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Two astronauts are taking a spacewalk outside the International Space Station. The first astronaut has a mass of 85.2 kg. The second has a mass of 65.4 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 1.2 m/s to the left. If the momentum of the system is conserved, what is the final velocity of the second person?
A. 1.3 m/s to the left
B. 1.6 m/s to the right
C. 1.6 m/s to the left
D. 1.3 m/s to the right

Answer :

Onyebuchinnajigo

The final velocity of the second person after the collision is determined as 1.6 m/s to the right.

Final velocity of the second person

The final velocity of the second person is determined by applying the principle of conservation of linear momentum.

m1u1 + m2u2 = m1v1 + m2v2

85.2(0) + 65.4(0) = 85.2(-1.2) + 65.4(v2)

0 = -102.24 + 65.4v2

65.4v2 = 102.24

v2 = 1.56 m/s to the right

Thus, the final velocity of the second person after the collision is determined as 1.6 m/s to the right.

Learn more about linear momentum here: https://brainly.com/question/7538238

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