Posted by Miso on .
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)

Physics momentum please helppp!!! 
Elena,
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.