A 0.2 kg plastic cart and a 20 kg lead cart can both roll without friction on a horizontal surface. Equal forces are used to push the two carts forward for a distance of 1 m, starting from rest. Which cart has the greater momentum, after the full 1m? After traveling 1 m, is the momentum of the plastic cart greater than, less than, or equal to the momentum of the lead cart? Explain.
2. Relevant equations
p = mv
Δs = (Vi+Vf) / 2 * (Vf-Vi) / a
F = ma
3. The attempt at a solution
Using kinematics, I found that the acceleration rate of each cart equals Vf^2 / 2, where Vf equals the final velocity of each cart.
Since Fnet = ma, I find that Fnet of the plastic cart = Fnet(c1) = 0.2kg * Vf(c1)^2 / 2 = Vf(c1)^2 / 10.
Fnet of the lead cart = Fnet(c2) = 20kg * Vf(c2)^2 / 2 = 10 * Vf(c2)^2
Since the problem tells us that the forces are equal for both carts, Fnet(c1) = Fnet(c2).
In other words, Vf(c1)^2 / 10 = 10 * Vf(c2)^2
Solving for Vf(c1), I find that Vf(c1) = 10*Vf(c2)
Since momentum P = mv, P(c1) = (0.2kg)(10*Vf(c2)) = 2 * Vf(c2).
P(c2) = 20kg*Vf(c2)
So I find that P(c1) = P(c2) / 10. Meaning P(c1) < P(c2).
There is no way to check my answer since there’s no answer key that I can find and I’m self-studying. Is my answer above correct? Did I go wrong anywhere in my thinking? Thanks a lot in advance.