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Momentum and impulse

  1. Feb 26, 2008 #1
    1. The problem statement, all variables and given/known data

    A bullet of mass 0.03kg is fired from a gun with a horizontal velocity of 400 ms^-1.
    Find the momentum of the bullet after it is fired. If the gun is then brought to rest in 1.2s by a horizontal force which rises uniformly from zero to B N and then falls uniformly to zero, find the value of B.

    2. Relevant equations

    Impulse = force * time.

    3. The attempt at a solution

    Momentum of the bullet = 0.03 * 400 = 12 kg ms^-1, taking the direction of the bullet to be positive. So momentum of the gun is also 12 kg ms^-1.

    Now, the problem, my friends have all jumped in and said as I = ft, then 12 = F*1.2, so F = 10 N. Why is this? How have we gone from the information given about rising uniformly to B and then back again, to just using that equation? Also, why is the impulse just the same as the momentum?

    Also, I was thinking, don't we need to double this? Is 10 N not the force to rise up, and the same is needed to come back down?

    Hope this post makes sense, thanks for help.
  2. jcsd
  3. Feb 26, 2008 #2
    I'm not sure if I am correct or not, but I'm giving my input.:smile:

    The deceleration of the gun is not constant. It increases to a certain max value and then decreases back to zero.
    If that max value is 'B' newtons, I guess the average force over the time interval is [tex]\frac{B}{2}[/tex] newton.

    avg. deceleration of gun = [tex]\frac{B}{2m_{gun}}[/tex]


    which gives B=20 N.

    But I doubt if I can use the average deceleration of gun like I did ? :uhh:
    I may be wrong
  4. Feb 26, 2008 #3
    Thanks for your input, although I thought you could only use v = u + at when acceleration is constant.

    Anyone else?
  5. Feb 26, 2008 #4
    What does "rises uniformly" and "falls uniformly" mean? I'm guessing it means the rate of change of this horizontal force F is constant for some interval of time. In other words, for [0, t'], dF/dt = c and for (t', 1.2], dF/dt = c' where c and c' are constants. Is c = c'? Is t' = 0.6?
  6. Feb 26, 2008 #5
    I would assume that too, but I'm not too sure. I'm not sure how to proceed with that.
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