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How can you find the rate of change of momentum?

  1. Oct 4, 2011 #1
    Hi guys. Need a little help with an IB paper(Just two questions!) and would really appreciate your help.

    1. The problem statement, all variables and given/known data

    A bus is travelling at a constant speed of 6.2m/s along a section of road that is inclined at an angle of 6.0 degrees to the horizontal.

    State the value of the rate of change of momentum of the bus.

    b) The total output power of the engine of the bus is 70kW and efficiency of the engine is 35%. Calculate the input power to the engine.

    c) The mass of the bus is 8.5 x 10^3 kg. Determine the rate of increase of gravitational potential energy of the bus.

    d) Using the answer to c and data in b, estimate the magnitude of resistive forces acting on the bus.

    2. Relevant equations


    a= g sin theta.

    F= ma.

    GPE = mgh.

    SUVAT equations.

    3. The attempt at a solution

    a) As it is constant speed, zero net force. Zero net force, zero momentum. Answer is zero.

    b) Output power is 70000/x * 100% = 35%. Solving for x yields x=200000W.

    c) Velocity is 6.2 m/s at a 6 degree incline. 6.2 sin 6 gives me the vertical component of velocity, 0.648 m/s. Rate of increase of GPE is mgh/t, so 8.5*10^3*9.8*0.648 = 54000 J.

    (Is this right?)

    d) No idea. @_@ Can someone lead me in the right direction?
  2. jcsd
  3. Oct 4, 2011 #2


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    Homework Helper
    Gold Member

    'Looks okay to me, with one exception. Your answer for c) isn't going to be Joules. A Joule is a unit of energy. You're looking for a unit of rate of change of energy.
    From part b), you know that the bus's engine outputs a certain amount of power. From part c) (after you correct the units) you know that part of that power is going into getting the bus up the hill, thus increasing its gravitational energy as time goes on. So where does the rest of the power go? The bus's momentum isn't changing, so it's not going into that.

    Once you figure out the above hint, here's another hint.

    [tex] W = \vec F \cdot \vec d [/tex]

    Where [itex] W [/itex] is work (a measure of energy), [itex] \vec F [/itex] is force, and [itex] \vec d [/itex] is displacement.

    Putting things in terms of "deltas" (changes):

    [tex] \Delta W = \vec F \cdot \Delta \vec d [/tex]

    Divide both sides by change in time, Δt.

    Look at the left hand side of the equation. What's change in energy per unit time?

    Look at the right hand side of the equation. What's change in displacement per unit time?
    Last edited: Oct 4, 2011
  4. Jan 15, 2012 #3
    Yes, so P = Fv, and that power is wasted as resistive forces. However, I am just confused about one thing. When solving for F do we substitute the vertical velocity (0.648) into the equation or the velocity along the plane as given (6.2)?
  5. Jan 15, 2012 #4
    momentum is not zero
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