Question

A 89.4 kg astronaut is working on the engines of a spaceship that is drifting through space with a constant velocity. The astronaut turns away to look at Earth and several seconds later is 40.6 m behind the ship, at rest relative to the spaceship.The only way to return to the ship without a thruster is to throw a wrench directly away from the ship. The wrench has a mass of 0.613 kg, and the astronaut throws the wrench with a speed of 24.9 m/s. How long does it take the astronaut to reach the ship? Answer in units of s.

Answer 1

Answer:

It will take the astronaut 239.55 seconds to get to the ship after he has thrown the wrench.

Explanation:

The astronaut and the wrench are an isolated system; therefore, the law of conservation of momentum applies:

$M_wv_w = M_a v_a$,

where, $M_w$,$v_w$ and $M_a,v_a$ are the mass & velocity of the the wrench and he astronaut respectively.

Putting in numbers we get:

$(0.613kg)(24.9) = (89.4kg)v_a$

$v_a = \dfrac{(0.613kg)(24.9)}{89.4kg}$

$v_a = 0.171m/s$

which is the velocity of the astronaut.

With this velocity, the time it takes the astronaut to cover a distance of 40.6 will be

$t = \dfrac{40.0m}{0.171m/s}$

$\boxed{t = 239.55 s}$

Thus, it will take the astronaut 239.55 seconds to get to the ship after he has thrown the wrench.

P.S: we discounted the constant velocity of the ship because the astronaut was at rest relative to it (he was moving at the same velocity)