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2D conservation of Energy and Momentum with *Three* unknowns

  1. Jan 26, 2015 #1
    1. The problem statement, all variables and given/known data
    A 2.0 kg ball moving with a speed of 3.0 m/s hits, elastically, an identical

    stationary ball as shown. If the first ball moves away with angle 30 ° to the

    original path, determine

    a. the speed of the first ball after the collision.

    b. the speed and direction of the second ball after the collision.

    2. Relevant equations
    m1v1 + m2v2 = m1v1' + m2v2'

    1/2m1v1^2 + 1/2m2v2^2 = 1/2m1v1'^2 + 1/2m2v2'^2

    3. The attempt at a solution

    m1 = m2 = 2.0kg
    v1 = 3.0m/s
    v2 = 0m/s
    theta1 = 30

    In the x direction

    m1v1 + m2v2 = m1v1'cos30 + m2v2'cos(theta2)
    (2.0)(3.0) + (2.0)(0) = (2.0)v1'cos30 + (2.0)v2'cos(theta2)
    masses cancel out to leave;
    3.0 = v1'cos30 + v2'cos(theta2)

    In the y direction

    m1(0) + m2(0) = m1v1'sin30 + m2v2'sin(theta2)
    masses cancel to leave:
    -v1'sin30 = v2'sin(theta2)

    Now here is where I am getting stuck. I know that I have to somehow combine the x and y equations as well as the conservation of NRG eqn to find the three unknowns, but I am lost as to how the algebra will work out. I think I have to square them somehow to eventually get cos^2(theta2) + sin^2(theta2) = 1 to get around the unknown theta... but everytime I have tried, it doesn't work out. Please help!
     
    Last edited by a moderator: Jan 27, 2015
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  3. Jan 26, 2015 #2

    Stephen Tashi

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    The balls each have a momentum vector after the collision and you can visualize those vectors as two sides of a triangle, with the third side being their sum. Conservation of energy says that the sum of the squares of the the magnitudes of the two momentum vectors is the square of the magnitude of the third side. So the triangle is a right triangle.

    To show this by algebra, you could argue from the law of cosines C^2 = A^2 + B^2 - 2 AB Cos(theta) that the term 2 AB cos(theta) is zero.
     
  4. Jan 26, 2015 #3
    So are you saying that:

    v1^2 = v1'^2 + v2'^2 - 2(v1')(v2')cos(theta) ??

    then would v1' and v2' just be the rearrangement of the x and y components of the momentum eqns in the Pythagorean theorem?
    i.e. v1' = sqrt(((v1 - v2'cos(theta))/cos30)^2 + ((v2'sin(theta))/sin30)^2)
    and v2' = sqrt(((v1-v1'cos30)/cos(theta))^2 +((v1'sin30)/sin(theta))^2) ???
     
  5. Jan 26, 2015 #4

    Stephen Tashi

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    No, not as I understand your notation. You aren't using "theta" for the angle between the two momentum vectors after the collision.

    I'm talking about the magnitudes of 3 vectors. If A,B,C are vectors with C = A + B then |C|^2 =|A|^2 + |B|^2 - 2|A||B| cos( alpha) where alpha is the angle between A and B and |V | denotes the magnitude of a vector V.

    In your notation, I think alpha = theta2 + 30 deg.

    That isn't a clear question.

    The magnitudes of the vectors can be computed from the magnitudes of their orthogonal components by using the Pythagorean theorem, but in this problem you can tell the magnitudes just by thinking of "the unit circle".

    For example one ball leaves the collision with a magnitude vector whose components are (m1 v'1 cos(30 deg), m1 v'1 sin(30 deg)) The magnitude of this vector is the absolute value of m1v'1.
     
  6. Jan 26, 2015 #5
    I am so very lost.

    I don't think I understand where you are trying to hint me towards using the law of cosines. Is it for the whole system, or just one isolated vector? i.e. C = v1 or am I just using it to find v1' and A and B will be the x and y component of v1'?

    Also is this referring to the momentum vectors of one ball, or the whole system?

