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Is this Elastic Collisions (Momentum) Right? watch

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    Just wanted to make sure im on the right tracks as the lesson confused me today.

    2. 'On an air track in the laboratory, a moving slider of mass 80g and initial velocity 0.20ms is in an elastic collision with a stationary slider mass 120g. calculate the velocity of both sliders after the collision.

    0.08 x 0.2 = 0.120 x 0

    0.0016kgms = 0.2kg x v
    0.016/0.2 = v

    v = 0.08ms?
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    Not quite.
    You have to equate total momentum before collision = total momentum after

    momentum before = 0.08 x 0.2 + 0.12 x 0 [the second slider is at rest]

    For the momentum after you say, after the collision

    let the velocity of the first slider be V1 and the second slider V2

    momentum after = 0.08V1 + 0.12V2
    so momentum equation gives

    0.08 x 0.2 = 0.08V1 + 0.12V2

    This gives you one equation with two unknowns.
    The fact that this was a (perfectly) elastic collision gives you a second equation with V1 and V2.
    Do you know it?
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    (Original post by Stonebridge)
    Not quite.
    You have to equate total momentum before collision = total momentum after

    momentum before = 0.08 x 0.2 + 0.12 x 0 [the second slider is at rest]

    For the momentum after you say, after the collision

    let the velocity of the first slider be V1 and the second slider V2

    momentum after = 0.08V1 + 0.12V2
    so momentum equation gives

    0.08 x 0.2 = 0.08V1 + 0.12V2

    This gives you one equation with two unknowns.
    The fact that this was a (perfectly) elastic collision gives you a second equation with V1 and V2.
    Do you know it?
    Thanks so far.

    Is it

    v1 = ((m1-m2)/(m1+m2))u1

    We've been given this sheet and it doesnt really help much if you dont understand. I was fine with the standard momentum questions.
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    Not that one. You need the so-called "relative velocity" equation using something called the coefficient of restitution.
    Have you got this one? It relates the velocity of approach of the two objects to the velocity with which they separate. The coefficient =1 for this collision.
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    Nothing to do with those only

    u1 = v2-v1
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    That's it, sort of.
    If u1 is the velocity of the first rider before the collision, then after the collision the velocity of separation is v2-v1
    The relative velocity equation gives e=(v2-v1)/u1 [e is the coefficient and equals 1]

    This gives 1=(v2-v1)/u1

    v2-v1=u1

    As you know the value of u1 from the question it means you now have another equation with v1 and v2.
    Just sub an expression for v1 or v2 from one equation into the other to solve it. It's just simultaneous equations.
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    (Original post by Oh my Ms. Coffey)
    2. 'On an air track in the laboratory, a moving slider of mass 80g and initial velocity 0.20ms is in an elastic collision with a stationary slider mass 120g. calculate the velocity of both sliders after the collision.
    First it is important to note that we do not know whether both sliders are moving with the same velocity, or even in the same direction. We have to work this out ourselves.



    Total MV before = total MV after
    M1U1+M2U2 = M1V1+M2V2

    Where M = mass, U = initial velocity (before collision), and V = final velocity (after the collision).
    Where object 1 is a mass of 0.08kg and initial velocity 0.2m/s, and object 2 is a mass of 0.12kg and initial velocity 0m/s.

    Substitute values in:

    (0.08)(0.2)+(0.12)(0)=(0.08)(V1)+(0.12)(V2)
    0.016 = 0.08V1 + 0.12V2




    During elastic collisions, both momentum and kinetic energy are conserved. We've just shown the conservation of momentum, but because we have two unknowns V1 and V2, we need to figure out a simultaneous equation to do in order to help us find one, then the other.




    Since kinetic energy is conserved,

    Total KE before = total KE after
    Remember kinetic energy is ½MV²

    ½M1U1²+½M2U2² = ½M1V1²+½M2V2²

    (It looks really ugly and complicated but just think about it in words in your head - total KE of both objects before has to equal total KE of both objects after.)

