Question

two astronauts are taking a spacewalk outside the International Space Station the first astronaut has a mass of 64 kg the second has a mass of 58.2 kg initially both astronauts push against each other giving the first astronauts final velocity of .8m/s to the left if the momentum of the system is conserved what is the final velocity of the second person

Answer 1

Approximately
`0.88\; {\rm m \cdot s^(-1)}` to the right (assuming that both astronauts were originally stationary.)

Step-by-step explanation:

If an object of mass
`m` is moving at a velocity of
`v`, the momentum
`p` of that object would be
`p = m\, v`.

Since momentum of this system (of the astronauts) conserved:

`\begin{aligned} &(\text{Total Final Momentum}) \\ &= (\text{Total Initial Momentum})\end{aligned}`.

Assuming that both astronauts were originally stationary. The total initial momentum of the two astronauts would be
`0` since the velocity of both astronauts was
`0\!`.

Therefore:

`\begin{aligned} &(\text{Total Final Momentum}) \\ &= (\text{Total Initial Momentum})\\ &= 0\end{aligned}`.

The final momentum of the first astronaut (
`m = 64\; {\rm kg}`,
`v = 0.8\; {\rm m\cdot s^(-1)}` to the left) would be
`p_(1) = m\, v = 64\; {\rm kg} * 0.8\; {\rm m\cdot s^(-1)} = 51.2\; {\rm kg \cdot m \cdot s^(-1)}` to the left.

Let
`p_(2)` denote the momentum of the astronaut in question. The total final momentum of the two astronauts, combined, would be
`(p_(1) + p_(2))`.

`\begin{aligned} & p_(1) + p_(2) \\ &= (\text{Total Final Momentum}) \\ &= (\text{Total Initial Momentum})\\ &= 0\end{aligned}`.

Hence,
`p_(2) = (-p_(1))`. In other words, the final momentum of the astronaut in question is the opposite of that of the first astronaut. Since momentum is a vector quantity, the momentum of the two astronauts magnitude (
`51.2\; {\rm kg \cdot m \cdot s^(-1)}`) but opposite in direction (to the right versus to the left.)

Rearrange the equation
`p = m\, v` to obtain an expression for velocity in terms of momentum and mass:
`v = (p / m)`.

`\begin{aligned}v &= (p)/(m) \\ &= \frac{51.2\; {\rm kg \cdot m \cdot s^(-1)}}{64\; {\rm kg}} && \genfrac{}{}{0}{}{(\text{to the right})}{} \\ &\approx 0.88\; {\rm m\cdot s^(-1)} && (\text{to the right})\end{aligned}`.

Hence, the velocity of the astronaut in question (
`m = 58.2\; {\rm kg}`) would be
`0.88\; {\rm m \cdot s^(-1)}` to the right.