Question

A(n) 55.3 kg astronaut becomes separated
from the shuttle, while on a space walk. She
finds herself 34.1 m away from the shuttle
and moving with zero speed relative to the
shuttle. She has a(n) 0.594 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? Answer in minutes.
1. 4.409 min
2. 4.850 min
3. 4.189 min
4. 5.291 min
5. 5.512 min
6. 5.071 min
7. 3.748 min
8. 4.630 min
9. 3.527 min
10. 3.968 min

Answer 1

By the law of conservation of momentum, $0 = - m_a v_a + m_c v_c$, where $m_a, v_a$ and $m_c, v_c$ are masses and velocities of astronaut and camera respectively (astronaut is moving in the opposite direction of camera).

From the last equation, $v_a = \frac{m_c v_c}{m_a}$. Assuming that astronaut is moving with constant speed, the time to reach the shuttle is $t = \frac{s}{v_a} = \frac{s m_a}{m_c v_c}$, where $s$ is the distance to the shuttle.

Hence, calculating, obtain $t = \frac{34.1 m \cdot 55.3 kg}{0.594 kg \cdot 12 \frac{m}{s}} \approx 264 s \approx 4.409 min$, so the correct answer is 1).