Question 705075

In the rest frame of $\Delta^{++}$, the total energy is $E_{\Delta} = 1200 \text{MeV}$. By conservation of energy, $E_{\pi} + E_p = 1200$. Using energy conservation in particle decay, $E_{\pi} = \frac{m_{\Delta}^2 + m_{\pi}^2 - m_p^2}{2m_{\Delta}} = \frac{1200^2 + 140^2 - 1000^2}{2(1200)} \approx 191.5 \text{MeV}$. Momentum $p = \sqrt{E_{\pi}^2 - m_{\pi}^2} = \sqrt{191.5^2 - 140^2} \approx 130.7 \text{MeV}$. Closest option is 190,130.