Elastic & Inelastic Collisions

Discussion in 'Physics & Math' started by kingwinner, May 18, 2006.

  1. kingwinner Registered Senior Member

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    796
    1) For inelastic collisions, total momentum before collision equals total momentum after collision, but the total kinetic energy before collision doesn't equal to total kinetic energy after collision. So momentum is conserved even if kinetic energy is not conserved. Why is this possible?

    2) Two balls collide head-on on the ground. The speed of ball 1 (1.00kg) before collision is 5.00m/s. Ball 2 (m=7.00kg) is initially stationary. Assume this is an elastic collision, find the velocities of ball 1 and ball 2 after the collision.


    [2 unknowns require 2 equations.
    p(total)=p'(total)
    m1v1+m2v2=m1v1'+m2v2'
    (1)(5)+0=(1)(v1')+(7)(v2')
    So v1'=5-7v2' (equation 1)

    Ek=Ek'
    0.5m1(v1)^2+0.5m2(v2)^2=0.5m1(v1')^2+0.5m2(v2')^2
    0.5(1)(5)^2+0=0.5(1)(v1')^2+0.5(7)(v2')^2
    12.5=0.5(v1')^2+3.5(v2')^2 (equation 2)

    Now, the thing that I am worrying is that for equation 1, v1 and v2 are VECTORS but for equation 2, v1 and v2 are SCALARS. They are not exactly the same thing because vectors can take on negative values and we don't know which is negative before calculating it. How can I deal with this problem? I have having a lot of trouble due to existance of vectors in equation 1...

    Can someone help me (especially for question 2)? I would appreciate!

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    Last edited: May 18, 2006
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  3. przyk squishy Valued Senior Member

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    Short answer: KE can be converted to other forms of energy. There's only one kind of momentum.

    There's also the slight issue that kinetic energy is a scalar and momentum is a vector. If you look at any obect or system, its really a collection of tiny masses or particles. These particles will have a centre of mass somewhere.

    The important point is that, while (total mass)*(velocity of centre of mass) gives you the total momentum of the system,

    ½*(total mass)*(velocity of centre of mass)² doesn't give you the total kinetic energy.

    The difference between the total energy and what you calculate using the centre of mass is usually just the heat for a solid object (unless its also rotating). It's easy to come up with examples of a few particles that have total kinetic energy, but no total momentum.
    Boring details: In the kinetic energy formula, the v² is really |v|², where |v| means the absolute value, or magnitude of the vector v (I'm using bold to represent vectors).

    If you've done vectors yet in maths, you can also write the formula as KE = ½m(v•v), where • denotes the scalar or dot product of the two vectors. When the 2 vectors being dot-producted are the same, like with v•v, its often just abbreviated as v². so you can also write KE = ½mv² (just as long as you remember its a dot product and not a scalar multiplication). Point is, you can leave vectors in the KE formula if you want.

    Practical info: In 1D, the scalar product is just the square of the 2 velocities. You don't have to remove the minus sign for the KE, because (-x)² = x². For your first set of equations you can just treat the vectors as scalars that are allowed to be negative. If you can rearrange eq.1 to get v1' = (whatever), you can substitute v1' for (whatever) in eq.2.

    Dunno if the info you needed was in there somewhere

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  5. mathman Valued Senior Member

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    For problem 2 as given, only one dimension is needed, so the vectors reduce to signed scalars. If the collision was not head on, the problem would be much more difficult.
     
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  7. kingwinner Registered Senior Member

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    796
    2) p(total)=p'(total)
    m1v1+m2v2=m1v1'+m2v2'
    (1)(5)+0=(1)(v1')+(7)(v2')
    So v1'=5-7v2' (equation 1)

    Ek=Ek'
    0.5m1(v1)^2+0.5m2(v2)^2=0.5m1(v1')^2+0.5m2(v2')^2
    0.5(1)(5)^2+0=0.5(1)(v1')^2+0.5(7)(v2')^2
    12.5=0.5(v1')^2+3.5(v2')^2 (equation 2)

    Sub. equation 1 in equation 2,
    This substituted part becomes, (v1')^2=(5-7v2')^2=25-70v2'+49v2'^2 <-------The square v2' is not a vector but the v2' in -70v2' is a vector. And in this case we know tht m2 will have a velocity of positive direction after collision (if we pick the driection of the initial velocity of m1 be positive) because m2 is stationary at the beginning, so v2' (vector) = v2' (scalar) because it's in the positive direction. We are just lucky in this case...

    What if both balls are moving and collide head-on? There is no way of knowing which direction each ball will move after the collision and we can't modify the signs of the velocities to transform them to speeds......
     
  8. Michalowski Registered Senior Member

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  9. kingwinner Registered Senior Member

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    796
    The simulation is cool...

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    but I still don't get the vector and scalar issue (i.e. combining a vector equation and a scalar equation without knowing whether each vector is in the positive direction or negative direciton...)

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  10. przyk squishy Valued Senior Member

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    In one dimension, you can consider the speed in the KE formula to be the same variable as the velocity in the momentum formula.

