if two carts, one twice as massive as the other, fly apart when a 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 and after an interaction remains the same if no external forces are involved.
Initially, the two carts are at rest, so their total momentum is zero. When the spring is released and the carts fly apart, they move in opposite directions. Let's denote the velocity of the lighter cart as v lighter and the velocity of the heavier cart as v heavier.
Since momentum is conserved, the momentum of the lighter cart after the spring is released is equal in magnitude but opposite in direction to the momentum of the heavier cart. Mathematically, we can express this as:
m lighter * v lighter = -m heavier * v heavier
where m lighter and m heavier are the masses of the lighter and heavier carts, respectively.
Given that the heavier cart is twice as massive as the lighter cart (m heavier = 2 * m lighter), we can substitute this into the equation:
m lighter * v lighter = -(2 * m lighter) * v heavier
Simplifying further, we find:
v lighter = -2 * v heavier
Therefore, the velocity of the lighter cart is twice the magnitude but opposite in direction to the velocity of the heavier cart. In other words, if the heavier cart rolls to the right with a certain speed, the lighter cart will roll to the left with twice that speed.
To summarize, the heavier cart rolls at half the speed of the lighter cart.
MV=mv
V=m/M v