Edits to “Statement”, “Solution”, “Answer”

jzmicer edited
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revision #19328 parent #19327 ← older
en/14.5.9.md +64 −4
@@ -1,11 +1,71 @@
### Statement
−$14.5.9.$ [Insert the problem statement]
+$14.5.9.$ Two particles with masses $m_1$ and $m_2$, moving with velocities $v_1$ and $v_2$ directed towards each other at an angle $\alpha$, merge into one particle. Determine the mass $M$ and velocity $v$ of the resulting particle.
### Solution
−1
+First, let us solve the problem in the framework of special relativity, since the condition does not specify that approximations are allowed, and the chapter is indeed called "Special Relativity"...
−#### Answer
+The gamma factor:
+$$
+\gamma(v) = \left(1 - \frac{v^2}{c^2}\right)^{-\frac{1}{2}}. \tag{1}
+$$
−[Insert a concise answer or boxed result]
+Energy conservation:
+$$
+\gamma_1 m_1 c^2 + \gamma_2 m_2 c^2 = \gamma M c^2,
+$$
+or, simplifying slightly,
+$$
+\gamma_1 m_1 + \gamma_2 m_2 = \gamma M. \tag{2}
+$$
+
+Write the momentum conservation along the axes (one axis is along $v_1$, the other perpendicular to it). Let the velocity of the final particle be directed at an angle $\varphi$ to the first axis:
+$$
+\gamma_1 m_1 v_1 + \gamma_2 m_2 v_2 \cos\alpha = \gamma M v \cos\varphi, \tag{3}
+$$
+$$
+\gamma_2 m_2 v_2 \sin\alpha = \gamma M v \sin\varphi. \tag{4}
+$$
+
+Writing the conservation laws in such problems is not difficult; the most important thing is to solve the resulting system carefully and quickly. One can do the following: square (3) and (4), add the resulting expressions, use the Pythagorean identity to eliminate $\varphi$, and substitute $\gamma M$ from (2):
+$$
+(\gamma_1 m_1 v_1 + \gamma_2 m_2 v_2 \cos\alpha)^2 + (\gamma_2 m_2 v_2 \sin\alpha)^2 = (\gamma_1 m_1 + \gamma_2 m_2)^2 v^2,
+$$
+$$
+(\gamma_1 m_1 v_1)^2 + 2\gamma_1\gamma_2 m_1 m_2 v_1 v_2 \cos\alpha + (\gamma_2 m_2 v_2)^2 = (\gamma_1 m_1 + \gamma_2 m_2)^2 v^2,
+$$
+$$
+v^2 = \frac{(\gamma_1 m_1 v_1)^2 + 2\gamma_1\gamma_2 m_1 m_2 v_1 v_2 \cos\alpha + (\gamma_2 m_2 v_2)^2}{(\gamma_1 m_1 + \gamma_2 m_2)^2}. \tag{5}
+$$
+
+Strictly speaking, this is the answer. Substituting (5) into (2) gives the mass as well.
+
+However, judging by the author's answer, it is assumed that for all velocities $\gamma \approx 1$. Then
+$$
+v = \frac{\sqrt{(m_1 v_1)^2 + (m_2 v_2)^2 + 2m_1 m_2 v_1 v_2 \cos\alpha}}{m_1 + m_2}, \tag{6}
+$$
+$$
+M = m_1 + m_2. \tag{7}
+$$
+
+If the author had stated this in the condition, one could obtain the same result much more simply:
+$$
+M = m_1 + m_2,
+$$
+$$
+(m_1 + m_2)\vec v = m_1 \vec v_1 + m_2 \vec v_2.
+$$
+Squaring and using the scalar product rules:
+$$
+(m_1 + m_2)^2 v^2 = m_1^2 v_1^2 + m_2^2 v_2^2 + 2m_1 m_2 v_1 v_2 \cos\alpha,
+$$
+and then express the answer.
+
+#### Answer
+$$
+M = m_1 + m_2,
+$$
+$$
+v = \frac{\sqrt{(m_1 v_1)^2 + (m_2 v_2)^2 + 2m_1 m_2 v_1 v_2 \cos\alpha}}{m_1 + m_2}.
+$$