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Homework Help: Ball Momentum

  1. Mar 23, 2005 #1
    I had this problem on a quiz today and I got 4.05m. I'm not sure if this is right though, can anyone verify this?

    A 3kg ball is situated on a bent frictionless wire at A as shown below (I hope this shows up) and a 1 kg ball is situated at B. After the two balls collide, how far does ball B rise up the other side of the wire? The height of A is 1.8m.

    -------\------------ /
    --------\---------- /
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    Last edited: Mar 23, 2005
  2. jcsd
  3. Mar 23, 2005 #2
    Conserve momentum in the collision to find the velocity of ball b. You should see that it is 3 times slower than ball A.

    The kinetic energy ball A gains from falling will be the same as the kinetic energy lost by ball B by climbing the ramp.
  4. Mar 23, 2005 #3
    I solved it and got 4.05m, I just don't know whether or not its right.
  5. Mar 25, 2005 #4
    As much as I hate to bump threads....

    I thought this was a simple problem, but does no one know for sure? I'm pretty sure between 3.9 and 4.05 is right, I just need some verification, because no one else got what I got.
  6. Mar 25, 2005 #5


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    You can't assume the first ball will stop. Use the initial potential energy to find the speed and so kinetic energy and momentum of the first ball just before the collision. Use conservation of momentum and energy to find the speed of both balls just after the collision. Use conservation of energy to find how high the second ball goes.
  7. Mar 25, 2005 #6
    I don't assume anything. I got that the speed of the first ball after the collision is 2.95 m/s and the speed of the second is 8.9 m/s. I just don't know whether or not my numbers are correct.
  8. Mar 25, 2005 #7
    Is there a reason why he cant assume hte ball will stop? Its just a simple conservation of momentum in an elastic collision. If you found the speed of the ball before impact, use [tex] m_1v_1 = m_2v_2 [/tex] to find the velocity of the second ball. Calculate the second ball's kinetic energy, which will all be lost to potential energy.

    [tex] KE = PE, \frac{mv^2}{2} = mgh [/tex]

    Solve that for h, and you should be able to find how high the second ball will climb.
  9. Mar 25, 2005 #8
    Yeah. Because it doesn't. :P
  10. Mar 25, 2005 #9
    Why not!!! How do you know how much energy was transferred?
  11. Mar 25, 2005 #10
    1: [tex] p_i = p_f [/tex]
    2: [tex] (3kg)(5.9m/s) = 3v_1_f + v_2_f [/tex]

    And, in an elastic collision
    3: [tex] v_1_i - v_2_i = -v_1_f + v_2_f [/tex]
    4: [tex] 5.9m/s - 0m/s = -v_1_f + v_2_f [/tex]
    5: [tex] 5.9m/s + v_1_f = v_2_f [/tex]

    Now I replaced the [tex] v_2_f [/tex] in line 2 with the work in line 5.

    6: [tex] (3kg)(5.9m/s) = 3v_1_f + 5.9m/s + v_1_f [/tex]
    7: [tex] 11.8 = 4v_1_f [/tex]
    8: [tex] v_1_f = 2.95m/s [/tex]
    9: [tex] v_2_f = 8.85m/s [/tex]

    So the speed of ball one after the collision is 2.95 m/s. It does still move.

    Now for the rest of my work so someone might verify it.....

    Line 9 is rearranging of [tex] mgh = 1/2mv^2[/tex]
    10: [tex] v_f^2 = v_i^2 + 2gh [/tex]
    11: [tex] (-8.85m/s)^2 = 2(-9.8)h [/tex]
    12: [tex] h = 4.0 [/tex]
    Last edited: Mar 25, 2005
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