A 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 10 m/s and then throws the battery at 7 m/s in the same direction. What's the resulting increase in his speed after each throw? speed after discarding the camera (m/s). speed after discarding the battery(m/s). Please provide answer and explanation.
That is a conservation of momentum problem. Momentum is mass times velocity. The momentum of the thrown TV is 1kg * 10 m/s 10 kgm/s. The astronaut plus the battery pack he is still holding have a combined mass of 100 kg and have to gain 10 kgm/s of momentum in the opposite direction. So we write an equation: 10 kgm/s (90+10)kg * V m/s And solve for the unknown velocity V 10100V 10/100V 1/10V 0.1 m/s The astronaut and battery pack together speed up by 0.1 m/s Then do the same with the battery momentum10 kg * 7 m/s 70 kgm/s The 90 kg astronaut will gain 70 kgm/s of momentum in the opposite direction. 70 90 * V 70 / 90 V 7/9 V 0.77777 V m/s So the astronaut speeds up another 0.777 m/s to end up with a final velocity of 0.1 + 0.777 0.87777 m/s.
