Outside the International Sapce Station, a 60 kg astronaut holding a 4.0 kg object (both initially at rest) throws the object at 10 m/s relative to the space station. A 50 kg astronaut, initially at rest, catches the object. What is the speed of the separation of the two astronauts?
because of the need to conserve momentum, the 60 kg astronaut acquires a velocity V1 such that
60 V2 = 4.0 * 10 m/s as soon as he throws the object. Thus V1 = 2/3 m/s
When the object is caught by the 50 kg astronaut, he or she acquires a velocity such that
4.0 x 2/3 = (50 + 4) V2
V2 = (8/3)/54 = 0.15 m/s
Check my numbers and reasoning
To find the speed of the separation of the two astronauts, we need to apply the principle of conservation of momentum.
The momentum of an object is given by the product of its mass and velocity. According to the conservation of momentum principle, the total momentum before the interaction should be equal to the total momentum after the interaction.
The initial total momentum of the system is zero since both astronauts are initially at rest.
After the object is thrown, the momentum of the first astronaut-object system is given by:
momentum1 = mass1 * velocity1 = (60 kg + 4.0 kg) * 10 m/s
After the second astronaut catches the object, the momentum of the second astronaut-object system is given by:
momentum2 = mass2 * velocity2 = (50 kg) * v
Since the total momentum before the interaction is equal to the total momentum after the interaction, we can set up an equation:
momentum1 = momentum2
(60 kg + 4.0 kg) * 10 m/s = (50 kg) * v
Simplifying the equation:
(64 kg) * 10 m/s = (50 kg) * v
640 kg*m/s = 50 kg * v
v = 640 kg*m/s / 50 kg
v ≈ 12.8 m/s
Therefore, the speed of the separation of the two astronauts is approximately 12.8 m/s.
To find the speed of the separation of the two astronauts after the object is thrown and caught, we can use the principle of conservation of momentum.
The conservation of momentum states that the total momentum of a system before an event is equal to the total momentum after the event, as long as no external forces act on the system.
Let's consider the system consisting of the two astronauts and the object before and after the throw.
Before the throw:
Total initial momentum = 0 kg*m/s (since both astronauts and the object are at rest)
After the throw:
The initial momentum of the astronaut who throws the object is given by:
Momentum of astronaut = mass of astronaut * velocity of astronaut
= 60 kg * 0 m/s (since the astronaut is at rest)
= 0 kg*m/s
The initial momentum of the object is given by:
Momentum of object = mass of object * velocity of object
= 4.0 kg * 10 m/s (since the object is thrown at 10 m/s)
= 40 kg*m/s
The total initial momentum of the system (astronaut + object) is:
Total initial momentum = 0 kg*m/s + 40 kg*m/s
= 40 kg*m/s
After the catch:
The object is caught by the second astronaut, who is initially at rest. Therefore, the final momentum of this astronaut is:
Final momentum of astronaut = mass of astronaut * final velocity of astronaut
= 50 kg * velocity of separation
According to the conservation of momentum, the total final momentum of the system should be zero.
Total final momentum = 0 kg*m/s
Since the momentum of the object is transferred to the astronaut who catches it, we can write:
Total final momentum = Final momentum of astronaut + Momentum of object
0 kg*m/s = 50 kg * velocity of separation + 40 kg*m/s
Now we can solve for the velocity of separation:
50 kg * velocity of separation = -40 kg*m/s
velocity of separation = (-40 kg*m/s) / 50 kg
velocity of separation ≈ -0.8 m/s (Note: The negative sign indicates that the astronauts are moving apart)
Therefore, the speed of the separation of the two astronauts is approximately 0.8 m/s.