Asked by Jay
A 75.0-kg paint cart with rubber bumpers is rolling 0.965 m/s to the right and strikes
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
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
Answers
Answered by
herp_derp
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.
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.
Answered by
Jay
thank you, similar but Had to divide by the 2nd mass instead of the 1st mass, Only difference
Answered by
Mary
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
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
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