Правка разделов «Statement», «Solution», «Answer»

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en/14.5.4.md +39 −4
@@ -1,11 +1,46 @@
### Statement
−$14.5.4.$ [Insert the problem statement]
+$14.5.4$. In a head‑on collision of protons, a particle with rest mass $k$ times the proton rest mass $m_p$ can be produced:
+$$
+p + p \to p + p + M, \quad M = k m_p.
+$$
+Determine the minimum mass of the moving protons for which this reaction is possible. What is the minimum speed of the protons?
+$Note:$ There seem to have been several misprints in the problem statement.
+
### Solution
−1
+It is important that Savchenko treats mass as a measure of total energy – the constancy of the proton rest mass has not been cancelled.
−#### Answer
+Let the protons collide head‑on with equal speeds (I believe the author meant this). Then, in the laboratory frame, the total momentum is zero. The reaction threshold is reached when all final particles are at rest in the centre‑of‑mass frame – no energy needs to be “spent” on accelerating them. Then energy conservation gives:
−[Insert a concise answer or boxed result]
+$$
+2 m c^2 = 2 m_p c^2 + M c^2 \tag{1}
+$$
+$$
+m = \left(1 + \frac{k}{2}\right) m_p. \tag{2}
+$$
+
+Now recall that the author’s mass is a measure of total energy, i.e.
+$$
+m c^2 = \gamma m_p c^2, \tag{3}
+$$
+where $\gamma = \frac{1}{\sqrt{1 - (v/c)^2}}$. Then we find the speed (naturally, the minimum speed corresponds to the minimum energy):
+
+$$
+\left(1 + \frac{k}{2}\right) = \frac{1}{\sqrt{1 - \left(\frac{v}{c}\right)^2}},
+$$
+
+$$
+v = c \sqrt{1 - \frac{4}{(2 + k)^2}}, \tag{4}
+$$
+$$
+v = c \frac{\sqrt{k(k+4)}}{k+2}.
+$$
+
+It seems that the author’s answer is incorrect.
+
+#### Answer
+$$
+\boxed{m = \left(1 + \frac{k}{2}\right) m_p, \qquad v = c \frac{\sqrt{k(k+4)}}{k+2}}.
+$$