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Homework Help: Magnitude of the change in momentum

  1. Jun 6, 2007 #1
    In a test the total braking force on the car of mass 1000kg was measured on a straight, level road. The figure below shows the resultant force on the car as a function of time from the instant the brakes were apploed (t=0s). The car came to a stop after 10s.

    http://img412.imageshack.us/img412/2132/asdfem1.png [Broken]

    a) what is the magnitude of the change in momentum of the car from 0 to 10 seconds?

    b) what was the intial speed of the car at t=0s?

    3. The attempt at a solution

    I'm not sure what to apply/where to start :confused:
    Somebody please guide me through?

    Thanks in advance!
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Jun 7, 2007 #2


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    The change in momentum is given by the impulse that the braking force (friction between tires and road) applies to the car.

    The impulse is generally given by

    [tex]I = F \Delta t = m_f v_f - m_i v_i[/tex]

    but in this case we have that the force changes with time so we have to sum all the products of the (changing) force and small time intervals. This exactly what is achieved by integration.

    [tex]I = \int{F\ dt}[/tex]

    but the integration above will actually give us the area under the [tex]F(t)[/tex] graph. So to obtain the total impulse that the car experienced during the ten seconds of braking you need to calculate the total area under the curve.
  4. Jun 7, 2007 #3
    Thanks, got it to be 22500kgm/s

    For part b) do I use the equation:
    deltap=m(v final - v initial)

    and hence

    22500 = 1000(0-v initial)

    and hence

    v initial = -22.5m/s?
  5. Jun 7, 2007 #4


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    Yes. That fine.

    Note that the direction of the braking force was implicitly taken as the positive direction in the graph. That is why you get the initial velocity in the opposite direction.
  6. Jun 7, 2007 #5
    I have a feeling I would be marked incorrect on that then..

    So to resolve this problem I should change deltaP to -22500 for the second part of the question so I get a positive initial velocity then?
  7. Jun 7, 2007 #6


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    You have a choice in choosing the positive direction. If one were to choose the direction of the braking force as negative the impulse would be in the negative direction, giving the total impulse as (- 22 500 kgm/s, which would then mean one would plot the force on the negative axis). This would then give you a positive initial velocity, that is the car went in the opposite direction as that of the braking force (again).
  8. Jun 7, 2007 #7
    Ah, so it would be smarter to just state that I'm taking the breaking force as positive and leave the answer as negative. Gotcha!
  9. Jun 7, 2007 #8
    use the are under the graph