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solong [7]
4 years ago
12

A 1.5 kg cart traveling at 1.2 m/s hits and bounces off a 0.75 kg cart that was at rest. The 0.75 kg cart moves forward at 2.0 m

/s. How fast does the first cart end up going?​
Physics
1 answer:
Andre45 [30]4 years ago
8 0

The law of conservation of momentum says that the total momentum in the system before and after the collision remains the same. Remember that <em>p = mv </em>(where p is momentum, m is mass, and v is velocity). To find the total momentum in the system, add up the momentum of each component.

Before the collision:

The momentum of the first cart is m*v = 1.5 * 1.2 = 1.8.

The momentum of the second cart is m*v = 0.75 * 0 = 0.

The total momentum is 1.8.

After the collision:

(where x is the unknown velocity):

The momentum of the first cart is m*v = 1.5x

The momentum of the second card is m*v = 0.75 * 2 = 1.5.

The total momentum is 1.5x + 1.5. Because of conservation of momentum, you know this is equal to the momentum before the collision:

1.8 = 1.5x + 1.5

Subtracting 1.5 from both sides:

0.3 = 1.5x

And dividing by 1.5:

x = 0.2 m/s forward (you know it is forward because it is positive)

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yvonne van gennip of the netherlands ice skated 10.0 km with an average speed of 10.8 m/s. suppose vang ennip crosses the finish
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By conservation of momentum, we will find that the mass is 4.97 kg.

So in the original system, we have two objects, the bouquet of flowers of mass M that is not moving and Yvonne, which has a mass of 63 kg and a speed of 10.8 m/s.

Then the total momentum of this system is:

P = (63kg)*(10.8m/s) + M*(0 m/s) = 680.4 kg*m/s.

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Then the final momentum is:

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If you want to learn more about momentum, you can read:

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