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The man of mass m₁ = 82 kg and the woman of mass m₂ = 80 kg are standing on opposite ends of the platform of mass m0 = 103 kg which moves with negligible friction and is initially at rest with s = 0. The man and woman begin to approach each other. Find the displacement s of the platform when the two meet if the displacement x₁ of the man relative to the platform is 9 m. The length l of the platform is 15.5 m.

Answer :

The displacement of the platform when the man and woman meet is -4.56 meters, which is in the opposite direction to the man's displacement, as derived from the conservation of momentum.

The conservation of momentum applies to the problem where a man and a woman are moving on a platform without friction. Since there are no external forces, the center of mass of the system must remain at rest. As the man and woman walk towards each other, the platform moves in the opposite direction to preserve the system's momentum.

The total displacement of the system (man, woman, and platform) is zero. Given the man moves 9 m relative to the platform, we can calculate the displacement of the platform by using the following equations, which stem from the conservation of momentum:

m1 * (s + x1) + m2 * s = 0

Solving for s:

s = -m1 * x1 / (m1 + m2)

Substituting the given values:

s = -(82 kg * 9 m) / (82 kg + 80 kg)

s = -738 kg ext{m} / 162 kg

s = -4.56 m

The negative sign indicates that the displacement of the platform, s, is in the opposite direction to the man's displacement.

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Rewritten by : Barada