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Linear momentum and impulse

  1. Nov 24, 2011 #1
    1. The problem statement, all variables and given/known data
    On a ride in a theme park, a carriage of mass 450 kg is travels horizontally at a speed of 18 m s–1. It passes through a shallow tank containing stationary water. The tank is 9.3 m long. The carriage leaves the tank at a speed of 13 m s–1.

    e) For the carriage in (b) passing through the water-tank, determine:
    i) its total loss in kinetic energy.
    ii) the gain in kinetic energy of the water that is moved in the direction of motion of the carriage.

    in previous questions ive worked out the water weight to 118.42kg and the speed it moves at after being hit is 19m/s

    2. Relevant equations
    KE = 1/2mv2 or kinetic energy = 0.5 x mass x velocity squared


    3. The attempt at a solution
    the bit that is confusing me is that it wants you to work it out in terms of the conservation of energy but when i put the maths in:

    0.5x118.42 x 19^2 = 21374.81J is the answer i get.
    the overall energy lost by the carriage is 34875J

    i was thinking that asking for how it relates to conservation means that it should be the same as the energy loss by the carriage but i cant get that no matter what i try? am i doing the maths right here?
    if those numbers are correct, is there a way to explain where the other 13500.19J of energy goes?
     
  2. jcsd
  3. Nov 24, 2011 #2
    Are you sure your numbers of 118.42 kg and 19 m/s are correct? Could you post calculations of those too?

    If the number are correct, then they are probably assuming that the energy was lost due to friction or something.
     
  4. Nov 24, 2011 #3
    19m/s was given as part of the 1st question. and 118kg came from part c. heres the full thing.

    4. a) Define linear momentum and impulse.

    On a ride in a theme park, a carriage of mass 450 kg is travels horizontally at a speed of 18 m s–1. It passes through a shallow tank containing stationary water. The tank is 9.3 m long. The carriage leaves the tank at a speed of 13 m s–1.


    As the carriage passes through the tank, it loses momentum and causes some water to be pushed forwards with a speed of 19 m s–1 in the direction of its motion.

    b) As the carriage passes through the water, show that the magnitude of its total change in momentum is 2250 N s.

    c) Use the answer from b) to deduce that the mass of water moved in the direction of motion of the carriage is approximately 120 kg.


    d) Calculate the mean value of the magnitude of the acceleration of the carriage in the water.

    e) For the carriage in (b) passing through the water-tank, determine:
    i) its total loss in kinetic energy.
    ii) the gain in kinetic energy of the water that is moved in the direction of motion of the carriage.

    f) By reference to the principles of conservation of momentum and of energy, explain your answers in e)
     
  5. Nov 24, 2011 #4
    OK.

    So, as you've discovered, there was energy unaccounted for in the simple model given to you. That energy might have been lost due to friction (which in turn causes heating in the carriage and the water), sound, turbulence in the water etc.

    There are no laws of physics broken here because in real world scenarios, energy is in fact lost this way.
     
  6. Nov 25, 2011 #5
    ok thanks part F just made it sound like the answer was meant to be that all of the energy was passed onto the water. and that the maths should apply that way.
     
  7. Nov 25, 2011 #6
    just a thought, is F) asking for the relationship between kintetic energy and momentum... in the way that if you double the speed you quadruple the KE? could that be the solution? that way i just need to calculate if loosing 5m/s fits that bill.
     
  8. Nov 25, 2011 #7
    What they expect from you in f) probably is to write down whether your answers for part e) agree with conservation of momentum and energy, and to give reasons if they don't.

    Calculate the total initial momentum of the system. Calculate the total final momentum of the system. Are they equal? If they do, that means conservation of momentum holds for this scenario. If the final momentum is less than the initial momentum, that means some momentum was lost. Explain how that momentum was lost.

    Do the same for energy.
     
  9. Nov 27, 2011 #8
    thanks.

    for part d) i need to calculate the ammount of time the carriage spends in the water tank. im not sure how to go about this or even what the equation would be to work that out because it is decelerating in that water. any ideas?
     
  10. Nov 27, 2011 #9
    You know the initial velocity, final velocity and the length of the tank. You can use those with one dimensional kinematic equations to find the average deceleration for part d, in the problem you've stated before.

    If you need to find the time the carriage spends in water tank, that can also be done with the same information and 1D kinematic equations.
     
  11. Nov 27, 2011 #10
    thanks that gives me something to go on but it seems i need the time to work out the average deceleration.. and i need the acceleration to get the time..... my brain hurts.
     
    Last edited: Nov 27, 2011
  12. Nov 27, 2011 #11
    The 1D kinematic equations are given in this link.

    Use this one
    [tex]v_f^2 = v_i^2 + 2 a d[/tex]

    The ability to choose the correct equation comes with experience as you solve more and more problems.
     
  13. Nov 27, 2011 #12
    i really needed to redo maths this year but wasnt allowed... do you know of a good guide for rearranging equations?
     
  14. Nov 27, 2011 #13

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    This is the only guide you really need when re-arranging equations:

    Anything that you do to one side of the equation, you must also do to the other side of the equation. ​


    If you do something to one side and not to the other, then it is no longer true that they are equal. So this rule is to preserve the equality. Generally what you end up doing is adding, subtracting, multiplying, or dividing things to/from both sides of the equation.
     
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