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conservation of momentum: a 1200-kg ore cart is rolling at 10.8 m/s across a flat friction-free surface. a crane suddenly drops 858 kg of ore vertically into the cart. how fast does the cart move just after being loaded with the ore?

User Randomblue
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The final velocity of the cart after being loaded with ore is approximately 628.2 m/s.

The conservation of momentum principle states that the total momentum of a system remains constant before and after a collision or interaction. In this case, we have an ore cart rolling across a friction-free surface with an initial momentum. When the crane drops ore into the cart, the combined mass of the cart and the ore increases, resulting in a final momentum.

To find the final velocity of the cart after being loaded with the ore, we can use the principle of conservation of momentum. The initial momentum of the cart is given by its mass multiplied by its initial velocity, and the final momentum is given by the sum of the initial momentum and the momentum contributed by the dropped ore. Since there is no external force or friction acting on the system, the total momentum before and after the collision is conserved.

We can use the equation:

Initial momentum = Final momentum

mass of the cart times initial velocity of the cart = (mass of the cart + mass of the ore) times final velocity of the cart

Substituting the given values:

Mass of the cart = 1200 kg
Initial velocity of the cart = 10.8 m/s
Mass of the ore = 858 kg

Plugging these values into the equation, we get:

1200 kg 10.8 m/s = (1200 kg + 858 kg) final velocity of the cart

Solving for the final velocity:

1080 m/s 1200 kg = 2058 kg final velocity of the cart

Final velocity of the cart = 1080 m/s 1200 kg / 2058 kg = approximately 628.2 m/s

User Brettcvz
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