1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Momentum and impulse of a bullet

  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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook