A 2.3-kg cart is rolling along a frictionless, horizontal tracktowards a 1.7-kg

Question

A 2.3-kg cart is rolling along a frictionless, horizontal tracktowards a 1.7-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 +3.3 m/s, and thesecond cart's velocity is -2.8 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

The total momentum is given bym1v1+m2v2=2.3*3.3+1.7*-2.8=2.83kgm/s b) We know momenum is conserved by the Law of Conservation ofMomentum, so 2.83 was the momentum of the carts initially. Thesecond cart wasn't moving, so the first cart had all of themomentum at that point. So solve the equation 2.83=2.3kg*v v=1.23m/s b) We know momenum is conserved by the Law of Conservation ofMomentum, so 2.83 was the momentum of the carts initially. Thesecond cart wasn't moving, so the first cart had all of themomentum at that point. So solve the equation 2.83=2.3kg*v v=1.23m/s

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