Let's solve each part step by step with explanations: --- ## (a) How long does the muon live as measured by an Earth-bound observer? ### **Step 1:** Identify the proper time (\(\Delta t_\)), which is the lifetime of the muon in its own frame: \[ \Delta t_ = 1.52 \ \mu s \] The velocity (\(v\)) is \(.950c\). ### **Step 2:** Use the time dilation formula \[ \Delta t = \gamma \Delta \] where \( \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} \). Calculate \( \gamma \): \[ \gamma = \frac{1}{\sqrt{1 - (.950)^2}} = \frac{1}{\sqrt{1 - .9025}} = \frac{1}{\sqrt{.0975}} \approx 3.207 \] Now, calculate the dilated lifetime: \[ \Delta t = 3.207 \times 1.52 \ \mu s \approx 4.87 \ \mu s \] **Explanation:** The muon lives longer (as measured from Earth) due to time dilation. **Final Answer:** \[ \boxed{\Delta t = 4.87 \ \mu s} \] --- ## (b) How far does the muon travel according to the Earth-bound observer? ### **Step 1:** Now, use the time measured by the Earth observer (\(\Delta t\)) and the muon's velocity (\(v\)): \[ L = v \Delta t \] ### **Step 2:** Plug in the values: - \(v = .950c\) - \(c = 3.00 \times 10^8 \ m/s\) - \(\Delta t = 4.87 \ \mu s = 4.87 \times 10^{-6} \ s\) \[ L = .950 \times 3.00 \times 10^8 \ m/s \times 4.87 \times 10^{-6} \ s \] \[ L = 2.85 \times 10^8 \ m/s \times 4.87 \times 10^{-6} \ s \approx 1,386 \ m \] **Explanation:** This is the distance the muon covers before decaying, as seen from Earth. **Final Answer:** \[ \boxed{L = 1,386 \ m} \] --- Would you like to proceed to the next parts (c) and (d)?

Sure! Let's continue solving the remaining parts of the problem. --- ## (c) How far does the muon travel as viewed by an observer moving with it? ### **Step 1:** For an observer moving with the muon, the distance it travels is given by the formula: \[ L' = v \Delta t' \] where \(\Delta t' = 1.52 \ \mu s\) (the proper time). ### **Step 2:** Substituting the values: - \(v = 0.950c = 0.950 \times 3.00 \times 10^8 \ m/s\) - \(\Delta t' = 1.52 \ \mu s = 1.52 \times 10^{-6} \ s\) Calculating \(L'\): \[ L' = (0.950 \times 3.00 \times 10^8 \ m/s) \times (1.52 \times 10^{-6} \ s) \] \[ L' = (2.85 \times 10^8 \ m/s) \times (1.52 \times 10^{-6} \ s) \approx 434 \ m \] **Explanation:** This distance is shorter than what an Earth observer measures due to the muon's proper time being used for calculation. **Final Answer:** \[ \boxed{L' = 434 \ m} \] --- ## (d) What is the value of \(\gamma\) from the length contraction? ### **Step 1:** We already computed \(\gamma\) in part (a) using the formula: \[ \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} \] ### **Step 2:** We found: \[ \gamma \approx 3.207 \] **Explanation:** This factor indicates how much time and length are dilated or contracted due to the muon's high velocity relative to the speed of light. **Final Answer:** \[ \boxed{\gamma \approx 3.207} \] --- All parts have been solved. If you have any further questions, feel free to ask!

Sure! Let's summarize and solve all parts (a, b, c, d) of the problem in a structured manner. --- # Cosmic Ray and Muon Problem ## Given Data - Velocity of the muon: \( v = 0.950c \) - Proper lifetime of the muon: \( \Delta t' = 1.52 \ \mu s \) --- ## (a) Muon's Lifetime as Measured by an Earth-bound Observer ### **Step 1: Calculate \(\gamma\)** Use the time dilation formula: \[ \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} \] Substituting \( v = 0.950c \): \[ \gamma = \frac{1}{\sqrt{1 - (0.950)^2}} = \frac{1}{\sqrt{1 - 0.9025}} = \frac{1}{\sqrt{0.0975}} \approx 3.207 \] ### **Step 2: Calculate Dilated Lifetime** Using the time dilation formula: \[ \Delta t = \gamma \Delta t' \] Substituting known values: \[ \Delta t \approx 3.207 \times 1.52 \ \mu s \approx 4.87 \ \mu s \] **Final Answer:** \[ \boxed{\Delta t \approx 4.87 \ \mu s} \] --- ## (b) Distance Traveled According to Earth-bound Observer ### **Step 1: Use the Dilated Lifetime** Calculate the distance \( L \) using: \[ L = v \Delta t \] ### **Step 2: Substitute Values** Convert velocity to meters per second and time to seconds: - \( v = 0.950 \times 3.00 \times 10^8 \ m/s = 2.85 \times 10^8 \ m/s \) - \( \Delta t = 4.87 \times 10^{-6} \ s \) Now calculate: \[ L = (2.85 \times 10^8 \ m/s) \times (4.87 \times 10^{-6} \ s) \approx 1,386 \ m \] **Final Answer:** \[ \boxed{L \approx 1,386 \ m} \] --- ## (c) Distance Traveled as Viewed by an Observer Moving with the Muon ### **Step 1: Use Proper Time** Calculate the distance \( L' \) using: \[ L' = v \Delta t' \] ### **Step 2: Substitute Values** Using the proper time: - \( \Delta t' = 1.52 \times 10^{-6} \ s \) Calculate: \[ L' = (2.85 \times 10^8 \ m/s) \times (1.52 \times 10^{-6} \ s) \approx 434 \ m \] **Final Answer:** \[ \boxed{L' \approx 434 \ m} \] --- ## (d) Value of \(\gamma\) from Length Contraction ### **Step 1: Recall \(\gamma\) Calculation** As previously computed: \[ \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} \approx 3.207 \] ### **Step 2: Final Result for \(\gamma\)** This value indicates the factor for time dilation and length contraction. **Final Answer:** \[ \boxed{\gamma \approx 3.207} \] --- All parts of the problem have been solved comprehensively. Let me know if you need any further assistance!