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 ma=113 kgma=113 kg and the bag of tools has a mass of mb=13.0 kg. mb=13.0 kg. if the astronaut is moving away from the space station at =1.20 m/svi=1.20 m/s initially, what is the minimum final speed ,fvb, f of the bag of tools with respect to the space station that will keep the astronaut from drifting away forever?
v = Velocity of the combined mass of astronaut and tools = 1.8 m/s
= Mass of astronaut = 124 kg
= Mass of tools = 16 kg
= Velocity of astronaut = 0
= Velocity of tools
As linear momentum is conserved
The velocity of the tools is 15.75 m/s
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
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.
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.
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 = = 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.
The velocity of the bag will be 16.08 m/s
Mass of the astronaut, M = 124 kg
Mass of the bag, m = 10 lg
Initial speed of the astronaut with bag, = 1.20 m/s
Applying the concept of conservation of momentum
(M + m) = Mv₁ + mv₂
v₁ is the final velocity of the astronaut = 0
v₂ is final velocity of the bag
thus, on substituting the values, we have
(124 + 10)1.20 = 124 × 0 + 10v₂
v₂ = 16.08 m/s
Hence, the velocity of the bag will be 16.08 m/s
a.) travels faster in solids because the particles are closer together.
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