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### Statement
−$14.5.11.$ [Insert the problem statement]
+$14.5.11$
+A $\pi^0$‑meson decays into two $\gamma$‑quanta: $\pi^0 \to \gamma + \gamma$. Find the kinetic energy of the $\pi^0$‑meson if a counter placed along its direction of motion registers a $\gamma$‑quantum with energy $E_1 = 270\ \text{MeV}$.
+
### Solution
−1
+$\textbf{Method 1}$
−#### Answer
+Recall that for a photon the momentum is $p = \frac{E}{c}$.
−[Insert a concise answer or boxed result]
+The counter detected only one photon; therefore, the second one must have flown in the strictly opposite direction (if it had a component not parallel to the meson's direction of motion, there would be nothing to compensate it). Moreover, momentum conservation must hold:
+$$
+p = \frac{E v}{c^2} = \frac{1}{c} (E_1 - E_2).
+$$
+Energy conservation:
+$$
+E = m_{\pi^0} c^2 + \mathcal{E}_K = E_1 + E_2.
+$$
+Solve this system:
+$$
+E_2 = E - E_1,
+$$
+$$
+E v = c (2E_1 - E).
+$$
+Denote $\beta = v/c$, write the definitions of energy and solve carefully:
+$$
+\frac{m_{\pi^0} c^2 (1 + \beta)}{\sqrt{1 - \beta^2}} = 2E_1,
+$$
+$$
+\sqrt{\frac{1 + \beta}{1 - \beta}} = \frac{2E_1}{m_{\pi^0} c^2} = s,
+$$
+$$
+\beta = \frac{s^2 - 1}{s^2 + 1},
+$$
+$$
+\mathcal{E}_K = E - m_{\pi^0} c^2 = m_{\pi^0} c^2 \left( \frac{1}{\sqrt{(1 + \beta)(1 - \beta)}} - 1 \right),
+$$
+$$
+\mathcal{E}_K = m_{\pi^0} c^2 \left( \frac{s^2 + 1}{2s} - 1 \right),
+$$
+$$
+\mathcal{E}_K = m_{\pi^0} c^2 \left( \frac{s}{2} + \frac{1}{2s} - 1 \right),
+$$
+$$
+\mathcal{E}_K = m_{\pi^0} c^2 \left( \frac{E_1}{m_{\pi^0} c^2} + \frac{m_{\pi^0} c^2}{4E_1} - 1 \right).
+$$
+Or, more compactly,
+$$
+\mathcal{E}_K = \frac{(2E_1 - m_{\pi^0} c^2)^2}{4E_1} \approx 152\ \text{MeV}.
+$$
+The formula in Savchenko's answer is incorrect, but the numerical answer is correct.
+
+$\textbf{Method 2}$
+
+The problem can also be solved by writing the invariant before and after decay:
+$$
+I = \frac{E^2}{c^2} - p^2 = m_{\pi^0}^2 c^2,
+$$
+$$
+I = \left( \frac{E_1 + E_2}{c} \right)^2 - \left( \frac{E_1 - E_2}{c} \right)^2 = \frac{4E_1 E_2}{c^2}.
+$$
+Equating:
+$$
+E_2 = \frac{m_{\pi^0}^2 c^4}{4E_1},
+$$
+$$
+\mathcal{E}_K = E - m_{\pi^0} c^2 = E_1 + E_2 - m_{\pi^0} c^2,
+$$
+$$
+\mathcal{E}_K = E_1 + \frac{m_{\pi^0}^2 c^4}{4E_1} - m_{\pi^0} c^2,
+$$
+which leads to the same result.
+
+#### Answer
+$$
+\boxed{\mathcal{E}_K = \frac{(2E_1 - m_{\pi^0} c^2)^2}{4E_1} \approx 152\ \text{MeV}}
+$$