An 79.5 kg astronaut is taking a space walk to work on the engines of his ship,

Question

An 79.5 kg astronaut is taking a space walk to work on the engines of his ship, which is drifting through space with a constant velocity. The astronaut, wishing to get a better view of the Universe, pushes against the ship and much later finds himself 30.7 m behind the ship. Without a thruster or tether, the only way to return to the ship is to throw his 0.507 kg wrench directly away from the ship. If he throws the wrench with a speed of 19.2 m/s relative to the ship, how long does it take the astronaut to reach the ship?

Explanation / Answer

momentum before = momentum after (79.5kg + 0.507kg)(0m/s) = (0.507kg)(-19.2m/s) + (79.5kg)v v- 9.7344/79.5 m/s v = 0.122m/s v = d/t so to go 30.7 m t = d/v t = 30.7m/0.122m/s t = 251 s note that the ship is moving at constant velocity and it is not given this would also have to be taken into account

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.