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Average force exerted by momentum

  1. Dec 13, 2012 #1
    1. The problem statement, all variables and given/known data
    A car of mass 10 kg hits another car at 10 m/s and they both come to rest at the same time. In the process, the car that was hit moved back 10 meters. What is the average force exerted on it?


    2. Relevant equations
    Average force = change in momentum divided by time


    3. The attempt at a solution
    I'm taking it so everything is going to the right, so positive rightwards.
    Final momentum is 0. Initial is 10kg * 10m/s so: Average force = -100kg*m/s (because final - initial) divided by the time which is 1 second I think. 10m = 10m/s * t = 1 second.
    So -100kg*m/s divided by 1 second = -100N.

    But this is wrong. I can't find my error but I'm told I calculated the time incorrectly. How do I get the correct time?
     
  2. jcsd
  3. Dec 13, 2012 #2

    haruspex

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    If it only took 1 sec, the speed of the first car would have to be 10m/s for the whole second (then suddenly 0). To get the time, you have to assume uniform deceleration after the impact. That's reasonable, but worse is that you are not told the mass of the second vehicle, so you don't know what the speed is just after impact. So to get an answer you'll have to assume they're the same mass.
    I've a nasty feeling that the questioner intends you to use energy = force * distance. That will give you the force averaged over distance, but that is not the same as the force averaged over time.
     
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