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Momentum, Completely Inelastic Collision

  1. Oct 2, 2007 #1
    1. The problem statement, all variables and given/known data

    Your friend has just been in a traffic accident and is trying to negotiate with the insurance company of the other driver to pay for fixing his car. he believes that the other car was speeding and therefore the accident was the other driver's fault. He knows that you have a knowledge of physics...(He must not know me very well :)...) and hopes that you can prove his conjecture. He takes you out to the scene of the crash and describes what happened. He was traveling North when he entered the fateful intersection. There was no stop sign, so he looked in both directions and did not see another car approaching. It was a bright, sunny, clear day. When he reached the center of the intersection, his car was struck by the other car which was travelling East. The two cars remained joined together after the collision and skidded to a stop. The speed limit on both roads entering the intersection is 50 mph. From the skid marks still visible on the street, you determine that after the collision the cars skidded 56 feet at an angle of 30 degrees N of E before stopping. (blah blah blah more facts): His car(1) is 2600 lbs and the other drivers car(2) is 2200 lbs (including weight of the drivers). Coefficient of Kinetic Friction= 0.80...I have to find both speeds of the drivers.

    m1=Mass of Car 1 = 2600 lbs
    m2=Mass Car 2 = 2200 lbs
    uk=Coeff. Kinetic Friction = 0.80
    dx=56 ft
    @=Theta=30 degrees



    2. Relevant equations

    mv=p where p is conserved.

    All of newton's laws of motion... vf=vi +at, dx=vit+1/2at^2, vf^2=vi^2+2ad

    KE=1/2mv^2 (even though its a completely inelastic collision, i might use it)

    friction=uN

    F=ma

    F(friction)*d=Work done by friction


    3. The attempt at a solution

    Since it is a completely inelastic collision, KE is not conserved; however, momentum is conserved.

    Here is the momentum equation:

    m1v1(j)+m2v2(i)=(m1+m2)*V

    Using dot product rules:

    Vx=(m2v2)/(m1+m2)

    Vy=(m1v1)/(m1+m2)

    V=(Vx^2+Vy^2)^.5

    Since F(friction)=uN, F=0.8*(2200+2600)*(g) =====>don't worry about converting, i'll do that.

    So here can I solve for (a) using F=ma? I would have:

    a=F(friction)/(2200+2600). Then using that, I could solve for t in above question. Is this correct or am I missing something important?

    Once I solve for (t) I can solve for V(initial). Then I can substitute for Vx, Vy. When Vx=Driver 1, and Vy=driver 2.
     
    Last edited: Oct 2, 2007
  2. jcsd
  3. Oct 3, 2007 #2

    learningphysics

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    Homework Helper

    most of it looks right. But why do you want to get t? Just get Vinitial directly. Then get Vx and Vy. then you can solve for v1 and v2.

    but what do you mean by this:

    ? This isn't right. v1 is not Vx. v2 is not Vy.
     
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