Question

A woman is 1.6 m tall and has a mass of 50 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 2.1 × 1010 kg·m/s. What does the observer measure for her height?

Answer 1

To solve this problem it is necessary to apply the concepts related to linear momentum, velocity and relative distance.

By definition we know that the relative velocity of an object with reference to the Light, is defined by

$V_0 = \frac{V}{\sqrt{1-\frac{V^2}{c^2}}}$

Where,

V = Speed from relative point

c = Speed of light

On the other hand we have that the linear momentum is defined as

P = mv

Replacing the relative velocity equation here we have to

$P = \frac{mV}{\sqrt{1-\frac{V^2}{c^2}}}$

$P^2 = \frac{m^2V^2}{1-\frac{V^2}{c^2}}$

$P^2 = \frac{P^2V^2}{c^2}+m^2V^2$

$P^2 = V^2 (\frac{P^2}{c^2}+m^2)$

$V^2 = \frac{P^2}{\frac{P^2}{c^2}+m^2}$

$V^2 = \frac{(2.1*10^10)^2}{\frac{(2.1*10^10)^2}{(3.8*10^8)^2}+50^2}$

$V = 2.81784*10^8m/s$

Therefore the height with respect the observer is

$l = l_0*\sqrt{1-\frac{V^2}{c^2}}$

$l = 1.6*\sqrt{1-\frac{(2.81*10^8)^2}{(3*10^8)^2}}$

$l = 0.56m$

Therefore the height which the observerd measure for her is 0.56m