A(n) 70.1 kg astronaut becomes separated from the shuttle, while on a space walk. She finds herself 41.4 m away from the shuttle and moving with zero speed relative to the shuttle. She has a(n) 0.916 kg camera in her hand and decides to get back to the shuttle by throwing the camera at a speed of 12 m/s in the direction away from the shuttle. How long will it take for her to reach the shuttle

4.4 min

Explanation:

To solve this, we use the law of conservation of momentum

0 = -m(a).v(a) + m(c).v(c), where

m(a) = mass of the astronaut

m(c) = mass of the camera

v(a) = velocity of the astronaut

v(c) = velocity of the camera

Making v(a) subject of the formula, we have

v(a) = m(c).v(c)/m(a)

Now, we make one last assumption, that the astronaut is moving at constant speed and time, thus the time needed will be

t = s/v -> since s = vt

Now, we already have our v from above, so we substitute

t = s/[m(c).v(c)/m(a)]

t = s.m(a)/m(c).v(c)

Applying the values, we have

t = (41.4 * 70.1) / (0.916 * 12)

t = 2902.14 / 10.992

t = 264 seconds or 4.4 minutes

falling water hydro = water .. spectcular places these plants are

increase masses, decrease separation