A 50 kg NASA astronaut is stranded, floating at rest, 100 m away from the international space station. Recalling Physics 101, she tosses her 3.0 kg wrench in the opposite direction from the space station, with a speed of 5.0 m/s.

a) How long does it take her to get to the space station?

b) How much energy did she have to expend in throwing the wrench?

a) Apply conservation of momentum to the (isolated) astronaut-wrench system; the inital momentum is zero, so we have (note that this is a one-dimensional problem):

p = 0 = mv + MV

where m and M are the masses of the wrench and astronaut, respectively, and v and V are their final velocities, respectively. Thus,

V = -mv/M = - (3.0 kg)(5.0 m/s)/(50 kg) = - 0.30 m/s

where the minus sign indicates the direction (toward the space station). The time of flight is thus

t = d/v

  = (100m)/(0.30 m/s) = 333 s

b) The intial mechanical energy of the system is zero (the astronaut and wrench are both at rest); the final mechanical energy is

E = ½ MV² + ½mv²

  = ½(50 kg)(0.3 m/s)² + ½(3.0 kg)(5.0 m/s)²

  = 2.25 J + 37.50 J

  = 39.75 J

All of this energy had to come from the astronaut's muscles (since there is no other source of energy) - not only does she throw the wrench away, but in doing so, she throws herself in the opposite direction!

Note: We did part a) as an example in the lecture (Monday, Oct 9th)...

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