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Momentum vs. Kinetic Energy

  1. Oct 7, 2013 #1
    1. The problem statement, all variables and given/known data
    Two iceboats hold a race on a frictionless horizontal lake. The two iceboats have masses m and 2m. Each iceboats has an identical sail, so the wind exerts the same constant force F on each boat. The two ice boats start from rest and cross the finish line a distance s away. The total work done to accelerate each of the boats from rest are the same (because the net force and displacement were the same for both). Hence both iceboats cross the finish line with the same kinetic energy.

    a)Which iceboat crosses the finish line with greater momentum?

    b)Can you show that the iceboat with mass 2m has √2 times as much momentum at the finish line as the iceboat of mass m?


    2. Relevant equations
    J = p2-p1=ƩFΔt
    p=mv
    K=(1/2)mv2
    W=Fds


    3. The attempt at a solution
    I know that the boat with mass 2m will have greater momentum crossing the finish line by realizing the boat with the larger mass will take a longer amount of time for it to travel from rest to a distance s. Thus the impulse from the larger boat will be bigger. Since the iceboat starts
    from rest, this equals the iceboat's momentum p at the finish line:
    P=FΔt.

    Im having trouble with part b of the problem. This is my thought process:

    Both boats will cross the finish line with the same kinetic energy
    ∴ (1/2)mv2=(1/2)(2m)[(1\2)v2

    Half, of the heavier boat's, square speed must be equal to the square of the lighter one for the kinetic relation to be true. If this is the case then,
    cant I make new relation of speeds? what I mean is:
    (1/2)vheavier2=vlighter2
    after some algebra → (vheavier)/√2 = vlighter


    so

    pheavier=2mvheavier
    plighter=mvlighter


    so to find how much larger the momentum of the heavier boat is to divide the two using the substitution of the lighter velocity:
    (2m)(vheavier)/[(mvheavier)/√2] = 2√2


    why is my quantity two times larger than it should be?
     
  2. jcsd
  3. Oct 8, 2013 #2
    Equating the kinetic energies was a right choice... But the simplification was a bit confusing...
    This is where it went wrong. It has to be √2vheavier=vlighter....
    Other than that everything is fine, check your calculations and you will be on your way...
    Regards
     
  4. Oct 8, 2013 #3

    haruspex

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    I think it went wrong a little earlier.
    To clarify
    (1/2)mvlight2=(1/2)(2m)vheavy2
    Take it forwards from there.
     
  5. Oct 8, 2013 #4
    Ah, i see now. I didnt need that extra (1/2) factor. I also understand why I didnt need it. Thanks guys
     
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