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mestny [16]
3 years ago
8

A 2.9-kg cart is rolling across a frictionless, horizontal track toward a 1.4-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 +4.7 m/s, and the second cart's velocity is −1.9 m/s. (Indicate the direction with the sign of your answer.)
Physics
1 answer:
VikaD [51]3 years ago
6 0

Answer:

The question is incomplete. I will assume you intend to find the total momentum of the two carts during collision. Therefore, we can use the conservation of momentum principle to get the total momentum at a certain instant before collision.

Explanation:

The conservation of momentum principle states that the initial net momentum of two bodies before collision is equal to the final net momentum after collision.

In this case, let's denote the rolling cart as <em>'a'</em> and the stationary cart as <em>'b'</em>.

Momentum, P = MaVa +MbVb

P=(2.9)(+4.7) + (1.4)(-1.9)\\P= 13.68 + (-2.66)\\P= 13.68-2.66\\P=+11.02 Kg.m/s

Therefore, the total momentum before collision is 11.02 Kg.m/s.

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Explanation:

In order to solve this question we need to know that  speed = \frac{distance}{time}. Then we need to convert 4 minutes into seconds and cm into m. We can do that by multiplying the number of minutes by 60 (because there is 60 seconds in one minute) and dividing the number of cm by 100 (because there is 100 cm in one m). So.......

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⇒ speed=\frac{distance}{time} = \frac{3m}{240s} = \frac{0.0125m}{s}

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