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Momentum intuition, had me thinking last night

  1. Nov 24, 2011 #1
    I came up with last night and I couldn't figure it out.

    Can you actually predict what happens in a collision?

    For instance, suppose a truck and a toy duck is coming from opposite ends, they both have a velocity, now suppose they crash, it should be obvious that the collison isn't going to be elastic. But how do we know this? Is it because of the significantly large momentum of the truck?

    Suppose I give you another situation

    Situation #2

    A truck and a car coming towards each other, with the same speed.

    And they clashed. What happens?

    1) The truck has a greater momentum and hence will inelastically collide with the car and become one
    2) Even though the truck has a greater momentum, they will collide and elastically bounce off each other

    How do we predict this? Please only use linear momentum. Use these

    M = mass of truck
    m = mass of car
    vtruck
    vcar
    vtruck + car
    vtruck'
    vcar'

    Situation # 3 (other)

    Suppose I throw a snowball at an 500-year-old oak tree and the ball sticks to the tree and eventually falls to the ground when it melts

    So that

    Mtree >> msnowball

    Now we have

    Mtreevtree + msnowballvsnowball = (Mtree + msnowball)v'


    But

    0 + msnowballvsnowball = (Mtree + msnowball)v'

    0 + msnowballvsnowball = (Mtree)v'

    For Mtree >> msnowball

    And that the tree doesn't move (maybe the roots sustained the net force impaled?)



    msnowballvsnowball = 0

    But this doesn't make sense seeing neither are 0

    Does this have to do with Newton's third law?
     
  2. jcsd
  3. Nov 24, 2011 #2

    Doc Al

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    Staff: Mentor

    No. You can only predict that the collision will be inelastic due to your experience with such materials. Momentum is always conserved.

    Since momentum is always conserved, that won't be enough to predict whether the collision is elastic.

    Since the tree is attached to the earth, other forces are involved besides that from the snowball. For practical purposes, the speed of the tree remains 0. (Of course, you know that the momentum of the earth+tree+snowball system must remain constant. )
     
  4. Nov 24, 2011 #3
    You are right in your 'maybe...'. There is another external force (besides that of the snowball) acting on the tree and so linear momentum of (snowball + tree) is not conserved.
     
    Last edited: Nov 24, 2011
  5. Nov 24, 2011 #4
    So physics can't tell us why trucks can run over toy ducks?

    Or inelastic either? Is what you are implying in the first quote?

    But what's wrong with my math though? How do I include the earth into my equation? Or is this entirely up to energy?
     
  6. Nov 24, 2011 #5

    Doc Al

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    Staff: Mentor

    There's a lot more to physics than simple conservation of momentum.
    Right.
    Your equation would be true if the tree were not attached to the earth.
     
  7. Nov 24, 2011 #6


    But I could use that equation again and arrive at

    msnowballvsnowball = 0

    Can't we still tell what would happen in a collison based on the material of the bodies slamming?

    In my duck and truck example, it isn['t just experience

    It is

    1) Truck mass/density >> toy duck mass/density
    2) Truck speed >> duck speed (0)
    3) Duck made of rubber? I don't know and truck made of metal
     
  8. Nov 24, 2011 #7

    Doc Al

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    Staff: Mentor

    Only by treating the mass of the tree as 'infinitely' larger than the mass of the snowball. A reasonable approximation. Just depends on how accurate you want to be.
    Of course, given knowledge of the physics of materials. That is not always trivial.
     
  9. Nov 24, 2011 #8
    But that's it, m and v aren't 0, how could I have gotten 0 from their product?
     
  10. Nov 24, 2011 #9

    Doc Al

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    Staff: Mentor

    I assumed you meant the speed after the collision. Of course, before the collision the speed of the snowball isn't zero.

    The momentum before equals the momentum after. (Assuming the tree is not attached to the earth and is free to slide, as if on ice.) Of course, since the mass of the tree >> the mass of the snowball, the speed of both after the collision is approximately zero. So what? Exactly zero? No. Total mass X final speed = mass of snowball X initial speed.
     
  11. Nov 24, 2011 #10
    So isn't the answer that msnowballvsnowball (Mtree)v' ? or is = correct?
     
  12. Nov 24, 2011 #11

    sophiecentaur

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    Science Advisor
    Gold Member

    I don't understand how anyone would think that elastic or inelastic collision is something to do with the relative masses or speeds. It's entirely to do with the construction of the objects. A ball bearing hitting the steel plate on the front of an Express train is a pretty well 100% elastic collision. Two equal masses of rice pudding coming together are an inelastic collision. A (real) duck and a truck is inelastic. Two ball bearings colliding is elastic. Need I go further?

    Also, it is pointless trying to find errors in the Momentum Conservation law. The law refers to the total system involved and always applies; you just have to take everything relevant into account if you want to apply the Law. When a 1kg duck hits a massive truck, which is freewheeling over the ground, it is only the masses of duck and truck that need to be taken into consideration for that collision. If the truck is exerting significant force on the Earth's surface (e.g. driving), the mass (or moment of inertia) of the Earth becomes, strictly speaking, part of the calculation. There is a tiny change in the Earth's rotation - as there was when the truck was initially accelerating - when anything happens to the truck. At the end of the process, the Earth will still be rotating at the same rate.
    The fact is that in collisions, any change in the Earths motion will be insignificant. If there is an elastic collision then the small colliding object will have its velocity of approach reversed.

    When I learned about 'Dynamics' we were taught about the Coefficient of Restitution. This was the ratio of speed of separation after to the speed of approach before collision. This can be used to calculate what will happen in partly elastic collisions (very common events). You could set up equations for Momentum Conservation and for velocities before and after and find out what would happen as a result of a collision. This Coefficient is not easy to measure and it will often depend upon the actual speeds and masses involved (most objects do not behave linearly) but it provided hours of 'fun' for us.
     
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