Updated 2.8.37 solution

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en/2.5.18.md +51 −53
@@ -51,58 +51,6 @@
<h3>Solution</h3>
<p>
− This phenomenon is also knowned as Compton scattering
−</p>
−<center>
− <figure>
− <img src="drawing1.png"
− loading="lazy" width="320" />
− <figcaption>
− Particle $m_1$ flew on a particle $m_2$
− </figcaption>
− </figure>
−</center>
−<p>
−
−Let's write the law of conservation of momentum:
−$$\left\{\begin{matrix}
−X\text{: }m_1v=m_1v_1\cos\beta+m_2v_2\cos\alpha \\
−Y\text{: }m_1v_1\sin\beta=m_2v_2\sin\alpha \\
−\end{matrix}\right.$$
−Let's
−$$\kappa=\frac{m_2}{m_1}$$</p><p>
−From where:
−$$v_1\sin\beta=\kappa v_2\sin\alpha$$
−$$v=v_1\cos\beta+\kappa v_2\cos\alpha$$
−We could find $\cos\beta$ from first expression:
−$$\cos\beta=\sqrt{1-\sin^2\beta}$$
−$$\cos\beta=\sqrt{1-(\kappa \frac{v_2}{v_1}\sin\alpha)^2}$$
−The law of conservation of energy:
−$$\frac{m_1v^2}{2}=\frac{m_1v_1^2}{2}+\frac{m_2v_2^2}{2}$$
−From where
−$$v_1^2=v^2-\kappa v_2^2$$
−From the other hand:
−
−$$(v-\kappa v_2\cos\alpha)^2=v_1^2\cos^2\beta$$
−$$(v-\kappa v_2\cos\alpha)^2=v_1^2\cdot(1-(\kappa \frac{v_2}{v_1}\sin\alpha)^2)$$
−$$(v-\kappa v_2\cos\alpha)^2=v_1^2-(\kappa v_2\sin\alpha)^2$$
−$$v^2 - 2\kappa vv_2\cos\alpha=v_1^2-\kappa^2 v_2^2$$
−From where
−$$v_2^2\kappa(1+\kappa)=2\kappa vv_2\cos\alpha$$
−$$v_2=v\frac{2\cos\alpha}{1+\kappa}$$
−And, finally
−$$\boxed{v_2=v\frac{2m_1\cos\alpha}{m_1+m_2}}$$
− </p>
−
− <h4>Answer</h4>
− <p>
− $$v_2=v\frac{2m_1\cos\alpha}{m_1+m_2}$$
− </p>
− <p style="text-align: right; font-style: italic; font-size: 14;">
− Yuldashev Ulugbek<br>
− </p>
− <h3>Alternative solution</h3>
− <p>
</p>
<center>
<figure>
@@ -130,7 +78,7 @@
Substituting into $(1)$ expression
$$m_2u\left(1+\frac{m_1}{m_2}\right)=2m_1v\cos\alpha$$
$$u(m_1+m_2)=2m_1v\cos\alpha$$
−$$\boxed{u=v\frac{2m_1\cos\alpha}{m_1+m_2}}$$
+$$\boxed{u=v\frac{2m_1\cos\alpha}{m_1+m_2s}}$$
</p>
<h4>Answer</h4>
@@ -140,6 +88,56 @@
<p style="text-align: right; font-style: italic; font-size: 14;">
Almaskhan Arsen<br>
+ </p>
+
+ <h3>Alternative solution</h3>
+ <p>
+ This phenomenon is also knowned as Compton scattering
+</p>
+<center>
+ <figure>
+ <img src="drawing1.png"
+ loading="lazy" width="320" />
+ <figcaption>
+ Particle $m_1$ flew on a particle $m_2$
+ </figcaption>
+ </figure>
+</center>
+<p>
+
+Let's write the law of conservation of momentum:
+$$\left\{\begin{matrix}
+X\text{: }m_1v=m_1v_1\cos\beta+m_2v_2\cos\alpha \\
+Y\text{: }m_1v_1\sin\beta=m_2v_2\sin\alpha \\
+\end{matrix}\right.$$
+Let's
+$$\xi=\frac{m_2}{m_1}$$</p><p>
+From where:
+$$v_1\sin\beta=\xi v_2\sin\alpha$$
+$$v=v_1\cos\beta+\xi v_2\cos\alpha$$
+We could find $\cos\beta$ from first expression:
+$$\cos\beta=\sqrt{1-\sin^2\beta}$$
+$$\cos\beta=\sqrt{1-(\xi \frac{v_2}{v_1}\sin\alpha)^2}$$
+The law of conservation of energy:
+$$\frac{m_1v^2}{2}=\frac{m_1v_1^2}{2}+\frac{m_2v_2^2}{2}$$
+From where
+$$v_1^2=v^2-\xi v_2^2$$
+From the other hand:
+
+$$(v-\xi v_2\cos\alpha)^2=v_1^2\cos^2\beta$$
+$$(v-\xi v_2\cos\alpha)^2=v_1^2\cdot(1-(\xi \frac{v_2}{v_1}\sin\alpha)^2)$$
+$$(v-\xi v_2\cos\alpha)^2=v_1^2-(\xi v_2\sin\alpha)^2$$
+$$v^2 - 2\xi vv_2\cos\alpha=v_1^2-\xi^2 v_2^2$$
+From where
+$$v_2^2\xi(1+\xi)=2\xi vv_2\cos\alpha$$
+$$v_2=v\frac{2\cos\alpha}{1+\xi}$$
+And, finally
+$$\boxed{v_2=v\frac{2m_1\cos\alpha}{m_1+m_2}}$$
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ $$v_2=v\frac{2m_1\cos\alpha}{m_1+m_2}$$
</p>
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