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Find the Impulse and Momentum

  1. Apr 15, 2007 #1
    1. The problem statement, all variables and given/known data
    Before a collision, a 25-kg object is moving at +12m/s. Find the impulse that acted on teh object if, after the collision, it moves at:
    a. 8.0m/s
    b.-8.0m/s
    2. Relevant equations
    (c)=changes in:
    To find impulse, we would use the F=ma, so F=m*(c)v/(c)t
    So, F(c)t = m(c)v
    Therefore, F(c)t=p2-p1 (impulse-momentum theorem).
    3. The attempt at a solution
    Ok, so here's what I did. Let's solve A first.
    GIVEN:
    m=25kg
    v1=12m/s
    v2=8m/s
    F=25*(8.0-12)/(c)t
    F= -100/(c)t
    So, we take the difference in time to the other side.
    F(c)t=-100
    Therefore, its impulse is -100. At first I got confused because I had left two variables, but I think that's the impulse because F(c)t=p2-p1, so:
    p2-p1=-100
    25(8)-(25)(12)=-100
    -100=-100
    Im assuming thats the impulse, so its -100 N*s.

    So, lets do this for B, which is -8.
    GIVEN:
    m=25kg
    v1=12m/s
    v2=-8.0m/s
    F=?
    F(c)t=m(v2-v1)
    Impulse=25(-8-12)
    Impulse=25(-20)
    Impulse=-500
    Seems to make sense, because to make the object thats moving go on its opposite direction it must be really strong. And also, both answers are negative as it moves in the opposite direction.

    I would appreciate if you could check over my work to see if it is correct.
     
  2. jcsd
  3. Apr 15, 2007 #2

    Chi Meson

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    The answer is 400 Ns.

    While you are correct, initially, in looking at the problem, the impulse you can see by looking at the initial and final velocities, that the (c)v [to use your symbols] is -16 m/s.

    Impluse IS F(c)t, so the answer is 25 kg * (-16m/s)

    m
     
  4. Apr 15, 2007 #3
    For Problem A, the c(v) is V2-V1, so 8-12 would equal -4.
    For Problem B, the c(v) is V2-V1, so -8-12 would equal -20. Where did you get -16? Did you use -20-(-4)? Wouldn't that be the change of the change in velocity?
     
  5. Apr 16, 2007 #4

    Chi Meson

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    Sorry, I misread the question. I had thought it was "8m/s to -8m/s." Your answers are correct, but still you did far more work than necessary.
     
  6. Apr 16, 2007 #5
    Hehe

    Im in my first year of physics so Im really careful step by step, hehe!! I handed in the homework today so thank god the answer was right hehehe!! thanks soo much for your help! :)

    'better safe than sorry'
     
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