Two skaters with masses of 65 kg and 45 kg, respectively, stand 8.0 m apart, eac

Question

Two skaters with masses of 65 kg and 45 kg, respectively, stand 8.0 m apart, each holding one end of a piece of rope. (a) If they pull themselves along the rope until they meet, how far does each skater travel? (Neglect friction.) (b) If only the 45-kg skater pulls along the rope until she meets her friend (who just holds the rope), how far does each skater travel?

I have found part (a) to be 3.27 m for the 65 kg skater and 4.72 m for the 45 kg skater. I'm pretty sure that it correct. However, I'm stuck on Part B. Thank you.

Explanation / Answer

part B: Mass of the first skater, m1 = 65 kg Mass of the second skater, m2 = 45 kg Distance, d = 8 m Solution: Both of them meet at the center of mass of the system. Center of mass wrt first skater: Xcm = (m1 x1 + m2 x2) / (m1 + m2)         = (65 * 0 + 45 * 8) / (65 + 45)         = 3.272 m Ans: Distance traveled by first skater ˜ 3.27 m Distance traveled by second skater = 8 - 3.272                                                       = 4.72 m Mass of the first skater, m1 = 65 kg Mass of the second skater, m2 = 45 kg Distance, d = 8 m Solution: Both of them meet at the center of mass of the system. Center of mass wrt first skater: Xcm = (m1 x1 + m2 x2) / (m1 + m2)         = (65 * 0 + 45 * 8) / (65 + 45)         = 3.272 m Ans: Distance traveled by first skater ˜ 3.27 m Distance traveled by second skater = 8 - 3.272                                                       = 4.72 m

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