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A man (mass = 50 kg) and his son (mass = 20 kg) are standing on a frictionless surface facing each other. The man pushes his son so that he starts moving at a speed of 0.70 m s⁻¹ with respect to the man. The speed of the man with respect to the surface is:

1. 0.28 m s⁻¹
2. 0.47 m s⁻¹
3. 0.20 m s⁻¹
4. 0.14 m s⁻¹

Correct answer: 0.20 m s⁻¹

Solution

Let man speed = v (backward), son speed = u (forward). Momentum: 20u = 50v -> u = 2.5v. Relative speed u + v = 0.70 -> 3.5v = 0.70 -> v = 0.20 m/s.

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