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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="../../#2.5">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#2.5">$\leftarrow$Back</a></h3> |
| | | |
| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $2.5.18^*.$ A particle of mass $m_1$ flew at a velocity $v$ on a stationary particle of mass $m_2$, which, after an elastic impact, flew at an angle $\alpha$ to the initial direction of movement of the incoming particle. Determine the velocity of a particle of mass $m_2$ after impact. | | $2.5.18^*.$ A particle of mass $m_1$ flew at a velocity $v$ on a stationary particle of mass $m_2$, which, after an elastic impact, flew at an angle $\alpha$ to the initial direction of movement of the incoming particle. Determine the velocity of a particle of mass $m_2$ after impact. |
| </p> | | </p> |
| | | |
| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| This phenomenon is also knowned as Compton scattering | | This phenomenon is also knowned as Compton scattering |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="drawing1.png" | | <img src="drawing1.png" |
| loading="lazy" width="320" /> | | loading="lazy" width="320" /> |
| <figcaption> | | <figcaption> |
| Particle $m_1$ flew on a particle $m_2$ | | Particle $m_1$ flew on a particle $m_2$ |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| | | |
| Let's write the law of conservation of momentum: | | Let's write the law of conservation of momentum: |
| $$\left\{\begin{matrix} | | $$\left\{\begin{matrix} |
| X\text{: }m_1v=m_1v_1\cos\beta+m_2v_2\cos\alpha \\ | | X\text{: }m_1v=m_1v_1\cos\beta+m_2v_2\cos\alpha \\ |
| Y\text{: }m_1v_1\sin\beta=m_2v_2\sin\alpha \\ | | Y\text{: }m_1v_1\sin\beta=m_2v_2\sin\alpha \\ |
| \end{matrix}\right.$$ | | \end{matrix}\right.$$ |
| Let's | | Let's |
| $$\kappa=\frac{m_2}{m_1}$$</p><p> | | $$\kappa=\frac{m_2}{m_1}$$</p><p> |
| From where: | | From where: |
| $$v_1\sin\beta=\kappa v_2\sin\alpha$$ | | $$v_1\sin\beta=\kappa v_2\sin\alpha$$ |
| $$v=v_1\cos\beta+\kappa v_2\cos\alpha$$ | | $$v=v_1\cos\beta+\kappa v_2\cos\alpha$$ |
| We could find $\cos\beta$ from first expression: | | We could find $\cos\beta$ from first expression: |
| $$\cos\beta=\sqrt{1-\sin^2\beta}$$ | | $$\cos\beta=\sqrt{1-\sin^2\beta}$$ |
| $$\cos\beta=\sqrt{1-(\kappa \frac{v_2}{v_1}\sin\alpha)^2}$$ | | $$\cos\beta=\sqrt{1-(\kappa \frac{v_2}{v_1}\sin\alpha)^2}$$ |
| The law of conservation of energy: | | The law of conservation of energy: |
| $$\frac{m_1v^2}{2}=\frac{m_1v_1^2}{2}+\frac{m_2v_2^2}{2}$$ | | $$\frac{m_1v^2}{2}=\frac{m_1v_1^2}{2}+\frac{m_2v_2^2}{2}$$ |
| From where | | From where |
| $$v_1^2=v^2-\kappa v_2^2$$ | | $$v_1^2=v^2-\kappa v_2^2$$ |
| From the other hand: | | From the other hand: |
