woman is 1.6 m tall and has a mass of 55 kg . She moves past an observer with t rection of the motion parallel to her height. The observer measures her relativistic omentum to have a magnitude of \( 2.0 \times 10^{10} \mathrm{~kg} \mathrm{~m} / \mathrm{s} \). What does the observer mea r her height?

We start with the fact that the relativistic momentum is given by \[ p = \gamma m v, \] where \[ \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}. \] The observer measures the momentum to be \[ p = 2.0 \times 10^{10} \, \mathrm{kg \, m/s}, \] and the rest mass is \[ m = 55 \, \mathrm{kg}. \] Because the velocity is along her height, the length measured by the observer (parallel to the motion) will be Lorentz contracted. In her rest frame her height is \[ L_0 = 1.6 \, \mathrm{m}, \] and the contracted length is given by \[ L = \frac{L_0}{\gamma}. \] Our goal is to compute \( \gamma \) and then find \( L \). ### Step 1. Express \(\gamma\) in terms of \(v\) We have \[ \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}. \] The momentum equation is \[ \gamma m v = 2.0 \times 10^{10}. \] Thus, \[ \gamma v = \frac{2.0 \times 10^{10}}{m} = \frac{2.0 \times 10^{10}}{55} \approx 3.636 \times 10^8 \, \mathrm{m/s}. \] ### Step 2. Write the equation in terms of \(\beta\) Define \[ \beta = \frac{v}{c}, \] \[ \gamma = \frac{1}{\sqrt{1-\beta^2}}. \] Then, \[ \gamma v = \frac{\beta c}{\sqrt{1-\beta^2}}. \] So the equation becomes \[ \frac{\beta c}{\sqrt{1-\beta^2}} \approx 3.636 \times 10^8 \, \mathrm{m/s}. \] Taking \( c \approx 3.0 \times 10^8 \, \mathrm{m/s} \), we have \[ \frac{\beta \, (3.0 \times 10^8)}{\sqrt{1-\beta^2}} \approx 3.636 \times 10^8. \] Dividing both sides by \(3.0 \times 10^8\) gives \[ \frac{\beta}{\sqrt{1-\beta^2}} \approx 1.212. \] ### Step 3. Solve for \(\beta\) Square both sides: \[ \frac{\beta^2}{1-\beta^2} \approx (1.212)^2 \approx 1.468. \] Multiply both sides by \((1-\beta^2)\): \[ \beta^2 \approx 1.468 (1-\beta^2). \] This gives \[ \beta^2 \approx 1.468 - 1.468 \beta^2. \] Collect like terms: \[ \beta^2 (1 + 1.468) \approx 1.468, \] \[ \beta^2 \approx \frac{1.468}{2.468} \approx 0.595. \] Thus, \[ \beta \approx \sqrt{0.595} \approx 0.772. \] So the speed is \[ v = \beta c \approx 0.772 \times 3.0 \times 10^8 \approx 2.316 \times 10^8 \, \mathrm{m/s}. \] ### Step 4. Calculate \(\gamma\) Now compute \[ \gamma = \frac{1}{\sqrt{1-\beta^2}} = \frac{1}{\sqrt{1-0.595}} = \frac{1}{\sqrt{0.405}} \approx \frac{1}{0.636} \approx 1.573. \] ### Step 5. Determine the contracted height The Lorentz contraction formula is \[ L = \frac{L_0}{\gamma}. \] Substitute \( L_0 = 1.6 \, \mathrm{m} \) and \( \gamma \approx 1.573 \): \[ L \approx \frac{1.6}{1.573} \approx 1.016 \, \mathrm{m}. \] Thus, the observer measures the woman's height to be approximately \[ \boxed{1.02 \, \mathrm{m}}. \]