Question

A 70-kg astronaut is space walking outside the space capsuleand is stationary when the tether line breaks. As a means of returning to the capsule he throws his 2.0 kg space hammer at a speed of 14 m/s away from the capsule. At what speed does the astronaut move toward the capsule?

Answer 1

Answer:

0.4 m/s

Explanation:

Law of conservation of momentum tell us that the change in momentum of the hammer will be equal to the change in momentum of the astronaut

change in momentum of hammer = change in momentum of astronaut

2 kg (14 m/s - 0 m/s) = 70 kg * (v-0)

                                v  = 0.4 m/s

Answer 2

Answer:

The speed of the astronaut toward the capsule is $v_{a}=0.4\frac{m}{s}$

Explanation:

We have a system of two "particles" which are the astronaut and the hammer.

Initially, they are together and their relative velocities are zero, therefore the initial linear momentum is zero.

As there are no external forces to this system, the momentum is constant. This means that the initial momentum is equal to the final momentum:

$0=p_{i}=p_{f}=m_{h}v_{h}-m_{a}v_{a}$

where the mass and velocity with h subscript corresponds to the hammer, and the ones with a subscript corresponds to the astronaut.

Then, we clear the velocity of the astronaut, and calculate

$m_{h}v_{h}-m_{a}v_{a}\Leftrightarrow v_{a}=\frac{m_{h}}{m_{a}}v_{h}\Leftrightarrow v_{a}=\frac{2kg}{70kg}*14\frac{m}{s}=0.4\frac{m}{s}$

which is the speed of the astronaut toward the capsule.