Question

suppose an astronaut (total mass of 250 kg, person and accessories) got into trouble and found themselves floating away from their space ship at 1 m/s. without thrusters she could not get back to the ship. thinking fast she took her air tank (50 kg) and threw it to slow her motion away from the ship. with what speed (in m/s) would she need to throw it to stop her from floating further away from the ship (so her final speed is zero)?

Explanation:

Step1: Apply the law of conservation of momentum

The initial momentum of the astronaut - air tank system is \(p_{i}=(m_{1} + m_{2})v_{i}\), where \(m_{1}=200\space kg\) (mass of astronaut without tank), \(m_{2} = 50\space kg\) (mass of tank), and \(v_{i}=1\space m/s\). The final momentum \(p_{f}=m_{1}v_{1f}+m_{2}v_{2f}\). Since \(v_{1f} = 0\space m/s\) (astronaut stops), by conservation of momentum \(p_{i}=p_{f}\), so \((m_{1}+m_{2})v_{i}=m_{1}v_{1f}+m_{2}v_{2f}\).

Step2: Solve for \(v_{2f}\)

Substitute the values into the equation \((200 + 50)\times1=200\times0+50\times v_{2f}\).
Simplify the left - hand side: \(250\times1 = 250\).
The equation becomes \(250=50v_{2f}\).
Solve for \(v_{2f}\): \(v_{2f}=\frac{250}{50}\).

Answer:

\(5\space m/s\)