Path between two touches:</p> | | <p>Path between two touches:</p> |
| <p class="exp"> | | <p class="exp"> |
| $$ L=v_{0}\cos\alpha t-\frac{g\sin\alpha t^{2}}{2} $$ | | $$ L=v_{0}\cos\alpha t-\frac{g\sin\alpha t^{2}}{2} $$ |
| </p> | | </p> |
| <p>When we receive the required time, we choose the one that is smaller, because we need to find the time when it will come out</p> | | <p>When we receive the required time, we choose the one that is smaller, because we need to find the time when it will come out</p> |
| <p class="exp"> | | <p class="exp"> |
| $$ t=\frac{v_{0}\cos\alpha-\sqrt{v_{0}^{2}\cos^{2}\alpha-2g\sin\alpha L}}{g\sin\alpha} $$ | | $$ t=\frac{v_{0}\cos\alpha-\sqrt{v_{0}^{2}\cos^{2}\alpha-2g\sin\alpha L}}{g\sin\alpha} $$ |
| </p> | | </p> |
| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $\begin{aligned}&t=\frac{2v}{g}\operatorname{ctg}\alpha\text{ with }v\cos\alpha<\sqrt{2gl\sin\alpha};\\&t=\frac vg\operatorname{ctg}\alpha\bigg(1-\sqrt{1-\frac{2gl\operatorname{tg}\alpha}{v^2\cos\alpha}}\bigg)\text{ with }v\cos\alpha>\sqrt{2gl\sin\alpha}.\end{aligned}$ | | $\begin{aligned}&t=\frac{2v}{g}\operatorname{ctg}\alpha\text{ with }v\cos\alpha<\sqrt{2gl\sin\alpha};\\&t=\frac vg\operatorname{ctg}\alpha\bigg(1-\sqrt{1-\frac{2gl\operatorname{tg}\alpha}{v^2\cos\alpha}}\bigg)\text{ with }v\cos\alpha>\sqrt{2gl\sin\alpha}.\end{aligned}$ |
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