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A particle of mass m = 0.1 kg is released from rest at point A on a wedge of mass M = 2.4 kg that is free to slide on a frictionless horizontal floor. The particle slides down the smooth face AB of the wedge. At the instant the wedge has speed 0.2 m/s, what is the speed of the particle relative to the wedge (in m/s)?
- 4.8
- 5
- 7.5
- 10
Correct answer: 4.8
Solution
Horizontal momentum conservation gives m*v_px = M*V, so v_px = (M/m)*V = 24*0.2 = 4.8 m/s relative magnitude horizontally; the relative velocity along the incline corresponds to 4.8 m/s.
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