Respuesta :
Answer:
15.8m/s
Explanation:
This problem can be solved by taking into account the conservation of the momentum. In this case the momentum of the astronaut and the bag of tools must equal the momentum of the astronaut and the bag of tool after the astronaut throws the bag.
Hence, we have
[tex]p_{before}=p_{after}\\(m_{a}+m_{b})v=m_{a}v_{a}+m_{b}v_{b}[/tex]
where ma and va are the mass and velocity of the astronaut, mb and vb are the mass and velocity of the bag, after the astronaut throw the bag. The velocity v is the velocity where the astronaut has the bag of tool
By taking into account that the velocity of the astronaut must be zero to keep him near of the space station, we have that vb = 0.
Thus
[tex](124 + 19.0)2.10=19.0v_{b}\\v_{b}=\frac{300.3}{19.0}=15.8m/s[/tex]
Answer:
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 15.81 m/s.
Explanation:
To solve the question, we list out the variables as follows,
Mass of astronaut, m₁ = 124 kg
Mass of the bag of tools, m₂ = 19.0 kg
Initial velocity of astronaut, v₁ = 2.10 m/s = Initial velocity of bag of tools v₂
Final velocity of astronaut, v₃ = 0 m/s, assuming the astronaut is brought to a stop
Velocity of the bag of tools = v₄
We can observe that the question is about conservation of linear momentum. Therefore we have, from the principle of conservation of linear momentum.
Initial total momentum = Final Total momentum
We then have
m₁v₁ + m₂v₂ = m₁v₃ - m₂v₄
Since v₁ = v₂ we have
(m₁ + m₂) × v₁ = m₁v₃ - m₂v₄
Plugging the values and solving for the required unknown variable we have
(124 kg + 19.0 kg) × 2.10 m/s = 124 kg × v₃ - 19.0 kg × v₄
Since v₃ = 0 m/s, we have
300.3 kg·m/s = 0 kg·m/s - 19.0 kg × v₄
∴ v₄ = [tex]\frac{300.3 kg\cdot m/s}{19,0 kg}[/tex] = 15.81 m/s
The minimum final speed of the bag of tools (with respect to the space station) that will keep the astronaut from drifting away forever = 15.81 m/s.
