4. A woman is 1.6 m tal and has a mass of 55 kg . She moves past an observer with the direction of the motion parallel to her height. The observer measures her relativistic momentum to have a magnitude of \( 2.0 \times 10^{10} \mathrm{~kg} \mathrm{~m} / \mathrm{s} \). What does the observer measure for her height?

We start by noting that in relativity the momentum is given by \[ p=\gamma m v, \] where the Lorentz factor is \[ \gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}. \] Since the woman’s rest height (proper length) is \( L_0=1.6~\mathrm{m} \) and she is moving in the direction parallel to her height, the observer measures her length as \[ L=\frac{L_0}{\gamma}. \] The observer measures a momentum of \[ p=2.0\times10^{10}~\mathrm{kg\,m/s}. \] It is useful to introduce \(\beta=\frac{v}{c}\) so that \[ p=\gamma m v=\gamma m c\,\beta. \] Let \[ A=\frac{p}{mc}. \] Then we have \[ \gamma\,\beta = A. \] Since \(\gamma=\frac{1}{\sqrt{1-\beta^2}}\), the above equation becomes \[ \frac{\beta}{\sqrt{1-\beta^2}}=A. \] Squaring both sides we obtain \[ \frac{\beta^2}{1-\beta^2}=A^2. \] Solving for \(\beta^2\): \[ \beta^2=A^2(1-\beta^2) \quad\Longrightarrow\quad \beta^2(1+A^2)=A^2, \] \[ \beta^2=\frac{A^2}{1+A^2}. \] Then the Lorentz factor is \[ \gamma=\frac{1}{\sqrt{1-\beta^2}}=\sqrt{1+A^2}. \] Now, substitute the known values. The mass is \( m=55~\mathrm{kg} \), and taking \( c=3.0\times10^8~\mathrm{m/s} \), we compute \[ A=\frac{p}{mc}=\frac{2.0\times10^{10}}{55\times3.0\times10^8}. \] First, calculate the denominator: \[ 55\times3.0\times10^8=1.65\times10^{10}, \] so \[ A=\frac{2.0\times10^{10}}{1.65\times10^{10}}\approx1.212. \] Then the Lorentz factor is \[ \gamma=\sqrt{1+A^2}=\sqrt{1+(1.212)^2}. \] Calculating, \[ (1.212)^2\approx1.468, \] thus \[ \gamma=\sqrt{1+1.468}=\sqrt{2.468}\approx1.57. \] Finally, the observed (contracted) height is \[ L=\frac{L_0}{\gamma}=\frac{1.6~\mathrm{m}}{1.57}\approx1.02~\mathrm{m}. \]