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Dynamics Problem

  1. Oct 8, 2005 #1
    I have tried long and hard on this question and it is the last one i have to do before the assignment is finished.

    Two identical spheres collide. One is at rest on a horizontal surface, the other drops vertically. The collision is as shown (see attatchment). After the collision, both spheres move horizontally. What is the apparrent coefficient of restitution.

    Now we cannot use conservation of momentum in the y direction, or the normal direction, as an external force from the surface will act. We can use it however in the x direction, and by doing this you get;

    v'A=-v'B

    furthermore, by using the equation for the tangential direction of the collision we get;

    tan(23) = (v'A)/(vA)

    here i have used a tangential normal sytem and my dynamics lecturer insists what i have above is all that is required to solve the problem when you apply it to the equation;

    e = (v'B-v'A)/(vA-vB)

    which simplifies to

    e = (v'B-v'A)/(vA)

    However, when i apply all that i have found as above, i get the top line of above equal to zero, and i can assure you this is not the correct answer.

    Clearly i have attmepted this question and i am unfortunately stuck. perhaps someone could help me....
     

    Attached Files:

  2. jcsd
  3. Oct 8, 2005 #2
    Addition

    Sorry, i may not have been clear in the last equation;

    when i say

    e=(v'B-v'B)/(vA)

    i am referring to the velocities initial and final of the balls in the NORMAL direction (along line joing the centres of the balls)

    this i was not clear on (though this is the actual definition of the coefficient of restitution)
     
  4. Oct 8, 2005 #3

    siddharth

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    Let the velocity of the sphere which is falling vertically (sphere 1) just before collision be [itex] v_0 [/itex]

    Let the velocity of sphere 2 be [itex] v [/itex] and let the velocity of sphere 1 after collision in direction tangential and normal to the line joining the two centers be [itex] v_t [/itex] and [itex] v_n [/itex]

    First, of all, velocity of sphere 1 in the tangential direction remains unchanged.
    So, [itex] v_0 sin(\theta) = v_t [/itex]

    Also, as you said, Linear momentum along the x-axis is unchanged. So,
    [tex] mv -mv_t \cos\theta - mv_n \sin\theta = 0 [/tex]

    And the final velocity of sphere 1 in the y-axis is zero. Therefore

    [tex] v_t \sin\theta = v_n\cos\theta [/tex]

    So, you have three equations and three unknowns ([tex] v_n,v_t,v [/tex]).

    Solve for them and then you can find the coeffiecient of restitution which would be
    [tex] (v\sin\theta + v_n)/(v_0\cos\theta) [/tex]
     
    Last edited: Oct 8, 2005
  5. Oct 8, 2005 #4
    Thanks, but i dont understand how you got the final equation you have used...

    i think you may have used conservation of momentum of the first ball in the y direction but im not sure you can...??
     
  6. Oct 8, 2005 #5
    Addition

    Again, sorry, i am very much thankful for your help, but we do not know what v0 is, so doesnt that make it 4 unknowns in only 3 equations...?
     
  7. Oct 8, 2005 #6

    siddharth

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    Whoops! You are right. The 'm' must not be there! I get the equation by taking the components of the tangential and normal velocity along the y-axis.

    Find the 3 unknowns in terms of v0. While finding the coefficient of restitution, the v0 will cancel out in the numerator and denominator.
     
    Last edited: Oct 8, 2005
  8. Oct 10, 2005 #7

    mukundpa

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    may this help you

    Consider the impulse Fdt between the two balls during collision. This is along the line joining the centers.

    For A
    vertically

    Fdt cos@ = 0 - mv_0 where v_o is the velocity of A just before collision.

    Horizontally

    Fdt sin@ = mv_x - 0 where v_o is the velocity of A just after collision.
    gives
    tan@ =v_x/v_0..................(1)

    For B
    Say in absence of any normal force due to the table, the ball would have been move with velocity v along the line joining the centers, then in horizontal direction

    v sin@ = v_x or
    v = v_x/sin@ = v_0 sec@............(2)

    Velocity of approach and separation is measured along the line joins the centers, hence velocity of approach = v_0 cos@ - 0
    and velocity of separation = v - (- v_x sin@ )

    so the coefficient of restitution between the balls
    e = (v + v_x sin@ )/(v_0 cos@)
     
  9. Oct 13, 2005 #8
    Thanks Guys...

    finished the assignment...and that question

    if you were wondering, the correct solution is in fact

    e = 2tan^2(theta)

    which i got after some intensivve rearrangement of the basic formulae....

    Cheers
     
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