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 13.0 kg. If the astronaut is moving away from the space station at 2.10 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?
the astronaut+bag has momentum (113+13.0)*2.1 = 264.6 kg-m/s relative to the station
We want him to have momentum < 0, so he will be drifting back toward the station.
So, the bag must have all the previous momentum.
13.0 * v = 264.6
v = 20.35 m/s
With this bag velocity, the astronaut comes to a halt. Anything greater causes him to drift backwards.
Well, it seems that the unfortunate astronaut is in quite a "tight" situation. Let's see if we can "orbit" around this problem and find a solution.
To prevent the astronaut from drifting away forever, the backward force exerted on the astronaut should be equal in magnitude but opposite in direction to the forward force exerted on the bag of tools. This is due to Newton's third law of motion, which states that for every action, there is an equal and opposite reaction.
Let's denote the final speed of the bag of tools as Vf. Since the astronaut is initially moving away from the space station at 2.10 m/s, the change in velocity of the astronaut, ΔVa, will be 2.10 m/s. The change in velocity of the bag of tools, ΔVb, will be the final velocity of the bag of tools, Vf, minus 0 m/s (since the bag was initially at rest).
According to the law of conservation of momentum, the total momentum before the astronaut throws the bag of tools should be equal to the total momentum after he throws it. The momentum, p, of an object can be calculated by multiplying its mass, m, by its velocity, v.
Before the astronaut throws the bag of tools:
Momentum of astronaut (Pa) = (Mass of astronaut) x (Initial velocity of astronaut)
Pa = (113 kg) x (2.10 m/s)
After the astronaut throws the bag of tools:
Momentum of astronaut (P'a) = (Mass of astronaut) x (Final velocity of astronaut)
P'a = (113 kg) x (2.10 m/s - ΔVa)
Momentum of bag of tools (Pb) = (Mass of bag of tools) x (Final velocity of bag of tools)
Pb = (13.0 kg) x (Vf - 0 m/s)
Since momentum is conserved, we can set up the equation:
Pa = P'a + Pb
(113 kg) x (2.10 m/s) = (113 kg) x (2.10 m/s - ΔVa) + (13.0 kg) x (Vf - 0 m/s)
Now, let's solve for ΔVa and Vf. Bring all the terms to one side of the equation and solve:
(113 kg) x (2.10 m/s - ΔVa) + (13.0 kg) x (Vf - 0 m/s) - (113 kg) x (2.10 m/s) = 0
Doing some calculations (and taking the liberty to round the values):
(113 kg) x (2.10 m/s - ΔVa) + (13.0 kg) x Vf - (238 kg·m/s) = 0
(237.30 kg·m/s - ΔVa x 113 kg) + (13.0 kg) x Vf = (238 kg·m/s)
237.30 kg·m/s + 13.0 kg·Vf - ΔVa x 113 kg = 238 kg·m/s
13.0 kg·Vf = ΔVa x 113 kg
Vf = (ΔVa x 113 kg) / 13.0 kg
Vf = (2.10 m/s x 113 kg) / 13.0 kg
Vf ≈ 18.34 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 approximately 18.34 m/s.
Now, I hope the astronaut can "orbit" back to safety with the help of those tools! Safe travels!
To solve this problem, we can make use of the principle of conservation of momentum.
The initial momentum of the astronaut is given by:
Momentum of the astronaut = mass of astronaut × initial velocity of the astronaut
Let's calculate the initial momentum of the astronaut:
Momentum of the astronaut = (mass of astronaut) × (initial velocity of the astronaut)
= 113 kg × 2.10 m/s
= 237.3 kg·m/s
According to the conservation of momentum, the total momentum before and after the astronaut throws the bag of tools must be the same.
After throwing the bag of tools, the astronaut will start moving in the opposite direction, so his final momentum will be negative:
Final momentum of the astronaut = (mass of astronaut) × (final velocity of the astronaut)
Let's calculate the final momentum of the astronaut:
Final momentum of the astronaut = 113 kg × (-final velocity of the astronaut) (note the negative sign)
= -113 kg × (final velocity of the astronaut)
The final momentum of the bag of tools will be:
Final momentum of the bag of tools = (mass of the tools) × (final velocity of the tools)
According to the conservation of momentum:
Initial momentum of the astronaut + Initial momentum of the bag of tools = Final momentum of the astronaut + Final momentum of the bag of tools
Substituting the values we have:
237.3 kg·m/s + 0 = -113 kg × (final velocity of the astronaut) + (13.0 kg) × (final velocity of the tools)
Simplifying the equation:
237.3 kg·m/s = 13.0 kg × (final velocity of the tools) - 113 kg × (final velocity of the astronaut)
To prevent the astronaut from drifting away forever, the final velocity of the astronaut must be zero. Therefore, we can simplify the equation further:
237.3 kg·m/s = 13.0 kg × (final velocity of the tools)
Now we can solve for the final velocity of the tools:
(final velocity of the tools) = (237.3 kg·m/s) / (13.0 kg)
= 18.28 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 18.28 m/s.
To solve this problem, we need to apply the law of conservation of momentum. According to this law, the total momentum before the throw is equal to the total momentum after the throw.
First, let's calculate the initial momentum of the astronaut before the throw:
Momentum of the astronaut = mass of the astronaut × initial velocity of the astronaut
= 113 kg × 2.10 m/s
= 237.3 kg·m/s
Since the astronaut is moving away from the space station, the direction of his momentum is positive.
Next, let's calculate the initial momentum of the bag of tools before the throw:
Momentum of the bag of tools = mass of the bag of tools × initial velocity of the bag of tools
Now, here's the key concept to understand: when the astronaut throws the bag of tools in the opposite direction (away from the space station), the momentum of the bag of tools becomes negative because the direction of its motion is opposite to that of the astronaut.
Now, let's assume that the final speed of the bag of tools is V (with respect to the space station) after the throw. The final velocity of the astronaut will be the sum of his initial velocity and the final velocity of the bag of tools, so it will be 2.10 m/s - V m/s.
Using the law of conservation of momentum, we can write the equation:
Initial momentum of the astronaut = Final momentum of the astronaut + Final momentum of the bag of tools
237.3 kg·m/s = 113 kg × (2.10 m/s - V m/s) + (13.0 kg) × (-V m/s)
Now we can solve for V.
First, multiply and distribute:
237.3 kg·m/s = 237.3 kg·m/s - 113 kg × V m/s + (-13.0 kg) × V m/s
Next, combine like terms:
0 = -113 kg × V m/s + (-13.0 kg) × V m/s
Now simplify further:
0 = -126.0 kg × V m/s
Dividing both sides by -126.0 kg gives us:
V = 0 m/s
This means 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. In other words, the bag of tools must come to a complete stop in order to stop the astronaut from drifting away.