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Homework Help: Collision between two balls physics problem

  1. Oct 28, 2007 #1
    Two identical steel balls, each of mass 4.3 kg, are suspended from strings of length 30 cm so that they touch when in their equilibrium position. We pull one of the balls back until its string makes an angle θ = 65° with the vertical and let it go. It collides elastically with the other ball. How high will the other ball rise?

    img:http://i242.photobucket.com/albums/ff106/jtdla/prob01a-1.gif [Broken]

    I have a few ideas as to how to tackle this:

    To get the velocity one ball travels couldn't I do this:

    mgh+0=1/2mv^2+0 (for the ball on the left)

    Then could I put that velocity into a momentum equation.

    My second idea was wouldn't the ball that is just sitting there go up to the same height as the other one's initial height. I remember something being demonstrated with so little motor cars when we were talking about collisions.
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Oct 28, 2007 #2


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    What you are describing is a simplified Newton's Cradle. Assuming, as in a ideal Newton's Cradle, that all of the first ball's momentum is transferred to the second, I would agree with you that the second ball should rise as high as the first.
  4. Oct 28, 2007 #3
    I got 17.32 cm, right on the first try!

    I have a next question that I'm sure I would solve in a similar way:

    In a pool game, the cue ball, which has an initial speed of 1.3 m/s, makes an elastic collision with the 8-ball, which is initially at rest. After the collision, the 8-ball moves at an angle θ = 35.1° with respect to the original direction of the cue ball, as shown in the figure. At what angle φ does the cue ball travel after the collision? Assume the pool table is frictionless and the masses of the cue ball and the 8-ball are equal.

    I'm sort of sure how to do this at all. It is currently the last problem that has stumped me.
  5. Oct 28, 2007 #4


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    Good Job! Yes, this problem is also a conservation of momentum problem.

    HINT: Remember that if momentum is conserved then the x and y components also have to be conserved. Tell me if you get stuck.
  6. Oct 28, 2007 #5
    I'm still stuck because I don't know how to interpret this in general equation form.
  7. Oct 28, 2007 #6
    well actually

    Momentum before= sin(theta)*1.3+sin(35.1)*1.3

    would that be about right?
  8. Oct 28, 2007 #7
    That doesn't look good to me.
    This is a question in which bodies move in a plane. (You cannot sum up whole motion of the two bodies in one direction.) So, we need two axes, say x and y. Let +ve x-axis be along the direction of the motion of initial cue-ball. And y-axis be in the plane of the table. Conserve momentum along each of the directions: x & y. Note that, initial momentum along x-axis was 1.3*M kg.m/s, where M is the mass of the cue ball; and that along y-axis, it was 0 kg.m/s!
  9. Oct 28, 2007 #8

    Momentum before= -M*sin(theta)*1.3+M*sin(35.1)*1.3

    Y- direction

  10. Oct 29, 2007 #9
    This is in reference to my original question:
    Suppose that instead of steel balls we use putty balls. They will collide inelastically and remain stuck together after the collision. How high will the balls rise after the collision?

    How would I set this up since KE is not conserved in an inelastic collision.
  11. Oct 29, 2007 #10


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    KE is not conserved but what about momentum?
  12. Oct 29, 2007 #11
    So momentum is conserved, but I'm not sure how to calculate velocity. Could I still use PE=KE at the bottom to get the velocity?
  13. Oct 29, 2007 #12


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    Yes. Mechanical energy is still conserved in that part of the problem. Then, how do you find the velocity of the big blob of clay after the collision?
  14. Oct 29, 2007 #13
    I assume I can just use the conservation of momentum, except the mass will be twice as big at the end.
  15. Oct 29, 2007 #14
    I've tried a few methods to no avail, could you give me a better hint?
  16. Oct 29, 2007 #15


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    Can you show me some work? I have no idea why your methods aren't working if you don't show them to me. If you want more detailed help, I'm going to need to see more details.
  17. Oct 29, 2007 #16
    So I took the PE=KE of just one ball and got the velocity as 1.8425 m/s.

    I setup up a few equations such as:

    1/2*4.3*1.8425^2=1/2*8.4*x^2 and then I got 1.30284 as the combined velocity.

    I then set up KE=KE+PE with the new velocity and got:

    .000067 M
  18. Oct 30, 2007 #17
    It was an inelastic collision, right? How come you are conserving energy for the duration between just before the collision and just after the collision?

    Basically, the question can be broken into 3 sub-parts:
    1. From the start of motion of the first ball to the moment just before collision. External work done by all other forces (except gravity) is zero, therfore you can use conservation of energy on the first ball.
    2. From the moment just before the collision starts to the moment just after the collision ends. Since the collision is inelastic, energy can't be conserved. But, net force on the two-balls system is zero in the direction of speed of the first ball at the moment, momentum can surely be conserved in that direction.
    3. From the moment after the collision takes place to (atleast) the moment(s) when each of the balls reach their topmost position. Again, one can apply conservation of energy, since on each of the balls, work done by external force (other than gravity) is zero.

    I hope you get the correct result now. Remember, being able to break a big problem into smaller problems is generally fruitful in Physics.
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