Question

A(n) 72.6 kg astronaut becomes separated from the shuttle, while on a space walk. She finds herself 48.1 m away from the shuttle and moving with zero speed relative to the shuttle. She has a(n) 0.704 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.
(a) How long will it take for her to reach the shuttle?

Answer 1

Answer:

The time taken to get the shuttle is 6.91 minutes.

Explanation:

Given that.

Mass of an astronaut, $m_1=72.6\ kg$

Distance between astronaut and shuttle, d = 48.1 m

Mass of the camera, $m_2=0.704\ kg$

Speed of camera, v = 12 m/s

We need to find the time taken by the astronaut to reach her shuttle. Both shuttle and astronaut are at rest, their initial velocities will be equal to 0. Using the conservation of linear momentum as :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\m_1(0)+m_2(0)=m_1v_1+m_2v_2\\\\m_1v_1+m_2v_2=0\\\\v_1=\dfrac{-m_2v_2}{m_1}\\\\v_1=\dfrac{0.704\times 12}{72.6}\\\\v_1=0.116\ m/s$

The time taken to get the shuttle is given by :

$t=\dfrac{d}{v_1}\\\\t=\dfrac{48.1}{0.116}\\\\t=414.65\ s$

or

t = 6.91 minutes

So, the time taken to get the shuttle is 6.91 minutes.