An Astronut is working in space station when he notices that his safety rope tying him to the station has come undone. He is 5.00 meters away and drifting further away at 0.100 m/s. Fortunately he is holding a 1.00 kg hammer as well as a 15.00 kg antenna (that costs $100000). The mass of the Astronut by himself is 95.00 kg. He can throw the hammer at 10.00 m/s and the antenna at 2.00 m/s.
a)- clearly explain how (why) he can get back to safety by throwing the object(s). Which direction should he throw them?
b)- how long (if ever) will take him to get back to the space station by throwing the hammer ?
c)- how long will it take him to get back if he throws away the hammer and the expensive antenna?
(neglect any time difference between throwing the two objects)
s =5 m, m = 95 kg, m1= 1 kg, m2 = 15 kg, v=0.1 m/s, v1 = 10 m/s, v2= 2 m/s.
1. Hammer
(m+m1+m2) •v = m1•v1 – (m+m2)•u1
u1 = {m1•v1 - (m+m1+m2) •v}/{m+m2} =
={1•10 –(95+1+15) •0.1}/{95+15} = - 0.01 m/s.
Astronaut will move in previous direction (away from the station)
but at smaller speed.
2. Antenna
(m+m1+m2) •v = m2•v2 – (m+m1)•u2,
u2 = {m1•v1 - (m+m1+m2) •v}/{m+m1} =
={15•2 –(95+1+15) •0.1}/{95+11} = 0.2 m/s.
t1= s/u2 = 5/0.2 = 25 s.
2. Hammer+Antenna
(m+m1+m2) •v = m1•v1+ m2•v2 – m•u3,
u3= {m1•v1+ m2•v2 - (m+m1+m2) •v}/m =
={1•10+15•2 –(95+1+15) •0.1}/95 = 0.3 m/s.
t2 = s/u3 = 5/0.3 = 16.7 s.
a) The astronaut can get back to safety by making use of Newton's third law of motion, which states that for every action, there is an equal and opposite reaction. When the astronaut throws an object in space, the force exerted on the thrown object in one direction will result in an equal and opposite force acting on the astronaut in the opposite direction. He needs to throw the object(s) in the direction opposite to his current drift, so that the resulting force propels him towards the space station.
b) To calculate how long it will take for the astronaut to get back to the space station by throwing the hammer, we need to use the equation of motion:
distance = initial velocity * time + (1/2) * acceleration * time^2
The initial velocity is 10.00 m/s and there is no acceleration acting on the hammer (we are considering the momentum transfer as instantaneous and neglecting the gravitational force since the astronaut is in space). The distance the astronaut needs to cover is 5.00 meters.
distance = (10.00 m/s) * time + (1/2) * 0 * time^2
By simplifying the equation, we get:
5.00 m = 10.00 m/s * time
Solving for time:
time = 5.00 m / 10.00 m/s = 0.50 seconds
So, it will take the astronaut 0.50 seconds to get back to the space station by throwing the hammer.
c) Similarly, to calculate how long it will take for the astronaut to get back by throwing away the hammer and the expensive antenna, we again use the equation of motion. The initial velocity for the antenna is 2.00 m/s, and the distance is still 5.00 meters.
distance = (2.00 m/s) * time + (1/2) * 0 * time^2
Simplifying the equation, we get:
5.00 m = 2.00 m/s * time
Solving for time:
time = 5.00 m / 2.00 m/s = 2.50 seconds
Therefore, it will take the astronaut 2.50 seconds to get back to the space station by throwing away the hammer and the expensive antenna.