    In 3 dimensions, you can use: E<sub>k</sub> = &frac12;m(v<sub>x</sub>&sup2; + v<sub>y</sub>&sup2; + v<sub>z</sub>&sup2

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    Where v<sub>x</sub>, v<sub>y</sub>, and v<sub>z</sub> are the x, y, and z components of the velocity, respectively. You don't have to worry about cutting off negative signs or anything - the squaring of the component velocities does this for you and ensures the KE is always positive. You just have to remember that there will be 2 solutions (which you get using the quadratic formula). One of these will correspond to the velocities before the collision, and the other to the velocities after the collision.
     
  11. kingwinner Registered Senior Member

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    796
    But v1' (velocity) = v1' (speed) if and only if v1' (velocity) is positive.

    v1' (velocity) = -v1' (speed) for the case that v1' (velocity) is negative.

    Basically, we have to convert the equation 1 from a vector equation into a scalar equation in order to substitute into the scalar equation 2.And for the question, since both v1' and v2' are not given, we don't know whether it's positive v1' or negative v1'. Oh, I don't understand...

    I have to find a way to convince myself...

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  12. przyk squishy Valued Senior Member

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    But how's the speed defined in general? It's the magnitude of the velocity vector, right?

    So v = |v| = sqrt(v<sub>x</sub>&sup2; + v<sub>y</sub>&sup2; + v<sub>z</sub>&sup2

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    .

    In one dimension, v<sub>y</sub> and v<sub>z</sub> disappear completely from the equation, so you get v = sqrt(v<sub>x</sub>&sup2

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    . It then follows that v is just v<sub>x</sub> with the minus sign cut off - It's not defined that way.

    So it should really be E = &frac12;m*sqrt(v<sub>x</sub>&sup2; + v<sub>y</sub>&sup2; + v<sub>z</sub>&sup2

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    &sup2; (substituting the above definition of v in the KE formula).

    This just gives E = &frac12;m(v<sub>x</sub>&sup2; + v<sub>y</sub>&sup2; + v<sub>z</sub>&sup2

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    , or E = &frac12;mv<sub>x</sub>&sup2; in 1D.
     
  13. kingwinner Registered Senior Member

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    796
    2) I have just tried calculating every possible specific case.
    e.g. a ball initially moving in the negative direction colliding a stationary ball.
    e.g. and also two moving balls colliding each other head-on...in this case, I just end up solving a quadratic equation.

    I have seen that in every case, the combination of the vector and scalar equations works (nice to know

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    ). I got reasonable answers in every single case. This nice result kind of surprises me...

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    For the original quesiton: Two balls collide head-on on the ground. The speed of ball 1 (1.00kg) before collision is 5.00m/s. Ball 2 (m=7.00kg) is initially stationary. Assume this is an elastic collision, find the velocities of ball 1 and ball 2 after the collision.

    After I substitute equation 1 into equation 2, I got 0=35v2'+28(v2')^2 if I let the the direction of original velocity before collision as begative. So v2'=-1.25m/s. Is this v2' actually representing velocity or speed? Why will it be negative? It seems that the scalar automatically becomes a vector again...it seems that it takes care of the direction of the velocity also...
     
  14. przyk squishy Valued Senior Member

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    It works because x&sup2; = (-x)&sup2; for any value of x.
     
  15. kingwinner Registered Senior Member

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    796
    When I attempt this question, nightmare occurs. The usual way of doing these type of problems works for all cases I have done so far, but doesn't seem to work here for Q3cd. The directions are really messed up.

    3)

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    Part a and b are ok, the problem here is Q3cd.

    What I did for Q3d,

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    But clearly, the result contradicts the information given by the question, the directions are wrong. Why is something like this happening? Please help me!

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  16. kingwinner Registered Senior Member

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    796
    Can anyone please help me with the above question? :bugeye:
     
  17. przyk squishy Valued Senior Member

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    3,203
    You've set East as your positive direction. You have one positive and one negative solution, and you're told that the 3 kg mass was initially heading west. Where's the problem?
     
  18. kingwinner Registered Senior Member

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    796
    Yes, east is positive direction.
    And I got, v2 = 2.0m/s[W] or 6.0 m/s[E], so is the mass moving 2.0m/s[W] initially? What does the 6.0m/s [E] represent, then?
     
  19. przyk squishy Valued Senior Member

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    3,203
    Suppose the question had asked for final velocities instead of initial velocities. What calculation would you have done instead?
     
  20. kingwinner Registered Senior Member

    Messages:
    796
    I would do the same thing. Isolate one variable for the first equation and then substitute for into the second equation. And in all cases I have done so far, this method works and the directions in the final answer are correct. But not the case for this question...
     
  21. przyk squishy Valued Senior Member

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    Would there be any differences, or would the calculations be identical?
     
  22. kingwinner Registered Senior Member

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    796
    They will be pretty much the same. Yes?
    Sorry, I don't see what this is leading to...
     
  23. przyk squishy Valued Senior Member

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    3,203
    The initial and final velocities are solutions of the same set of simultaneous equations. Now do you think you know how to interpret the 6.0 m/s[E] result?
     

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