DUE TONIGHT PLEASE HELPA 90 kg astronaut floating out in space is carrying a 1.0 kg TV camera and a 10 kg battery pack. He's drifting toward his ship but, in order to get back faster, he hurls the camera out into space (away from the space ship) at 18 m/s and then throws the battery at 11 m/s in the same direction. What's the resulting increase in his speed after each throw?speed increase after discarding camera.?speed increase after discarding battery?
The key here is to remember that momentum is conserved. Momentum is mass times velocity. Are you sure you've got the masses right, surely the TV camera is heavier than the battery pack? Since (relative to each other), neither astronaut, nor camera nor battery is moving, the total momentum can be taken to be zero. So after throwing the camera away, the total momentum should be zero m/s. That means the astronaut should have equal momentum in magnitude, but opposite in direction, to the camera. Let's take the direction the camera was thrown in to be negative. Then the momentum of the camera after it is released is (1.0 kg)(-18 m/s) -18 kg m/s. So the momentum after discarding the camera must be 18 kg m/s. Dividing by the mass of the astronaut (plus battery pack, as he's still got it at this point), which is 100 kg, gives an increase in speed of 0.18 m/s. Now we do the same thing with the battery pack. Since, relative to each other, neither are moving, we can say that the total momentum before releasing it is zero, so the total momentum after releasing it must be zero also. The momentum of the battery pack, since it is thrown in the negative direction also, is (10 kg)(-11 m/s) -110 kg m/s. Therefore the total momentum of the astronaut must increase by 110 kg m/s also, and dividing by the mass of the astronaut (90 kg) gives a speed increase of 1.22 m/s.
