two carts are on a frictionless surface. a compressed spring acts upon te carts, and the 1.5 kg cart moves with a velocity 27 cm/s to the left. what is the velocity of the 4.5 kg cart?
To solve this problem, 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.
1. Let's assume that the positive direction is to the right, and the negative direction is to the left.
2. The momentum of an object is calculated by multiplying its mass by its velocity. Therefore, the momentum of the 1.5 kg cart before the spring is released is given by:
Momentum_1 = mass_1 * velocity_1
= 1.5 kg * (-27 cm/s)
= -40.5 kg·cm/s
3. Since the total momentum is conserved, the total momentum after the spring is released is also -40.5 kg·cm/s.
4. Let's assume the velocity of the 4.5 kg cart after the spring is released is v₂.
5. The momentum of the 4.5 kg cart is given by:
Momentum_2 = mass_2 * velocity_2
= 4.5 kg * v₂
6. According to the principle of conservation of momentum, the total momentum before and after should be equal. Hence, we can express this as an equation:
Momentum_1 + Momentum_2 = Total Momentum
7. Substituting the values we have:
-40.5 kg·cm/s + 4.5 kg * v₂ = -40.5 kg·cm/s
8. Solving the equation:
-40.5 kg·cm/s + 4.5 kg * v₂ = -40.5 kg·cm/s
4.5 kg * v₂ = 0 kg·cm/s (subtracting -40.5 kg·cm/s from both sides)
v₂ = 0 cm/s (dividing by 4.5 kg)
Therefore, the velocity of the 4.5 kg cart is 0 cm/s. This means that the 4.5 kg cart does not move in either direction after the spring is released.