    Substitute values in:

    ½(0.08)(0.2²)+½(0.12)(0²)=½ (0.08)(V1²)+½(0.12)(V2)

    0.0016+0 = 0.04V1² + 0.6V2²




    Now we have two equations:
    1. 0.016 = 0.08V1 + 0.12V2
    2. 0.0016= 0.04V1² + 0.6V2²




    Can you go on from there?
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    (Original post by Chelle-belle)
    During elastic collisions, both momentum and kinetic energy are conserved.
    This is true, but it is much easier to use Newton's Law of Restitution (which is another way of looking at the KE side of things) as Stonebridge has suggested.
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    Urgh, why dont teachers ever tell me its simultaneous equations, second time i've completely over-complicated something.
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    (Original post by DFranklin)
    This is true, but it is much easier to use Newton's Law of Restitution (which is another way of looking at the KE side of things) as Stonebridge has suggested.
    Ah - in my school for my examination board we never learnt the term or method formally so this is what I would advise for the solution (since all the examples in the textbook would have been this way) :yep:

    For me it's easier to just remember KE before = KE after for elastic collisions; while it may take a little longer, if I just remember this definition of elastic collisions (which would be needed anyway) I shouldn't really ever get it wrong.

    But OP go for the one you find easier
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    With
    1. 0.016 = 0.08V1 + 0.12V2
    2. 0.0016= 0.04V1² + 0.6V2²

    Dont they just come out as really messy quadratics with me having to rearrange 1 in terms of V1 and putting that in V1² and solving the quadratic?
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    (Original post by Oh my Ms. Coffey)
    With
    1. 0.016 = 0.08V1 + 0.12V2
    2. 0.0016= 0.04V1² + 0.6V2²

    Dont they just come out as really messy quadratics with me having to rearrange 1 in terms of V1 and putting that in V1² and solving the quadratic?
    What you can also do is square the first equation and then you can use the 2 equations for find V2 in terms of V1.

    Newton's Law of Restitution (approach speed = separation speed) is *much* easier though.
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    (Original post by Oh my Ms. Coffey)
    With
    1. 0.016 = 0.08V1 + 0.12V2
    2. 0.0016= 0.04V1² + 0.6V2²

    Dont they just come out as really messy quadratics with me having to rearrange 1 in terms of V1 and putting that in V1² and solving the quadratic?
    Not if you do it the way I suggested.
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    (Original post by DFranklin)
    What you can also do is square the first equation and then you can use the 2 equations for find V2 in terms of V1.

    Newton's Law of Restitution (approach speed = separation speed) is *much* easier though.

    So in squaring its just basic elimination after that? As I did it the quadratic way and im getting 2 pages of working out.

    My calculator gets a math error and goes really slow when I try work it out.
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    Im going to go get my teacher to show me how to do this, its coming out really horrible.
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    (Original post by Oh my Ms. Coffey)
    So in squaring its just basic elimination after that? As I did it the quadratic way and im getting 2 pages of working out.
    It's fairly tedious elimination, but it should be well under a page. (It's about 6 steps, depending on how little you skip).
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    Why doesnt

    v1 = ((m1-m2)/(m1+m2))u1 = -0.2

    then plugging u1 = v2-v1 = 0 work?
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    (Original post by Oh my Ms. Coffey)
    Why doesnt

    v1 = ((m1-m2)/(m1+m2))u1 = -0.2

    then plugging u1 = v2-v1 = 0 work?
    I take it you mean "u1=v2-v1" not "u1=v2-v1=0" (since u1 does *not* equal 0).

    There's nothing wrong with doing it that way; that is Newton's Law of Restitution, which I've said (several times) is the best way to do this.
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    (Original post by DFranklin)
    I take it you mean "u1=v2-v1" not "u1=v2-v1=0" (since u1 does *not* equal 0).

    There's nothing wrong with doing it that way; that is Newton's Law of Restitution, which I've said (several times) is the best way to do this.

    How come we didnt get taught Newton's Law of Restitution for elastic and inelastic collisions?
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    (Original post by Oh my Ms. Coffey)
    How come we didnt get taught Newton's Law of Restitution for elastic and inelastic collisions?
    No idea. I did my A-levels 20 years ago; I know the maths exams pretty well (this question could also be set as a Maths question), but I'm not as well up on Physics.

    I don't know if either of the resident Physics experts have a thought on whether or not you can quote the result or if maybe you have to do it the "conservation of KE" way if NLR isn't taught.

    (Original post by Mr M)
    ..
    (Original post by Stonebridge)
    ..
 
 
 
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