| | | |
| $$(v-\kappa v_2\cos\alpha)^2=v_1^2\cos^2\beta$$ | | $$(v-\kappa v_2\cos\alpha)^2=v_1^2\cos^2\beta$$ |
| $$(v-\kappa v_2\cos\alpha)^2=v_1^2\cdot(1-(\kappa \frac{v_2}{v_1}\sin\alpha)^2)$$ | | $$(v-\kappa v_2\cos\alpha)^2=v_1^2\cdot(1-(\kappa \frac{v_2}{v_1}\sin\alpha)^2)$$ |
| $$(v-\kappa v_2\cos\alpha)^2=v_1^2-(\kappa v_2\sin\alpha)^2$$ | | $$(v-\kappa v_2\cos\alpha)^2=v_1^2-(\kappa v_2\sin\alpha)^2$$ |
| $$v^2 - 2\kappa vv_2\cos\alpha=v_1^2-\kappa^2 v_2^2$$ | | $$v^2 - 2\kappa vv_2\cos\alpha=v_1^2-\kappa^2 v_2^2$$ |
| From where | | From where |
| $$v_2^2\kappa(1+\kappa)=2\kappa vv_2\cos\alpha$$ | | $$v_2^2\kappa(1+\kappa)=2\kappa vv_2\cos\alpha$$ |
| $$v_2=v\frac{2\cos\alpha}{1+\kappa}$$ | | $$v_2=v\frac{2\cos\alpha}{1+\kappa}$$ |
| And, finally | | And, finally |
| $$\boxed{v_2=v\frac{2m_1\cos\alpha}{m_1+m_2}}$$ | | $$\boxed{v_2=v\frac{2m_1\cos\alpha}{m_1+m_2}}$$ |
| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$v_2=v\frac{2m_1\cos\alpha}{m_1+m_2}$$ | | $$v_2=v\frac{2m_1\cos\alpha}{m_1+m_2}$$ |
| </p> | | </p> |
| <p style="text-align: right; font-style: italic; font-size: 14;"> | | <p style="text-align: right; font-style: italic; font-size: 14;"> |
| Yuldashev Ulugbek<br> | | Yuldashev Ulugbek<br> |
| </p> | | </p> |
| <h3>Alternative solution</h3> | | <h3>Alternative solution</h3> |
| <p> | | <p> |
| </p> | | </p> |
| <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> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| Following from the law of conservation of energy | | Following from the law of conservation of energy |
| $$\frac{p_0^2}{2m_1}=\frac{p_1^2}{2m_1}+\frac{p_2^2}{2m_2}$$ | | $$\frac{p_0^2}{2m_1}=\frac{p_1^2}{2m_1}+\frac{p_2^2}{2m_2}$$ |
| From where | | From where |
| $$p_1^2=p_0^2-p_2^2\frac{m_1}{m_2}$$ | | $$p_1^2=p_0^2-p_2^2\frac{m_1}{m_2}$$ |
| From the cosine theorem | | From the cosine theorem |
| $$p_1^2=p_2^2+p_0^2-2p_2p_0\cos\alpha$$ | | $$p_1^2=p_2^2+p_0^2-2p_2p_0\cos\alpha$$ |
| From where | | From where |
| $$p_2\left(1+\frac{m_1}{m_2}\right)=2p_0\cos\alpha\quad (1)$$ | | $$p_2\left(1+\frac{m_1}{m_2}\right)=2p_0\cos\alpha\quad (1)$$ |
| By definition of momentum | | By definition of momentum |
| $$\left\{\begin{matrix} | | $$\left\{\begin{matrix} |
| p_0=m_1v \\ | | p_0=m_1v \\ |
| p_2=m_2u | | p_2=m_2u |
| \end{matrix}\right.$$ | | \end{matrix}\right.$$ |
| Substituting into $(1)$ expression | | Substituting into $(1)$ expression |
| $$m_2u\left(1+\frac{m_1}{m_2}\right)=2m_1v\cos\alpha$$ | | $$m_2u\left(1+\frac{m_1}{m_2}\right)=2m_1v\cos\alpha$$ |
| $$u(m_1+m_2)=2m_1v\cos\alpha$$ | | $$u(m_1+m_2)=2m_1v\cos\alpha$$ |
| $$\boxed{u=v\frac{2m_1\cos\alpha}{m_1+m_2}}$$ | | $$\boxed{u=v\frac{2m_1\cos\alpha}{m_1+m_2}}$$ |
| </p> | | </p> |
