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A railroad cart with a mass of m1 = 11.9 t is at rest at the top of an h = 10.2 m high hump yard hill. After it is pushed very slowly over the edge, it starts to roll down. At the bottom it hits another cart originally at rest with a mass of m2 = 20.1 t. The bumper mechanism locks the two carts together. What is the final common speed of the two carts? (Neglect losses due to rolling friction of the carts. The letter t stands for metric ton in the SI system.)

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Answer:

Explanation:

m1 = 11.9 metric ton = 11.9 x 1000 = 11900 kg

h = 10.2 m

m2 = 20.1 metric ton = 20100 kg

Let u be the initial velocity of the cart 1 as it reaches the bottom and v be the velocity of both the carts as they coupled together.

According to the energy conservation

m1 x g x h = 1/2 m1 x u²

u² = 2 x g x h

u² = 2 x 9.8 x 10.2

u = 14.14 m/s

Use conservation of momentum

m1 x u = (m1 + m2) xv

11900 x 14.14 = (11900 + 20100) x v

v = 5.26 m/s

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