Physics: Two ice skaters stand at rest in the center of an ice rink.?
Two ice skaters are resting at the center of an ice rink. When they push off against each other, the first skater, who weighs 70 kg, reaches a speed of 0.66 m/s. Meanwhile, the second skater accelerates to a speed of 0.83 m/s. What is the mass of the second skater?
3 Answers
Momentum before = momentum after
(M1+M2)*V = M1V1 + M2V2
0 = M1V1 + M2V2
M1v1 = - M2V2 [Taking 1 to represent 70kg skater, and 2 the other]
M2 = - (M1V1)/V2
M2 = - (70x0.66)/0.83
M2 = ~ - 55.7 [Minus represents motion in opposite direction]
Jan 19, 2025
conservation of momentum: m1v1 = m2v2. m1 = sixty six kg; v1 = 0.sixty 4 m/s. m2 = x; v2 = 0.80 3 m/s. m1v1 = (sixty six kg)(0.sixty 4 m/s) = 40 two kg·m/s (to 2 significant figures). x·v2 = m1v1, so x = m1v1 / v2 = (40 two kg·m/s) / (0.80 3 m/s) = fifty one kg.
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