A86.8 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 42.1 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 0.515 kg, and the astronaut throws the wrench with a speed of 19.6 m/s. how long does it take the astronaut to reach the ship

362 seconds, or 6 minutes, 2 seconds.

   This is an exercise in the conservation of momentum. For this problem, I'll use the initial coordinates and velocity of the astronaut as my frame of reference because it makes the math easier. Due to the law of conservation of momentum, the momentum before and after the astronaut throws the wrench has to remain the same. And since I'm using the starting situation of the astronaut as my frame of reference, that value is 0. But thankfully, momentum is a vector quantity and we can save the astronaut.
 The momentum of an object is mass times velocity. So the momentum of the wrench after being thrown is:
  -19.6 m/s * 0.515 kg = -10.094 kg*m/s

   Now to balance that, we need the astronaut to have a momentum of 10.094 kg*m/s which just happens to be the case (can't break the laws of physics). So let's do some division to get the velocity.
 10.094 kg*m/s / 86.8 kg = 0.116290323 m/s

   Yay! The astronaut is moving back to the shuttle at a reasonable velocity. But has 42.1 m to travel. Another situation for division: 42.1 m / 0.116290323 m/s = 362.0249653 s

    So rounding to 3 significant digits gives a travel time of 362 seconds, or just a couple of seconds longer than 6 minutes.

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