Question

A railroad cart with a mass of m1 = 11.6 t is at rest at the top of an h = 10.9 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 = 23.2 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.)

Answer 1

The final common speed of the two carts will be 69.3 m/sec.The momentum conservation principle is applied.

What is the law of conservation of momentum?

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

Unit conversion:

1 metric tons = 1000 kg

Given data;

m₁= 11.6 metric ton =11600 kg

m₂ = 23.2 metric ton =  23200 kg

Let v represent the combined velocity of the two carts once they are connected, and let u represent the starting velocity of cart 1 when it reaches the bottom.

Considering energy conservation;

$\rm m_1 g h = \frac{1}{2} m_1 \times u^2 \\\\ u^2 = 2gh\\\\ u^2 = 2 \times 9.8 \times 10.6 \\\\ u = 207.972 \ m/s$

From the conservation of momentum principle;

$\rm m_1 \times u = (m_1 + m_2) v\\\\ 11600 \times 207.972 = (11600 + 23200) \times v \\\\ v = 69.3 \ m/s$

Hence, the final common speed of the two carts will be 69.3 m/sec.

To learn more about the law of conservation of momentum refer;

https://brainly.com/question/1113396

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