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Homework Help: Oblique Collision Trigonometry problem

  1. Aug 15, 2011 #1
    1. The problem statement, all variables and given/known data
    A smooth sphere collides with an equal sphere which is at rest. Before the impact the line of motion of the first sphere makes an angle A with their line of centers, after the impact this is angle B. The coefficient of restitution is 1/4. Show that 8TanA = 3TanB

    2. Relevant equations
    -e = (V1 - V2) / (U1 - U2)

    M1U1 + M2U2 = M1V1 + M2V2

    3. The attempt at a solution

    Initial Velocity 1: (XcosA i + XsinA j)
    Initial Velocity 2: (0i + 0j)

    Final Velocity 1: (YcosB i + YsinB j)
    Final Velocity 2: (P i + 0 j)

    XsinA = YsinB. J components are equal.

    Coefficient of restitution
    -1/4 = [YcosB - p] / [XsinA]
    -XsinA = 4YcosB - 4p

    Conservation of momentum in i direction

    masses are equal.
    U1 + U2 = V1 + V2

    XcosA = YcosB + p
    XcosA - YcosB = p

    -XsinA = 4YcosB - 4( XcosA - YcosB )
    -XsinA = 8YcosB - 4XcosA

    -XsinA / -XcosA = 8YcosB /-XcosA -4XcosA /-XcosA
    Tan A = (8YcosB) / (-XcosA) - 4
  2. jcsd
  3. Aug 16, 2011 #2


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    This equation holds for one-dimensional motion. In general, the coefficient of restitution is (speed of separation)/(speed of aproach). You need to calculate with the magnitude of relative velocities before and after the collision.

    Was it given that the second ball moves in the direction of the line connecting their centres?

  4. Aug 16, 2011 #3
    I thought that was the equation for relative velocity? According to my book anyway. It shows the derivation.

    As for moving only on the i axis, according to my book "the bounce / impact takes place along the i axis therefore the j component of the velocity remains unchanged.

    I'm not having a problem with the physics, just the trig. I can't whittle it down to 8Tan A = 3TanB
  5. Aug 16, 2011 #4


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    You took the line connecting the centres of the spheres as x axis and the y axis is perpendicular to this line. Nothing ensures that the second ball will move along the x axis.
    In case of bounce from the ground or from a wall, the force during impact is normal to the wall, so only the normal component of the velocity changes. It is not the case here.

    The coefficient of restitution gives the relation between the magnitudes of the relative velocities, not only of the x components. See:https://www.physicsforums.com/library.php?do=view_item&itemid=270

  6. Aug 16, 2011 #5
    "You took the line connecting the centres of the spheres as x axis and the y axis is perpendicular to this line. Nothing ensures that the second ball will move along the x axis."

    But the j component of the velocity never changes. If it's originally 0 doesn't it stay 0?

    This is what my book says. See picture attached.

    Attached Files:

  7. Aug 16, 2011 #6


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    OK, it is an approximation, which is not always true in reality. Can you kick a ball so as it starts to spin? It is because the interaction between your foot and the ball has tangential component to, not only radial. When friction is negligible between the colliding balls, however, the approximation is valid. And the balls in this problem are perfectly smooth. So you are right, the y components do not change.
    As for the coefficient of restitution, I see, there are different definitions in different books.

    You made a mistake when writing up the coefficient of restitution:
    It has to be X cosA.

    And use the equation for the y components of velocity, too: XsinA=YsinB

  8. Aug 16, 2011 #7
    Yeah, all the problems I'm doing at the minute are High School and are under "Ideal conditions"

    Yes, that mistake messed me up a bit. With that error overcome, I have retried the problem.

    Attempt 2

    Ball 1 (XcosA i + XsinA j) ----> (YcosB i + YsinB j)
    Ball 2 (0i + 0j) ----> (Pi + 0j)

    Momentum is conserved in the i direction. Masses are equal.

    u1 + u2 = v1 + v2
    XcosA = YcosB + P
    P = XcosA - YcosB

    Coefficient of restitution.

    -e = [itex]\frac{v1 - v2}{u1 - u2}[/itex]

    [itex]\frac{-1}{4}[/itex] = [itex]\frac{YcosB - P}{XcosA}[/itex]

    -XcosA = 4YcosB - 4P

    P = [itex]\frac{-XcosA -4YcosB}{-4}[/itex]

    So now I have 3 equations.

    1. XsinA = YsinB

    2. P = XcosA - YcosB

    3. P = [itex]\frac{-XcosA - 4YcosB}{-4}[/itex]

    From that I get

    XcosA - YcosB = [itex]\frac{-XcosA - 4YcosB}{-4}[/itex]

    -4XcosA + 4YcosB = -XcosA - 4YcosB

    3XcosA = 8YcosB.

    And that is as far as I can go.
  9. Aug 16, 2011 #8


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    And you have the equation for the y components:

    XsinA = Y sinB**

    Just divide eq. ** with eq * what you get ? You know, do not you, that sinA/cosA=tanA?

  10. Aug 17, 2011 #9
    Yeah I understand that Tan A = Sin A / Cos A

    But I have never come across dividing one equation with another.

    Would you mind explaining please.
  11. Aug 17, 2011 #10


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    It is possible to multiply, divide add or subtract two equations.
    If A=B and C=D then A+C=B+D, A-C=B-D, A*C=B*D, and A/C=B/D (here you must be sure that neither C nor D are zero)

    Just a primitive example: 2=2, 3=3, dividing the equations you get 2/3=2/3.

    Your equations are:
    xsinA=ysinB and
    3xcosA = 8ycosB





    Multiply the whole equation by 24, and you get the solution.

    If you do not like to divide equations, you can express sinA from the first equation, cosA from the second one, and find tanA=sinA/cosA.

  12. Aug 18, 2011 #11
    Excellent, thank you!

    I never realised you could do that.
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