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| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $2.5.17^*.$ A heavy particle of mass $m_1$ collides with a resting light particle of mass $m_2$. What is the greatest angle that a heavy particle can deflect as a result of an elastic impact? | | $2.5.17^*.$ A heavy particle of mass $m_1$ collides with a resting light particle of mass $m_2$. What is the greatest angle that a heavy particle can deflect as a result of an elastic impact? |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
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| <center> | | <center> |
| <figure> | | <figure> |
| <img src="draw.png" | | <img src="draw.png" |
| loading="lazy" width="280" /> | | loading="lazy" width="280" /> |
| <figcaption> | | <figcaption> |
| Direction of balls momentum | | Direction of balls momentum |
| </figcaption> | | </figcaption> |
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| <p> | | <p> |
| Following from the law of conservation of energy | | Following from the law of conservation of energy |
| $$\frac{p_1^2}{2m_1}+\frac{p_2^2}{2m_2}=\frac{p_0^2}{2m_1}$$ | | $$\frac{p_1^2}{2m_1}+\frac{p_2^2}{2m_2}=\frac{p_0^2}{2m_1}$$ |
| From where | | From where |
| $$p_2^2=\frac{m_2}{m_1}\left(p_0^2-p_1^2\right)\quad\text{(1)}$$ | | $$p_2^2=\frac{m_2}{m_1}\left(p_0^2-p_1^2\right)\quad\text{(1)}$$ |
| From the cosine theorem | | From the cosine theorem |
| $$p_2^2=p_1^2+p_0^2-2p_1p_0\cos\alpha$$ | | $$p_2^2=p_1^2+p_0^2-2p_1p_0\cos\alpha$$ |
| Substituting into $\text{(1)}$: | | Substituting into $\text{(1)}$: |
| $$\frac{m_2}{m_1}p_0^2-\frac{m_2}{m_1}p_1^2=p_1^2+p_0^2-2p_1p_0\cos\alpha$$ | | $$\frac{m_2}{m_1}p_0^2-\frac{m_2}{m_1}p_1^2=p_1^2+p_0^2-2p_1p_0\cos\alpha$$ |
| Let's regroup this into a quadratic equation | | Let's regroup this into a quadratic equation |
| <div class="scroll-wrapper">$$p_1^2\left(1+\frac{m_2}{m_1}\right)-2p_0p_1\cos\alpha+p_0^2\left(1-\frac{m_2}{m_1}\right)=0$$</div> | | <div class="scroll-wrapper">$$p_1^2\left(1+\frac{m_2}{m_1}\right)-2p_0p_1\cos\alpha+p_0^2\left(1-\frac{m_2}{m_1}\right)=0$$</div> |
| Solving relative to $p_1$ | | Solving relative to $p_1$ |
| <div class="scroll-wrapper">$$p_1=\frac{2p_0\cos\alpha\pm \sqrt{4p_0^2\cos^2\alpha-4p_0^2\left(1-\frac{m_2^2}{m_1^2}\right)}}{2\left(1+\frac{m_2}{m_1}\right)}$$</div> | | <div class="scroll-wrapper">$$p_1=\frac{2p_0\cos\alpha\pm \sqrt{4p_0^2\cos^2\alpha-4p_0^2\left(1-\frac{m_2^2}{m_1^2}\right)}}{2\left(1+\frac{m_2}{m_1}\right)}$$</div> |
| Since $\Delta = 0$ | | Since $\Delta = 0$ |
| $$\cos^2\alpha=1-\frac{m_2^2}{m_1^2}$$ | | $$\cos^2\alpha=1-\frac{m_2^2}{m_1^2}$$ |
| From where we find the required ratio | | From where we find the required ratio |
| $$\frac{m_2^2}{m_1^2}=1-\cos^2\alpha=\sin^2\alpha$$ | | $$\frac{m_2^2}{m_1^2}=1-\cos^2\alpha=\sin^2\alpha$$ |
| $$\boxed{\sin\alpha=\frac{m_2}{m_1}}$$ | | $$\boxed{\sin\alpha=\frac{m_2}{m_1}}$$ |
| </p> | | </p> |
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| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$\sin\alpha=m_2/m_1$$ | | $$\sin\alpha=m_2/m_1$$ |
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| <p style="text-align: right; font-style: italic; font-size: 14;"> | | <p style="text-align: right; font-style: italic; font-size: 14;"> |
| Almaskhan Arsen<br> | | Almaskhan Arsen<br> |
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