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.)

Question

07-12-2019
Physics

contestada

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.)

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segunfarotade segunfarotade · Answer 1 · 2019-12-07T15:56:13+01:00

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 'a' and the stationary cart as 'b'.

[tex]Momentum, P = MaVa +MbVb[/tex]

[tex]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[/tex]

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