In pairs figure-skating competition, a 65-kg man and his 45-kg female partner stand facing each other both at rest on the ice. If they push apart and the woman has a velocity of 1.5 m/s eastward, what is the velocity of her partner? (Neglect friction.)

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Answer:

The final velocity of her partner is approximately -1.04 m/s or 1.04 m/s in the opposite direction to her direction of motion

Explanation:

The given parameters are;

The mass of the man, m₁ = 65 kg

The mass of the woman, mβ‚‚ = 45 kg

Taking the relative initial velocity of the man and the woman as 0 m/s, we have;

The initial velocity of the man, v₁₁ = 0 m/s

The initial velocity of the man, v₁₂ = 0 m/s

The final velocity of the woman, vβ‚‚β‚‚ = 1.5 m/s

The final velocity of the man = v₂₁

Therefore, we have, by the conservation of momentum principle;

The total initial momentum = The total final momentum

Which gives;

m₁ Γ— v₁₁ + mβ‚‚ Γ— v₁₂ = m₁ Γ— v₂₁ + mβ‚‚ Γ— vβ‚‚β‚‚

Substituting the known values;

65 Γ— 0 + 45 Γ— 0 = 65 Γ— v₂₁ + 45 Γ— 1.5

∴ 65 Γ— v₂₁ + 45 Γ— 1.5 = 0

45 Γ— 1.5 = - 65 Γ— v₂₁

v₂₁ = 45 Γ— 1.5/(-65) β‰ˆ -1.04 m/s

The final velocity of the man, her partner = v₂₁ β‰ˆ -1.04 m/s.