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Newton's Law of Motion

Two skaters are in the exact center of a circular frozen pond. Skater 1 pushes skater 2 off with a force of 100 N for 1.69 seconds. If skater 1 has a mass of 34 kg and skater 2 has a mass of 72 kg, what is the relative velocity (v1 - v2) after the push to the nearest hundredth of a m/s?

After reaching the other shore, how fast, to the nearest tenth of a m/s, must skater 1 run around the lake to meet skater 2 at the opposite shore?

Solution Preview

F = 100 N = the force of the push
t = 1.69 s = the duration of the push
m_1 = 34 kg = the mass of the first skater
m_2 = 72 kg = the mass of the second skater

Let axis X be in the direction of motion of skater 1 after the push, so that v_1 > 0 and v_2 < 0.

The two skaters get momenta of the same absolute value but opposite in direction:

F*t = m_1*v_1 = -m_2*v_2
So we get:
v_1 = F*t/m_1 ...