A 2.1-kg cart is rolling along a frictionless, horizontal track towards a 1.3-kg

Question

A 2.1-kg cart is rolling along a frictionless, horizontal track towards a 1.3-kg cart that is held initially at rest. The carts are loaded with strong magnets that cause them to attract one another. Thus, the speed of each cart increases. At a certain instant before the carts collide, the first cart's velocity is +5.6 m/s, and the second cart's velocity is -1.2 m/s. (a) What is the total momentum of the system of the two carts at this instant? (b) What was the velocity of the first cart when the second cart was still at rest?

Explanation / Answer

m1 = 2.1 , m2 = 1.3 kg , v1 = +5.6 m/s , v2 = -1.2 m/s

total momentum of the system = m1*v1+m2*v2 = 2.1*5.6 + 1.2 * (- 1.2) = 10.32 kg.m/s in right

since momentum is a vector quantity so we have to consider direction of carts in which they are moving and taking right as positive .

apply momentum conservation

Pi = Pf

10.32 = 2.1 *v1f + 0

v1f = 10.32/ 2.1 = 4.91 m/s in right

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