A 75.0-kg paint cart with rubber bumpers is rolling 0.965 m/s to the right and strikes

Question

a second cart of mass 85.0 kg moving 1.30 m/s to the left. After the collision, the
heavier cart is traveling 0.823 m/s to the right. What is the velocity of the lighter cart
after the collision?

I'm confused to how to set this one up since two of the velocity is positve

Answer 1

whada-whada-wha?
This problem is basically the same setup as the last one you had previously asked.

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'

(75.0 kg)(0.965 m/s) + (85.0 kg)(-1.30 m/s) = (75.0 kg)v₁' + (85.0 kg)(0.823 m/s)

Solving for v₁', you should get -1.44 m/s. This means the lighter cart was moving at a rate of 1.44 m/s to the left.

Answer 2

thank you, similar but Had to divide by the 2nd mass instead of the 1st mass, Only difference

Answer 3

A small rectangular tank 5.00 in. By 9.00 in is filled with mercury.
a) if the total force on the bottom of the tank is 165lb, how deep is the mercury?
b) find total force on the large side of the tank