# Particle Decay Problem Solution Given: - **Initial particle:** Rest mass energy \( E_{} = 50 \) MeV Velocity \( v = .c \) in \( +x \) direction - **Decay products:** - **Particle 1:** Rest mass energy \( m_{1}c^2 = 10 \) MeV Velocity \( v_1 = .1c \) in \( -x \) direction - **Particle 2:** Rest mass energy \( m_{2}c^2 = ? \) Speed \( v_2 = ? \) Direction \( = ? \) We need to find the rest mass energy, speed, and direction of particle 2. --- ## Step 1: Initial Particle's Energy and Momentum ### Lorentz Factor \[ \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} = \frac{1}{\sqrt{1 - (.4)^2}} = \frac{1}{\sqrt{.84}} \approx 1.0911 \] ### Total Energy \[ E_{\text{initial}} = \gamma m_ c^2 = 1.0911 \times 50 = 54.56~\text{MeV} \] ### Momentum \[ p_{\text{initial}} = \gamma m_ v = 1.0911 \times 50/c^2 \times .4c = (1.0911 \times 50 \times .4)/c \] \[ p_{\text{initial}} = 21.82~\text{MeV}/c \] --- ## Step 2: Particle 1's Energy and Momentum ### Lorentz Factor \[ \gamma_1 = \frac{1}{\sqrt{1 - (.1)^2}} = \frac{1}{\sqrt{.99}} \approx 1.005 \] ### Total Energy \[ E_1 = \gamma_1 m_1 c^2 = 1.005 \times 10 = 10.05~\text{MeV} \] ### Momentum \[ p_1 = \gamma_1 m_1 v_1 = 1.005 \times 10/c^2 \times .1c = (1.005 \times 10 \times .1)/c \] \[ p_1 = 1.005~\text{MeV}/c \] Direction is \( -x \), so \( p_1 = -1.005~\text{MeV}/c \). --- ## Step 3: Conservation Equations ### Conservation of Energy \[ E_2 = E_{\text{initial}} - E_1 = 54.56 - 10.05 = 44.51~\text{MeV} \] ### Conservation of Momentum \[ p_2 = p_{\text{initial}} - p_1 = 21.82 - (-1.005) = 22.83~\text{MeV}/c \] --- ## Step 4: Rest Mass Energy of Particle 2 The rest mass energy is: \[ E_2^2 = (p_2 c)^2 + (m_2 c^2)^2 \] \[ \Rightarrow (m_2 c^2)^2 = E_2^2 - (p_2 c)^2 \] \[ = (44.51)^2 - (22.83)^2 = 1981.26 - 521.33 = 1459.93 \] \[ m_2 c^2 = \sqrt{1459.93} \approx 38.22~\text{MeV} \] --- ## Step 5: Speed and Direction of Particle 2 \[ p_2 = \gamma_2 m_2 v_2 \] \[ E_2 = \gamma_2 m_2 c^2 \] So, \[ \frac{p_2 c}{E_2} = \frac{\gamma_2 m_2 v_2 c}{\gamma_2 m_2 c^2} = \frac{v_2}{c} \] \[ v_2 = c \cdot \frac{p_2 c}{E_2} \] \[ v_2 = c \cdot \frac{22.83}{44.51} = c \cdot .5133 \] So, \( v_2 \approx .51c \). Since \( p_2 > \), **direction is \( +x \)**. --- ## Summary Table | Quantity | Value | |---------------------------|-----------------| | Rest mass energy \( m_2c^2 \) | \( 38.2~\text{MeV} \) | | Speed \( v_2 \) | \( .51c \) | | Direction | \( +x \) | --- ## Final Answers - **Rest mass energy of particle 2:** \(\boxed{38.2~\text{MeV}}\) - **Speed of particle 2:** \(\boxed{.51c}\) - **Direction:** \(\boxed{+x}\) (same as original particle)