A 2.3 kg cart is rolling across africtionless, horizontal track toward a 1.5 kg

Question

A 2.3 kg cart is rolling across africtionless, horizontal track toward a 1.5 kg cart that is heldinitially at rest. The carts are loaded with strong magnets thatcause them to attract one another. Thus, the speed of each cartincreases. At a certain instant before the carts collide, the firstcart’s velocity is +4.5 m/s, and the second cart’svelocity is -1.9 m/s. (a) What is the total momentum of thesystem of the two carts at this instant? (b) What was the velocityof the first cart when the second cart was still at rest?

Explanation / Answer

Find the momentum for each cart and add them together: 1st cart: p = mv = 2.3 kg * 4.5 m/s = 10.35 kg m/s 2nd cart: p = mv = 1.5 kg * -1.9 m/s = -2.85 kg m/s Adding together (note that one is negative): 10.35 kg m/s + -2.85 kg m/s = 7.5 kg m/s This is the total momentum of the system. To find the speed of the cart before the other cart started tomove, we must remember that momentum is always conserved, so thefinal momentum is equal to the initial momentum. Since theother cart's velocity is zero, we know that all the momentum is inthe first cart. So, p = mv v = p / m v = 7.5 / 2.3 = 3.3 m/s

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