An astronaut with mass 84kg is at rest in space, before firing her water pack to move toward the shuttle. If the amount of water shot out is 2kg and it is fired at a speed of 10m/s, what will the speed of the astronaut be?

The astronaut is moving at a speed of 0.238 m/s in a direction opposite the direction of the water shot out.

Explanation:

We are given;

Mass of astronaut; m1 = 84 kg

Mass of water shoot out; m2 = 2 kg

Initial speed of astronaut; u1 = 0 m/s

Initial speed of water shoot out; u2 = 0 m/s

Final speed of shoot out; v2 = 10 m/s

From law of conservation of momentum, we can say that;

Initial momentum = final momentum

Thus;

m1•u1 + m2•u2 = m1•v1 + m2•v2

Where v1 is the final speed of the astronaut

Plugging in the relevant values, we get;

(84 × 0) + (2 × 0) = (84 × v1) + (2 × 10)

0 = 84v1 + 20

-20 = 84v1

v1 = -20/84

v1 = -0.238 m/s

The negative sign indicates that the astronaut is moving 0.238 m/s in a direction opposite the direction of the water shot out.

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