A 70 kg woman and her 35 kg son are standing at rest on an icerink. They push ag

Question

A 70 kg woman and her 35 kg son are standing at rest on an icerink. They push against eachother for a time of 0.6 seconds,causing them to glide apart. The speed of the womanimmediately after they seperate is 0.55m/s . Assume thatduring the push, friction is negligible compared with the forcesthe people exert on eachother.
D)d. After the initial push, the friction that the ice exerts cannot be considered negligible, and the mother comes to rest after moving a distance of 7.0 m across the ice. If their coefficients of friction are the same, how far does the son move after the push?

Explanation / Answer

From the conservation of linear momentum we can say Mass * velocity of mother = mass * velocity of son velocity of son = 70 * 0.55 / 35 = 1.1 m/s Impulse = force * time = change of momentum. So Ft = 35 * 1.1 F = 35 * 1.1 / 0.6 = 64.17 N in a direction opposite to that of the mother. From the conservation of energy we can say work done by mother = work done by son Since the force acting on mother and son is the same magnitude Force * distance traveled by mother = Force * distance traveled by son So the son travels 7.0 m too.

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