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Conservation Of Linear Momentum Question

  1. Aug 21, 2012 #1
    1. The problem statement, all variables and given/known data
    A 750 kg car is traveling north at 54 km/h while a 1200 kg truck travels south at 83 km/h. The two vehicles are about to have a head-on collision.

    1.1) Convert the speed's of both car's into m/s.
    1.2) After impact, the car rebounds backwards with a velocity of 18 km/h. Calculate the velocity of the truck after the collision.

    2. Relevant equations
    Pbefore = Pafter
    P = impulse
    MaUa + MbUB = MaVa + MbVb
    U = initial velocity
    V = final velocity
    M = mass in kilograms

    3. The attempt at a solution

    1.1) km/h to m/s (kilometers per hour to metres per second): You divide the value in km/h by 3.6

    Therefore the speed of the car in m/s:
    54 / 3.6 = 15 m/s

    And the speed of the truck in m/s:
    83 / 3.6 = 23.06 m/s

    1.2) So if we take north to be positive then the car is moving in the positive direction and the truck is approaching in a negative direction.

    So Ua = +15 m/s
    Ub = -23.06 m/s

    And if car A rebounds in the opposite direction than Va is -5 m/s (18/3.6 = 5) and Vb will be +.

    So if I then plug in all my values into the formula:
    Pbefore = Pafter
    MaUa + MbUB = MaVa + MbVb

    750 * 15 + 1200 * -23,06 = 750 * -5 + 1200 * Vb
    Therefore Vb = -10,56 or 10,56 in the original direction of the car (north and we took north to be positive) (+10,56 m/s)

    The answer in the memo is 35,56 m/s
    They say that both Ua and Ub are positive so Ua = 15 and Ub = 23,06 which conceptually does not make sense as they are heading towards each other from opposite directions and velocity is a vector so it has both magnitude and direction.

    Is my answer correct or am I making a mistake in my logic or calculations?
    Any help is appreciated, Thanks!
     
    Last edited: Aug 21, 2012
  2. jcsd
  3. Aug 21, 2012 #2
    If the truck is going 83 Kmph which is 23.1 m/sec before being hit head on by a car going in the opposite direction, how can it be going faster after the impact? The truck imparts momentum to the car. That momentum must be 'paid for' from the truck's momentum. I get your answer.
     
  4. Aug 21, 2012 #3
    Ok thanks, the memo said otherwise and I really couldn't figure out how it was getting Ua and Ub both as positive and implying that they were moving in the same direction...
     
  5. Aug 21, 2012 #4

    CAF123

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    Gold Member

    The answer in the book must be wrong. I too got -10.6 m/s.
    However, the negative indicates that after the collision, the truck has a velocity of 10.6 m/s in it's original direction of travel.
     
  6. Aug 21, 2012 #5
    -10.6m/s has to be wrong too as the truck would now overtake the "rebounding" car.

    A sanity check assuming truck and car stick together has the speed of the merged car+truck south at greater than the questions rebound speed of 18km/h.

    Not at all sure why the question asks in step 1.1 to convert from km/h to m/sec . It also seems a bit odd that a truck would be travelling faster than the car or weigh so very little. The car would have to be a "Smart" two seater with a very light driver. Not sure what kind of truck weighs only 1200kg

    An interesting question for all the wrong reasons.
     
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