5. A 70 kg astronaut floating in an orbiting space station throws a 1.0 kg water bottle across the room at a
speed of 8.0 m/s. What is the magnitude of the astronaut's recoil velocity? (Hint: use mV = -M2V2
where v2 is the astronaut's recoil velocity – you know m1 = 1.0 kg, m2 = 70 kg and V1 = 8.0 m/s)

We can solve this problem by using the law of conservation of momentum. In fact, in absence of external forces, the total momentum of the astronaut+water bottle is conserved.

Before the astronaut throws the bottle, the total momentum is zero:

p = 0 (1)

After throwing it, the total momentum is:

[tex]p=mv+MV[/tex] (2)

Where

m = 1.0 kg is the mass of the water bottle

v = 8.0 m/s is the velocity of the water bottle

M = 70 kg is the mass of the astronaut

V is the recoil velocity of the astronaut

Since the total momentum is conserved, (1) = (2), so we can write:

[tex]0=mv+MV[/tex]

And solving for V, we find:

[tex]V=-\frac{mv}{M}=-\frac{(1.0)(8.0)}{70}=-0.11 m/s[/tex]

where the negative sign indicates that the direction is opposite tot he direction of the water bottle.

Learn more about momentum:

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