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Homework Help: Conservation of Momentum in an Inelastic Collision

  1. Jan 24, 2008 #1
    1. The problem statement, all variables and given/known data
    A bullet of mass m and speed v passes completely through a pendulum bob of mass M. The bullet emerges with a speed of v/2. The pendulum bob is suspended by a stiff rod of length L and negligible mass.

    What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle.

    Basically, what occurs is a bullet m is shot through a mass M which is suspended by a rod and pinned on the wall (it pivots in a circle). The problem asks for the minimum value of v so that M will spin in a circle once if the bullet emerges with a speed of v/2.

    2. Relevant equations


    Maybe a work/energy equation?

    3. The attempt at a solution

    I understand that the momentum of the bullet is transferred to the pendulum which is why the pendulum swings around and the bullet leaves with half the velocity it entered with, but I don't know how to find the amount of energy it takes to spin the pendulum.

    Any help would be greatly appreciated.
  2. jcsd
  3. Jan 24, 2008 #2
    Do you know what that statement means?
  4. Jan 24, 2008 #3
    It means that the pendulum swings around in an perfect circle (it stops exactly where it starts).
  5. Jan 24, 2008 #4
    Yes, but it can swings with different velocities. What would be the minimum one, at the upper vertical position?

    P.S. Actually the pendulum never stops!
  6. Jan 24, 2008 #5
    I don't think it uses the equation I listed above with a set velocity. They give me L (the radius of the spin), which leads me to believe that I need to calculate the work done by the pendulum in the spin (W=FD and d is 2Lpi (circumference), but I'm not sure about force) and correlate the work done to the energy lost by the bullet when it leaves the pendulum. Also, there really is no displacement, so I'm not sure if I can even use the work equation for that.

    What do you mean by the pendulum never stops? I'm assuming there is gravity even though they don't mention it.
    Last edited: Jan 24, 2008
  7. Jan 24, 2008 #6
    You don't have to use a set velocity in the conservation of momentum!

    But you do have to understand from the problem's statement that the pendulum must have zero velocity in the upper vertical position.
    That's why it states minimum and barely.
  8. Jan 24, 2008 #7
    Are you sure? I understand a=0 at that point, but the pendulum is moving the entire rotation.
  9. Jan 24, 2008 #8
    You are asked to calculate the minimum value, so think like this.

    Let A be the point where the collision took place and let B the upper point.

    In order for the pendulum to make the entire rotation it must reach point B. The velocity of M at B depends from the velocity at A, [itex]v_B[/itex] increases if [itex]increases[/itex]. Thus if you find the [itex]v_A[/itex] in order [itex]v_B=0[/itex], then you can say that this [itex]v_A[/itex] is the mimimum one, in the sence that every velocity above that ensures that the pendulum makes an entire rotation.

    P.S. At point B the accerelation is not zero.
    Last edited: Jan 24, 2008
  10. Jan 24, 2008 #9
    Ohhh, I'm sorry if I incorrectly explained the problem, but the bullet is traveling at v and it asks for the minimum speed of the bullet as it passes through the pendulum in order for the pendulum to go around one rotation.
  11. Jan 24, 2008 #10
    :smile: My English are pretty poor! I will try again to explain.

    • From the conservation of momentum you can find the velocity [itex] V_A[/itex] of M, in terms of the velocity v of m.
    • From the conservation of energy for the mass M, from A to B, you can find the velocity [itex]V_B[/itex] in terms of [itex]V_A[/itex], i.e. in terms of v.
    • Now demanding [itex]V_B=0[/itex] in order the pendulum barely swings you can calulate v.

    Does it makes any sence?
  12. Jan 24, 2008 #11
    Doesn't Va of M is equal to zero since the pendulum is initially not moving? I don't quite understand what to do when the initial and final velocities of an object is zero, but it moves in between that period.
  13. Jan 24, 2008 #12
    Mass M:
    Velocity before collision zero
    Veloity after collision [itex]V_A[/itex] not zero! (zero at point B)

    Mass m:
    Velocity before collision v
    Velocity after collision v/2

    Can you find [itex]V_A[/itex] from the conservation of momentum?
  14. Jan 24, 2008 #13
    Ahhhh, I simply solve for the variable Va. Let me try that! :) I really appreciate your help Rainbow Child.
  15. Jan 24, 2008 #14
    Ok! The equation you wrote (conservation of momentum) reads

    [tex]p_{initial}=p_{final}\Rightarrow m\cdot u+M\cdot 0=m\cdot \frac{u}{2}+M\cdot V_A[/tex]

    so you have for [itex]V_A[/itex] ...
  16. Jan 24, 2008 #15
    Correct, that is what I have. I solved for Va and got Va=mv/2M. I'm not sure what to do after this. How would I find Vb and v from this?
  17. Jan 24, 2008 #16
    Now write the conservation of energy for M, from the point A to the point B, to calculate [itex]V_B[/itex] in terms of v.
  18. Jan 24, 2008 #17
    I don't quite understand. Is that the 1/2mv^2 equation?
  19. Jan 24, 2008 #18
    The conservation of energy is

    Potential EnergyA+Kinetic EnergyA=Potential EnergyB+Kinetic EnergyB (1)

    For mass M,

    [tex]PE_A=0,\, KE_A=\frac{1}{2}\,M\,V_A^2,\,PE_B=M\,g\,2\,L,\, KE_B=0[/tex]

    Plug the above values in (1) in order to find [itex]V_A[/itex]
    Last edited: Jan 24, 2008
  20. Jan 24, 2008 #19
    I just solved for Va and got sqrt(2gL) (the mass canceled out). This equation gives me the speed of if the pendulum before the collision. I also have the speed of the pendulum after collision, which I found earlier from the conservation of momentum equation. Is this correct?
  21. Jan 24, 2008 #20
    Oupps! My mistake! [tex]PE_B=M\,g\,2\,L[/tex]. The mass M is at height [itex]2\,L[/itex].
    Now equate the two results for [itex]V_A[/itex] in order to find v.
  22. Jan 24, 2008 #21
    Ahhh! I should've caught that mistake. I now have Va=sqrt(4gL). I understand how to get v, but I am still unclear why mv/2M and sqrt(4gL) are equal? Are they just two different ways to write Va?
  23. Jan 24, 2008 #22
    We have computed [itex]V_A[/itex] by two ways.

    The velocity [itex]V_A[/itex] is the final velocity of M after the collision, and simultaneously is the initial velocity for the movement of the pendulum.
  24. Jan 24, 2008 #23
    Ah, everything finally makes sense now. I re-read everything you've said and it's all very clear now. :)

    Thanks again Rainbow Child! You're a physics genius! I extremely appreciate you sticking with me the 3 or so hours it took for me to complete this fairly simple problem.
  25. Jan 24, 2008 #24
    Every problem fails to the category "fairly simple problem" when it is solved!!! :rofl:

    Glad I helped! :smile:
  26. Feb 17, 2010 #25
    Why don't you get the right answer if you only use conservation of energy->

    initial KE = final KE + final U

    [tex]\frac{1}{2} mv^2 = 2LMG + \frac{1}{2} m(v/2)^2[/tex]
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