An astronaut on a space walk bumps the shuttle and starts moving away at a velocity of 0.02m/s. The astronauts mass is 100kg. He has takes a 1kg "safety week" and shoves it away in exactly the direction of his motion at a speed of 6m/s. What speed does the astronaut move back toward the space shuttle?
To find the speed at which the astronaut moves back toward the space shuttle, we need to apply the Law of Conservation of Momentum. According to this law, the momentum before the collision is equal to the momentum after the collision.
Let's define the following variables:
- Mass of the astronaut (m₁) = 100 kg
- Mass of the safety weight (m₂) = 1 kg
- Initial velocity of the astronaut (v₁i) = 0.02 m/s (moving away from the shuttle)
- Initial velocity of the safety weight (v₂i) = 6 m/s (shoved away in the same direction)
Using the formula for momentum (p = mv), we can calculate the initial momentum of the astronaut (p₁i) and the safety weight (p₂i) separately:
p₁i = m₁ * v₁i
p₂i = m₂ * v₂i
Now, since the system is isolated and there are no external forces acting on it, the total momentum before the collision is equal to the total momentum after the collision. Therefore, we have:
p₁i + p₂i = p₁f + p₂f
Since the safety weight is shoved away in the same direction as the astronaut's motion, the final velocity of the safety weight (v₂f) will be the same as its initial velocity (v₂i). Let's assume the final velocity of the astronaut (v₁f) is in the opposite direction.
Now, applying the conservation of momentum equation:
m₁ * v₁i + m₂ * v₂i = m₁ * v₁f + m₂ * v₂f
Plugging in the given values:
100 kg * 0.02 m/s + 1 kg * 6 m/s = 100 kg * v₁f + 1 kg * 6 m/s
Simplifying the equation:
2 kg m/s = 100 kg * v₁f
Finally, solving for v₁f (the velocity at which the astronaut moves back toward the space shuttle):
v₁f = 2 kg m/s / 100 kg
v₁f = 0.02 m/s
Therefore, the astronaut moves back toward the space shuttle at a velocity of 0.02 m/s.