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Inelastic collision question

  1. Oct 4, 2004 #1


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    Hi everyone. I need some help with the following question.

    In a traffic accident, a car of mass 2000kg travelling south collided in the middle of an intersection with a truck of mass 7000kg travelling west. The vehicles locked and skidded off the road long a line pointing southwest. A witness claimed that the truck had entered the intersection at 80km/hour.
    a) Do you believe the witness? Why/why not?
    b) Find the velocities of the vehicles before and after the collisions in the centre of mass frame.
    c) How much kinetic energy is lost in the collision?

    My problem is that I think there are too many values that are not given. I tried using the equation m1v1+m2v2 = (m1+m2)v' but then I didn't know what values to use for v1 and v'. Does anybody have any thoughts on this?
  2. jcsd
  3. Oct 4, 2004 #2
    There is enough information as long as the reported SW direction after the accident is precisely accurate. If it is, then both vehicles contributed an equal amount of momentum to the collission, so the speed of the lighter vehicle had to be 280 km/hr. That's awfully fast, but not impossibly so.

    There isn't enough information to say how much energy was lost in deforming the metal and al the othe nasty effects of such a high speed collission, but we can calculate a lower limit assuming all energy was completely conserved in the vehicles. The combined mass is then 9,000 kg, and so 1/2*2000*280^2 + 1/2+7000*80^2 = 1/2*9000*v'^2 gives a post collission velocity of almost 150 km/hr (actually 149.666). Since energy is always lost in collission, the real world velocity would be somewhat less than that.

    It's marginally credible if the car was a Ferrari, though 2000 Kg is a bit heavy for a high performance car, isn't it? My Civic sure couldn't hit that speed.
  4. Oct 4, 2004 #3


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    LOL@ your Civic :wink:

    Ok one question, why can we take the velocity to be (7000/2000)*80? Somebody tried explaining that part to me today but I really didn't get it.
  5. Oct 5, 2004 #4


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    Look at the momentum of the system car+truck prior and after the collision.
    Since this momentum is conserved, so must the momentum's direction.
    Hence, if its direction is southwest after the collision, it must have been so also prior to collision.
    How does that help you?
  6. Oct 5, 2004 #5
    If both the vehicles are locked after collision and moved in the SW direction then the contribution of the momentum by both is equal,
    m1.v1=m2.v2 .....(1)
    if v' is the speed of combined vehicle than by conservation of momentum
    m1.v1 =m2.v2=(m1+m2)v'
    by (1)
    if the witness is true then
    speed of car= (9000/4000)*80
    speed of truck=(9000/14000)*80 before collision

    NOw you know the velocity calculate the kinetic energy easily.
  7. Oct 5, 2004 #6
    Exactly. Maybe it would help phy to think about what the final direction of the mangled cars would be if one of the vehicls (say the lighter car) were moving at different speeds. If it were stopped, sitting in the road right in the truck's path, it'd just get pushed in the same line same direction as the truck, so everything would move straight west. If the car were going south at a slow speed, it'd hardly deflect the truck at all as it got creamed, and the resultant angle would be displaced only a little south of west. If the car were mving incredibly fast, say at supersonic speeds so that the truck was almost standing still, the resulting path would be much more southern than western. Thinking like this helps to bracket the problem and home in on a method of solution.
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