    If it was for one ball, would it be

    v1'^2 = (m1v1'cos30)^2 + (m1v1'sin30)^2

    or using momentum;

    v1'^2 = ((v1-v2'cos(theta))/cos30)^2 + ((v2'sin(theta))/sin30)^2

    If it was for the system, would it look like this;

    v1^2 = v1'^2 + v2'^2
    v1^2 = (sqrt((m1v1'cos30)^2+(m1v1'sin30)^2))^2 + (sqrt((m2v2'cos(theta))^2 +(m2v2'sin(theta))^2))^2

    or if using the conservation on momentum eqns in x and y;

    v1^2 = (((v1 - v2'cos(theta))/cos30)^2 + ((v2'sin(theta))/sin30)^2)^2 +(((v1-v1'cos30)/cos(theta))^2 +((v1'sin30)/sin(theta))^2)^2

    Oh dear.... as you can see, I am very lost as to what to use where... is any of this right?
     
  7. Jan 26, 2015 #6

    Stephen Tashi

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    C is the vector sum of the momentum vectors of the two balls after the collision. A is the vector of momentum of one ball after the collisionl, B is the vector of momentum for the other ball. [Remember that momentum is a vector quantity, not a scalar quantity.] You haven't defined your variables. You seem to be using v1 to mean both a vector and the magnitude of a vector. That isn't good notation.

    Because of conservation of momentum C is also equal to the momentum vector of the first ball before the collision.

    To apply the law of cosines you must have a variable or an expression that represents the angle between the momentum vectors of the two balls after the collision.

    You wrote several equations that have velocity on the left hand side and momentum on the right hand side, so they don't make sense.
     
  8. Jan 27, 2015 #7
    OK, I think I understand a little better now?
    Just like in kinematics, adding vectors A and B tip to tail, will be equal to vector C, and like you said, C will be equal to the vector of the first ball because of conservation of momentum.

    So in the x direction

    (v1i)^2 = (v1fcos30)^2 + (v2fcos(alpha))^2 - (2(v1fcos30)(v2fcos(alpha)) cos theta)

    and in the y direction

    0 = (v1fsin30)^2 + (v2fsin(alpha))^2 - (2(v1fsin30)(v2fsin(alpha)) cos theta)

    (where alpha is the angle between v2f and the horizontal and theta is the angle between v1f and v2f)

    so would this become something like theta= alpha + 30?
     
  9. Jan 27, 2015 #8

    Stephen Tashi

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    Yes, but you are still thinking of things component-by-component instead of acknowledging that the magnitude of a 2-D vector involves two components.

    Write both components of the vectors.

    |(0, mv1)|^2 = |(m v'1 cos(30 deg), mv'1 sin(30 deg) )| ^2 + |(mv'2 cos (alpha), mv'2 sin(alpha)|^2 - 2 |(m v'1 cos(theta), mv'1 sin(theta) )| |(mv'2 cos (alpha), mv'2 sin(alpha)| cos(30 deg + alpha))

    This give you one equation , not two equations.

    You know the formula for the magnitude of a vector V = (a,b) in terms of it's components - yes ?
     
  10. Jan 27, 2015 #9
    That would be Pythagorean theorem... c^2 = a^2 + b^2
    so the first term (on the right) would become

    |sgrt((mv'1cos(30deg))^2 + (mv'1sin(30deg))^2)|^2... and so on...

    also in the second half of your equation (the -2AB...), is the theta referring to the 30 deg? or the 30 deg +alpha?
     
  11. Jan 27, 2015 #10

    Stephen Tashi

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    Yes. It comes out to be (mv'1)^2 since sin^2(angle) + cos^2(angle) = 1.

    [/QUOTE}

    Yes. I should have used 30 deg for theta.
     
  12. Jan 27, 2015 #11
    OK,
    so since sin^2(angle) + cos^2(angle) =1 is in both A and B terms, then everything would simplify to:

    mv1^2 = mv'1^2 + mv'2^2 - 2(mv'1)(mv'2) cos (30deg +theta)
     
  13. Jan 27, 2015 #12

    Stephen Tashi

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    Yes. Then you can apply conservation of energy to show the last term must be zero.
     
