Suppose now that the air carts below are both moving to the right initially. The cart to the left has a mass m1 and an initial speed v0; the cart to the right has an initial speed v0/2. If the center of mass of this system moves to the right with a speed 6/7 multiplied by v0, what is the mass of the cart on the right?
The answer is to be expressed in terms of mass 1.
I got .45 but that isn't correct.
m1*vo+m2(vo/2)=(m1+m2)6vo/7
solve for m2
To solve this problem, we need to apply the conservation of linear momentum. The center of mass will move to the right with a speed of 6/7 multiplied by v0 only if the total momentum of the system remains constant.
Let's denote the mass of the cart on the left as m1 and the mass of the cart on the right as m2.
The initial momentum of the system can be calculated as:
Initial momentum = (m1 * v0) + (m2 * (v0/2))
The final momentum of the system can be calculated as:
Final momentum = (m1 + m2) * (6/7 * v0)
According to the conservation of linear momentum, the initial and final momenta must be equal. Therefore, we have:
(m1 * v0) + (m2 * (v0/2)) = (m1 + m2) * (6/7 * v0)
Simplifying this equation further, we get:
m1 + m2 = (3/7) * m2
We can now solve for m2 in terms of m1:
m2 = (7/3) * m1
Therefore, the mass of the cart on the right, m2, is equal to (7/3) times the mass of the cart on the left, m1.
Note: The final answer is to be expressed in terms of mass 1, so the mass of the cart on the right is (7/3) * m1.