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Homework Help: Momentum question involving two objects of varying mass

  1. Feb 16, 2014 #1
    1. The problem statement, all variables and given/known data

    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.
  2. jcsd
  3. Feb 16, 2014 #2


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    Hello, goraemon. Welcome to PF!

    Your work looks correct.

    If you have studied the concept of impulse and also the impulse-momentum theorem, then you can use them to get the answer more quickly.

    [The work-energy theorem is another approach.]
  4. Feb 16, 2014 #3
    Thanks for your response TSny. I do recall reading about the impulse-momentum theorem...change in momentum = definite integral of the Force over some time interval, right?

    So, let me see if I got this...since the plastic cart is lighter in mass than the lead cart, the same force exerted on the plastic cart will cause it to accelerate much faster, meaning it will cover the same distance in less time.

    So this will result in a shorter time interval for the application of the force, meaning the definite integral of the force will be smaller...therefore, the change in momentum will likewise be smaller for the plastic cart. Is this the right reasoning to use?
  5. Feb 16, 2014 #4


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    Beautiful! :smile:
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