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Spheres hanging by strings - Finding impulse of tension

  1. Sep 26, 2013 #1
    1. The problem statement, all variables and given/known data
    Two equal spheres of mass m are suspended by vertical strings so that they are in contact with their centres at the same level. A third equal sphere of mass m falls vertically and strikes elastically the other two simultaneously so that their centres at the instant of impact form an equilateral triangle in a vertical plane. If u is the velocity of m just before impact, find the velocities just after impact and the impulse of tension of the strings.

    2. Relevant equations

    3. The attempt at a solution
    After collision, the hanging spheres will gain velocity along the direction of common normal of spheres. (Common normal is the line joining the centres of sphere)

    See attachment 2 (I haven't shown the strings and rotated the figure).
    Conserving momentum in x-direction,
    From the equation of coefficient of restitution,
    $$e=\frac{\text{Relative speed after collision}}{\text{Relative speed after collision}}$$
    The relative speed before collision is ##u\cos(\pi/6)## and after collision is ##v''-v'\cos(\pi/6)## and for elastic collision, coefficient of restitution is 1. Plugging and rearranging,
    Solving the two equations I have doesn't give me the right answer. :(

    Any help is appreciated. Thanks!

    Attached Files:

  2. jcsd
  3. Sep 26, 2013 #2
    Clearly the hanging spheres are constrained to move horizontally immediately after the collision. So their strings must have impulses preventing any vertical motion, just like the problem indicates.
  4. Sep 26, 2013 #3
    So do I assume that after collision, the hanging spheres have velocity in x direction and conserve momentum again?
  5. Sep 26, 2013 #4
    On your diagram, the x axis is vertical. As I said, the hanging spheres have no motion vertically: it is prevented by the strings.
  6. Sep 26, 2013 #5
    Sorry about that but what should I do now? I don't really know how to begin.
  7. Sep 26, 2013 #6
    One thing to observe is that momentum is not conserved in vertical direction because of the impulsive tension in the strings; this yields an equation. Momentum is conserved in the horizontal direction, which has an immediate consequence (which is also obvious from symmetry). Energy is conserved, too.
  8. Sep 26, 2013 #7
    And which equation is that? :confused:

    Conserving energy,
    $$\frac{1}{2}mu^2=\frac{1}{2}mv'^2+2 \times \frac{1}{2}mv''^2$$
    where v'' is the horizontal velocity of hanging balls after collision.
  9. Sep 26, 2013 #8
    Vertical momenta and impulses.
  10. Sep 26, 2013 #9
    Change in vertical momentum i.e impulse is ##mv'-mu## but how does it help? Impulse is force multiplied by time but I don't have the time of collision here. I am sorry if I miss something obvious.
  11. Sep 26, 2013 #10
    You are not required to find the force, you are required to find the impulse, so not knowing the time is OK.

    Knowing the impulse by one string should yield another equation that relates the momentum of the striking sphere, the impulse of the string, and the momentum of the hanging ball.
  12. Sep 26, 2013 #11
    I still don't understand what to do.

    Let me clear something first. I found the impulse to be ##mv'-mv##. Is this the net impulse by "two" strings? How do I find the impulse by one string? Divide the above by 2? And how do I relate it to the momentum of hanging ball? The momentum of hanging ball is always zero in the vertical direction so I don't understand what am I supposed to do here.
  13. Sep 26, 2013 #12
    The falling ball's momentum is purely vertical. Yet the impact is via two off-vertical directions. So the exchange of the momenta happens via these two directions. Can you decompose the incoming momentum into those two directions? Can you decompose the reaction from the hanging balls into those two directions? The hanging balls acquire strictly horizontal momenta. Add to that the impulsive tensions delivered vertically. What do you get?
  14. Sep 26, 2013 #13
    Is it ##mv\cos(\pi/6)##? Just to be sure, you are talking about the line joining the centres of the spheres. Right?
    Why you ask me to decompose the reaction force? I mean the reaction force is already along the line joining the centres of the spheres.
  15. Sep 26, 2013 #14
    The direction is correct, but the magnitude is not. The sum of these two momenta must be ##mv##, vertically downward.

    I did not say "force". But I did say "reaction", which was wrong. I should have said "the recoil" of the falling ball after the collision.
  16. Sep 26, 2013 #15

    I still don't get this. Can you please elaborate?
  17. Sep 26, 2013 #16
    The ball falls down. You decomposed its initial momentum into two directions.

    After the collision, it goes up. So its has some upward momentum now. Decompose it into those same directions.

    Then for each direction you can set up a vector equation relating the incoming momentum, the recoil momentum, the momentum of the hanging ball, and the impulse by the string.
  18. Sep 26, 2013 #17
    Let the sphere has velocity v' in the vertical direction after collision.

    Are you talking about direction AB? Change in momentum of striking sphere in that direction is ##mv'/\sqrt{3}+mu/\sqrt{3}##. The tension is in vertical direction and momentum of hanging balls in the horizontal direction. How do I relate them? :confused: :cry:

    Attached Files:

  19. Sep 26, 2013 #18
    I = incoming momentum (AB direction)
    R = recoil momentum (AB direction)
    S = hanging ball momentum (horizontal)
    T = impulse of tension (vertical)

    Total change of momentum: R + S - I.
    Total impulse: T.

  20. Sep 26, 2013 #19
    Some observations-

    1)The hanging masses moves in horizontal direction after collision (say v).
    2)The falling mass moves vertically (say w).
    3)The net impulse in vertical direction on hanging mass is zero .

    Now these are the steps you need to take

    1) Apply impulse momentum theorem in horizontal direction on left(or right) hanging mass .
    2) Apply impulse momentum theorem in vertical direction on left(or right) hanging mass .
    3) Apply restitution equation along the normal on falling mass and left(or right) hanging mass .
    4) Apply impulse momentum theorem on either falling mass or left(or right) hanging mass or both .

    Be careful with the signs.

    What do you get with step 1 ?
  21. Sep 26, 2013 #20
    Okay, I am redefining the coordinate system to make things easier. The vertical direction is y-axis and horizontal as x-axis. Sorry for the inconvenience, I should have done this from the beginning.

    I have drawn a vector diagram showing R,S and I. Change in momentum is:
    $$\vec{T}=\left(mv''-\frac{mv'}{2\sqrt{3}}-\frac{mu}{2\sqrt{3}}\right) \hat{i}+\left(\frac{mv'}{2}+\frac{mu}{2}\right) \hat{j}$$
    What should I do now? :confused:

    Attached Files:

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