Updated spacing between @ latex expressions

astrosander правка от
Посмотреть решение История
правка #10648 предыдущая #10123 GitHub c702ebe ← раньше позже →
en/2.5.18.md +12 −12
@@ -111,29 +111,29 @@
Let's write the law of conservation of momentum:
$$\left\{\begin{matrix}
−X\text{: }m_1v=m_1v_1\cos\beta+m_2v_2\cos\alpha \\
−Y\text{: }m_1v_1\sin\beta=m_2v_2\sin\alpha \\
+X\text{: }m_1v=m_1v_1\cos\beta +m_2v_2\cos\alpha\\
+Y\text{: }m_1v_1\sin\beta =m_2v_2\sin\alpha\\
\end{matrix}\right.$$
Let's
−$$\xi=\frac{m_2}{m_1}$$</p><p>
+$$\xi =\frac{m_2}{m_1}$$</p><p>
From where:
−$$v_1\sin\beta=\xi v_2\sin\alpha$$
−$$v=v_1\cos\beta+\xi v_2\cos\alpha$$
+$$v_1\sin\beta =\xi v_2\sin\alpha$$
+$$v=v_1\cos\beta +\xi v_2\cos\alpha$$
We could find $\cos\beta$ from first expression:
−$$\cos\beta=\sqrt{1-\sin^2 \beta}$$
−$$\cos\beta=\sqrt{1-(\xi \frac{v_2}{v_1}\sin\alpha)^2}$$
+$$\cos\beta =\sqrt{1-\sin^2 \beta}$$
+$$\cos\beta =\sqrt{1-(\xi\frac{v_2}{v_1}\sin\alpha )^2}$$
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:
−$$(v-\xi v_2\cos\alpha)^2=v_1^2\cos^2 \beta$$
−$$(v-\xi v_2\cos\alpha)^2=v_1^2\cdot(1-(\xi \frac{v_2}{v_1}\sin\alpha)^2)$$
−$$(v-\xi v_2\cos\alpha)^2=v_1^2-(\xi v_2\sin\alpha)^2$$
−$$v^2 - 2\xi vv_2\cos\alpha=v_1^2-\xi^2 v_2^2$$
+$$(v-\xi v_2\cos\alpha )^2=v_1^2\cos^2 \beta$$
+$$(v-\xi v_2\cos\alpha )^2=v_1^2\cdot(1-(\xi\frac{v_2}{v_1}\sin\alpha )^2)$$
+$$(v-\xi v_2\cos\alpha )^2=v_1^2-(\xi v_2\sin\alpha )^2$$
+$$v^2 - 2\xi vv_2\cos\alpha =v_1^2-\xi^2 v_2^2$$
From where
−$$v_2^2\xi(1+\xi)=2\xi vv_2\cos\alpha$$
+$$v_2^2\xi (1+\xi )=2\xi vv_2\cos\alpha$$
$$v_2=v\frac{2\cos\alpha}{1+\xi}$$
And, finally
$$\boxed{v_2=v\frac{2m_1\cos\alpha}{m_1+m_2}}$$
ещё строк без изменений 19