A 2.3-kg cart is rolling along a frictionless, horizontal tracktowards a 1.4-kg

Question

A 2.3-kg cart is rolling along a frictionless, horizontal tracktowards a 1.4-kg cart that is held initially at rest. The carts areloaded with strong magnets that cause them to attract one another.Thus, the speed of each cart increases. At a certain instant beforethe carts collide, the first cart's velocity is +3.7 m/s, and thesecond cart's velocity is -2.6 m/s. (a) What isthe total momentum of the system of the two carts at this instant?(b) What was the velocity of the first cart whenthe second cart was still at rest?

Explanation / Answer

Let mA = 2.3 kg and mB = 1.4 kg To find the total momentum of the system use: mAvA + mBvB = totalmomentum (2.3)(3.7) + (1.4)(-2.6) = 4.87 N*s The total momentum of the system is conserved: mAvA + mBvB = 4.87N*s In this situation, vB = 0. You are left with: mAvA = 4.87 N*s (2.3)vA = 4.87 N*s Solve for vA. vA = 2.12 m/s

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