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 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
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.
thank you, similar but Had to divide by the 2nd mass instead of the 1st mass, Only difference
To solve this problem, you can use the principle of conservation of momentum. According to this principle, the total momentum of an isolated system remains constant before and after a collision.
Let's assume the initial velocity of the lighter cart (75.0 kg) is v1, and the initial velocity of the heavier cart (85.0 kg) is v2. The final velocity of the heavier cart is given as 0.823 m/s to the right. We need to find the final velocity of the lighter cart.
The principle of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision. Mathematically, this can be expressed as:
(mass of lighter cart * initial velocity of lighter cart) + (mass of heavier cart * initial velocity of heavier cart) = (mass of lighter cart * final velocity of lighter cart) + (mass of heavier cart * final velocity of heavier cart)
Substituting the given information into this equation, we have:
(75.0 kg * v1) + (85.0 kg * (-1.30 m/s)) = (75.0 kg * final velocity of lighter cart) + (85.0 kg * 0.823 m/s)
Now, we can rearrange the equation to solve for the final velocity of the lighter cart:
75.0 kg * v1 + (-110.5 kg·m/s) = 75.0 kg * final velocity of lighter cart + 69.755 kg·m/s
Combining like terms, we have:
75.0 kg * v1 - 75.0 kg * final velocity of lighter cart = 69.755 kg·m/s + 110.5 kg·m/s
75.0 kg * v1 - 75.0 kg * final velocity of lighter cart = 180.255 kg·m/s
Now, we can solve for the final velocity of the lighter cart by isolating the term:
-75.0 kg * final velocity of lighter cart = 75.0 kg * v1 - 180.255 kg·m/s
Dividing both sides by -75.0 kg:
final velocity of lighter cart = (75.0 kg * v1 - 180.255 kg·m/s) / -75.0 kg
Simplifying further, we have:
final velocity of lighter cart = -2.404 m/s + v1
Therefore, the final velocity of the lighter cart after the collision will be -2.404 m/s plus the initial velocity of the lighter cart (v1).