An unfortunate astronaut loses his grip during a spacewalk and finds himself floating away from the space station, carrying only a rope and a bag of tools. First he tries to throw a rope to his fellow astronaut, but the rope is too short. In a last ditch effort, the astronaut throws his bag of tools in the direction of his motion (away from the space station). The astronaut has a mass of 113 kg and the bag of tools has a mass of 16.0 kg. If the astronaut is moving away from the space station at 1.20 m/s initially, what is the minimum final speed of the bag of tools (with respect to the space station) that will keep the astronaut from drifting away forever?

Question

An unfortunate astronaut loses his grip during a spacewalk and finds himself floating away from the space station, carrying only a rope and a bag of tools. First he tries to throw a rope to his fellow astronaut, but the rope is too short. In a last ditch effort, the astronaut throws his bag of tools in the direction of his motion (away from the space station). The astronaut has a mass of 113 kg and the bag of tools has a mass of 16.0 kg. If the astronaut is moving away from the space station at 1.20 m/s initially, what is the minimum final speed of the bag of tools (with respect to the space station) that will keep the astronaut from drifting away forever?

Answer 1

m1v1 = m2v2

solve for v2

Answer 2

To solve this problem, we can apply the principle of conservation of momentum. According to this principle, the total momentum before the throw should be equal to the total momentum after the throw.

Let's denote the velocity of the astronaut after throwing the bag as Va (final velocity of the astronaut) and Vb (final velocity of the bag).

Before the throw, the momentum of the astronaut-bag system is given by:

Initial momentum = (mass of astronaut) * (initial velocity of the astronaut) + (mass of bag) * 0

Now, after throwing the bag, the momentum of the astronaut-bag system is given by:

Final momentum = (mass of astronaut) * (final velocity of the astronaut) + (mass of bag) * (final velocity of the bag)

Since the astronaut wants to have a minimum final speed, we can assume that the velocity of the bag is in the opposite direction of the astronaut's motion, which means the final velocity of the bag is negative.

Now let's set up the equation using the principle of conservation of momentum:

(mass of astronaut) * (initial velocity of the astronaut) = (mass of astronaut) * (final velocity of the astronaut) + (mass of bag) * (final velocity of the bag)

Plugging in the given values:
(113 kg) * (1.20 m/s) = (113 kg) * (Va) + (16.0 kg) * (Vb)

Simplifying the equation:
135.6 kg m/s = 113 kg * Va - 16.0 kg * Vb

As we assumed that the final velocity of the bag (Vb) is negative, we can rewrite the equation as:
135.6 kg m/s = 113 kg * Va + 16.0 kg * (-Vb)

Now let's rearrange the equation to solve for Vb:

135.6 kg m/s = 113 kg * Va - 16.0 kg * Vb
16.0 kg * Vb = 113 kg * Va - 135.6 kg m/s
Vb = (113 kg * Va - 135.6 kg m/s) / 16.0 kg

Substituting the given value of Va (1.20 m/s):
Vb = (113 kg * 1.20 m/s - 135.6 kg m/s) / 16.0 kg

Calculating the value:
Vb = (135.6 kg m/s - 135.6 kg m/s) / 16.0 kg

The final velocity of the bag, Vb, is 0 m/s. Therefore, the minimum final speed of the bag of tools (with respect to the space station) that will keep the astronaut from drifting away forever is 0 m/s.