Question

Two astronauts are taking a spacewalk outside the International Space

Station. The first astronaut has a mass of 64.0 kg. The second has a mass of
58.2 kg. Initially, both astronauts have zero velocity relative to each other.
Then, the astronauts push against each other, giving the first astronaut a final
velocity of 0.8 m/s to the left. If the momentum of the system is conserved,
what is the final velocity of the second person?
OA. 1.1 m/s to the left
OB. 0.9 m/s to the right
C. 0.9 m/s to the left
D. 1.1 m/s to the right

Answer 1

The correct answer is D. 1.1 m/s to the right.

Step-by-step explanation:

Since the momentum of the system is conserved, the total momentum before and after the push must be equal. Since both astronauts initially have zero velocity, the total momentum before the push is zero.

Total momentum before push = Total momentum after push

0 = m1 * v1 + m2 * v2

where:

m1 is the mass of the first astronaut (64.0 kg)

v1 is the final velocity of the first astronaut (-0.8 m/s)

m2 is the mass of the second astronaut (58.2 kg)

v2 is the final velocity of the second astronaut

Solving for v2, we get:

v2 = -(m1 * v1) / m2

v2 = -(64.0 kg * -0.8 m/s) / 58.2 kg

v2 = 0.879 m/s

Since the first astronaut moves to the left, the second astronaut must move to the right to conserve momentum. Therefore, the final velocity of the second astronaut is 1.1 m/s to the right.

Answer: D