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Irina18 [472]
3 years ago
6

An 80-kg astronaut becomes separated from his spaceship. He is 15.0 m away from it and at rest relative to it. In an effort to g

et back, he throws a 500 gram object with a speed of 8.0 m/s in a direction away from the ship. How long does it take him to get back to the ship?
Physics
1 answer:
Allushta [10]3 years ago
3 0

Answer: 300 s

Explanation:

The momentum p is given by the following equation:

p=m.V

Where m is the mass of the object  and V is the velocity.

In addition, according to the conservation of linear momentum, we have:

p_{1}=p_{2} (1)

Where:

p_{1}=0 is the initial momentum of the astronaut, which is initially at rest

p_{2}=m_{object}V_{object}+m_{astronaut}V_{astronaut} is the final momentum, being m_{object}=500g=0.5 kg, V_{object}=8 m/s and  m_{astronaut}=80 kg

Then (1) is rewritten as:

0=m_{object}V_{object}+m_{astronaut}V_{astronaut} (2)

Finding te velocity of the astronaut V_{astronaut}:

V_{astronaut}=-\frac{m_{object}V_{object}}{m_{astronaut}} (3)

V_{astronaut}=-\frac{(0.5 kg)(8 m/s)}{80 kg} (4)

V_{astronaut}=-0.05 m/s (5) The negative sign of the velocity indicates it is directed towards the spaceship, however its speed (the magnitude of the velocity vector) is positive 0.05 m/s

On the other hand we have the following:

V_{astronaut}=\frac{d}{t} (6)

Where d=15 m the distance between the astronaut and the spacheship and t the time. So, we have to find t:

t=\frac{d}{V_{astronaut}} (7)

t=\frac{15 m}{0.05 m/s}

Finally:

t=300 s

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