  14. Jan 27, 2015 #13

    Stephen Tashi

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    mv1^2 = mv'1^2 + mv'2^2 - 2(mv'1)(mv'2) cos (30deg +alpha)

    In my notation the angle is 30 deg + alpha.
     
  15. Jan 27, 2015 #14
    Yes, sorry, just a typo.

    So since the 2(mv'1)(mv'2) cos (30deg + alpha) is 0.... then that just leaves me with mv1^2 = mv'1^2 + mv'2^2.... but then I still have two unknowns. So I would need another equation to substitute into that to figure out one of the unknowns.

    Now I am confused again... what can I possibly use to figure this out?
     
  16. Jan 27, 2015 #15

    Stephen Tashi

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    Don't forget to explain why 2(mv'1)(mv'2) cos (30 deg + alpha) = 0 in your work.

    2(mv'1)(mv'2) cos (30 deg + alpha) = 0 should tell you something about the possibilities for alpha
     
  17. Jan 28, 2015 #16
    Well I was thinking that the 2ABcos(angle) term had to be 0 because the rest of the equation (C^2 = A^2 + B^2) is analogous to the conservation of NRG eqn ( (1/2)mv1^2 + (1/2)mv2^2 = (1/2)mv'1^2 + (1/2) mv'2^2 ) and because of that, 2ABcos(angle) had to be zero.

    But the only for sure way that 2ABcos(angle) could be 0 is if the cos(angle) term is 0, which would mean that alpha is 60deg?
     
  18. Jan 28, 2015 #17

    Stephen Tashi

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    That's correct reasoning. You can show it by algebra if you write the conservation of kinetic energy equation , multiply it by 2 and subtract that equation from the above equation.


    Yes, if after the collision we draw the first balls momentum vector in the first quadrant and the second ball's momentum vector in the fourth quadrant, the angle of the second balls momentum vector measured counter clockwise to the x-axis is 60 deg. You would call the angle of the second ball's momentum vector -60 deg or 300 deg when describing it in the conventional way as an angle in the fourth quadrant.

    cos(angle) = 0 when angle = 90 deg or angle = 270 deg. You can say the 270 deg solution won't solve the problem since the second ball's momentum vector must be in the fourth quadrant in order to conserve y-momentum.
     
  19. Jan 28, 2015 #18
    OK, now that I have the angle of v'2, I think I know how to figure out v'1 and v'2;
    Using conservation of momentum...

    In the x direction
    m1v1 + m2v2 = m1v'1 + m2v'2
    (2)(3) + (2)(0) = (2)v'1cos(30) + (2)v'2cos(60)
    3 = v'1cos(30) + v'2cos(60)

    In the y direction
    m1v1 + m2v2 = m1v'1 + m2v'2
    (2)(0) + (2)(0) = (2)v'1sin(30) + (2)v'2sin(60)
    v'1sin(30) = v'2sin(60)
    v'1 = (v'2sin(60))/sin(30)

    Sub y eqn into x eqn

    3 = ((v'2sin(60))/sin(30))cos(30) + v'2cos(60)
    3 = v'2(1.5) + v'2(0.5)
    3 = 2v'2
    v'2 = 1.5

    Sub v'2 in y eqn

    v'1sin(30) = (1.5)sin(60)
    v'1 = 2.6

    So v'1 = 2.6m/s and v'2 = 1.5m/s

    Thank you so much for taking some of your time to help another having serious struggles! You are a champ for being patient with someone not so fluent in physics.
    Thanks again. :)
     
  20. Jan 29, 2015 #19

    Stephen Tashi

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    You need to subtract 2v'2 sin(60) or treat the angle as -60 deg, which will have the same effect. The y-components point in the opposite directions.
     
  21. Jan 29, 2015 #20
    Right! Missed that!

    Thanks again for all your help.
     
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