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Force exerted in inelastic collision

  1. Dec 15, 2008 #1
    Hi, this is my first post, I've had a look at some examples to try and use them to help with my answer, but I've had no luck so far. I've just started doing physics and not spent much time on this board, so if my formatting/layout anything is a bit dodgy, then please tell me.
    I Would really appreciate it if somebody could tell me what I'm doing wrong here, as the answer doesn't seem right to me, so here goes...

    1. The problem statement, all variables and given/known data
    A 7000kg wagon travelling at 8m/s runs into the back of a 5000kg stationary van.
    a)Calculate their combined speed immediately after the collision
    b)What is the force exerted on the wagon if the time taken to reach their combined speed after the collision is 0.3s?

    2. Relevant equations
    (m1 u1) + (m2 u2) = (m1+m2)v

    F=m(v-u)/t


    3. The attempt at a solution
    a)Momentum of the truck =56,000kg m/s
    I've worked out the combined speed
    (m1 u1) + (m2 u2) = (m1+m2)v
    v=61,000/12,000
    v=5.08m/s (Combined speed after the collision)

    b)The best I can work out is F=m(v-u)/t (I'm guessing the masses of the 2 vehicles are added together?)
    so f=m(5.08-8)/t
    F=(mv-mu)/t
    F=60,960-96,000/0.3
    F=-116,800 N

    Am I right here, or way off?
    Any help here will be much appreciated.
     
  2. jcsd
  3. Dec 15, 2008 #2

    Doc Al

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    Staff: Mentor

    Where did the 61,000 come from? (What's the initial speed of the van? What's its initial momentum?)
    That equation is fine, but why add the masses? Since you want the force on the wagon, use the change in momentum of the wagon. (Of course, you may also take advantage of Newton's 3rd law and look at the van.)
     
  4. Dec 16, 2008 #3
    Well, the initial velocity of the van is 0m/s before the collision
    the 61,000 came from
    7000kgx8m/s(the wagon) + 5000kgx0m/s(the truck) =61,000

    So, I'm thinking now...
    F=mass of the wagon x final velocity (7000 x 5.08) - mass x initial velocity (7000x8)
    giving me f=35,560-56,000 /0.3

    F=18760 N

    or does m(v-u)/0.3 mean 'final mass - initial mass'/0.3? (I'm sorry for the daft question -it's 10 years since I last did physics...)
     
  5. Dec 16, 2008 #4

    Doc Al

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    Staff: Mentor

    Good, except that your answer to part a for the final speed was wrong. (Redo it--see my earlier comment.)

    How did you get this number?

    No, you had it right. v is the final speed; u is the initial speed.
     
  6. Dec 16, 2008 #5
    The way I worked out part a) was this
    (m1 u1) + (m2 u2) = (m1+m2)v
    (7000x8) + (5000x0) = (7000+5000) ie, the 7000kg truck moving at 8m/s and 5000kg van at 0m/s
    so 56,000 + 5000 = 12,000v
    v=61,000/12,000
    v=5.08m/s after the collision

    so it gives me 5.08m/s as the velocity after the collision, which is where i put it into
    the equation for the final velocity for part b)
    I got F=18,760 by doing the calculation wrong. (not sure what happened there...)
    F=(7000x5.08) - (7000x8)
    F=35,560 -56,000 /0.3
    F= -20440/0.3
    F=-68133.33N
     
    Last edited: Dec 16, 2008
  7. Dec 16, 2008 #6

    Doc Al

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    Staff: Mentor

    What's 5000x0? :wink:
     
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