Suppose two carts, one twice as massive as the other, fly apart when the compressed spring that joins them is released. How fast does the heavier cart roll compared to the lighter cart?

To determine how fast the heavier cart rolls compared to the lighter cart, we can use the principle of conservation of momentum. According to this principle, the total momentum before the spring is released is equal to the total momentum after the spring is released.

Let's assume the initial velocities of both carts are zero, and the lighter cart has a mass of m and the heavier cart has a mass of 2m.

Initially, the total momentum is zero since both carts are stationary:
Total momentum before = momentum of the lighter cart + momentum of the heavier cart
= 0 + 0
= 0

After the spring is released, the carts move in opposite directions. Let's assume the lighter cart moves to the right with a velocity v1, and the heavier cart moves to the left with a velocity v2.

The momentum of an object is given by the product of its mass and velocity. So we have:
Total momentum after = momentum of the lighter cart + momentum of the heavier cart
= (m * v1) + (2m * v2)

According to the conservation of momentum, the total momentum before and after the spring is released should be equal:
0 = (m * v1) + (2m * v2)

Simplifying the equation, we can divide both sides by m:
0 = v1 + 2v2

Now we can solve this equation to find the relationship between the velocities of the two carts.

Since the masses of the carts are in a ratio of 1:2, we can write v2 = 2v1.

Substituting this value into the equation,
0 = v1 + 2(2v1)
0 = v1 + 4v1
0 = 5v1

So v1 = 0.

From the equation v2 = 2v1, we can conclude that v2 = 2 * 0 = 0.

Therefore, both carts come to rest after the spring is released. The velocity of the heavier cart rolling compared to the lighter cart is zero.