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Velocity of bullet before hitting a pendulum

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  1. Apr 1, 2017 #1
    <Mentor's note: moved from a technical forum and thus no template.>

    A pendulum consists of a bob of mass M and a rigid rod of length ℓ with negligible mass. The bob is struck by a bullet of mass m. The bullet passes through the bob and emerges with a speed half that of impact. The shot causes the pendulum to swing through a complete circle, but the bob only barely passes the top of the circle. Find v, the impact speed of the bullet.

    v1 is speed of bullet before impact, v2 is speed of bob
    This is what I done so far:
    (at beginning position h=0 therefore potential energy =0)
    1/2mv11 = 1/2m(v2/2)2+Mg2l (since h at top = 2l)
    For pendulum to go in a verticular circular motion at min velocity Fc = Fg
    then: mv2/l = mg => v2 = gl

    so plugging this in makes:
    1/2mv12 = mv12/8+M2v22
    cleaning this up we have
    v12 = 2(mv12 + 16Mv22)/8m

    Am I right in my thinking?. If not can someone explain where I went wrong and how to fix it?
     
    Last edited: Apr 1, 2017
  2. jcsd
  3. Apr 1, 2017 #2

    PeroK

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    I'm not sure I follow your notation. That is correct for the velocity of the pendulum at the top of the circle.

    What you need next is the velocity of the pendulum at the bottom of the circle (immediately after the impact of the bullet).

    I don't follow this at all. You are looking for an expression for the initial speed of the bullet, ##v##, in terms of the other quantities: ##m, M, l, g##.

    PS things like mv12/8+M2v2 are fairly unreadable. You need to sort that out.
     
  4. Apr 1, 2017 #3
    I'm sorry but what do you mean by unreadable, do you not follow the terms in the equation or is it that you do not follow how I formulated the equations? also If I found the velocity for the top of the ball what would I have to do to find it for the bottom?
     
    Last edited: Apr 1, 2017
  5. Apr 1, 2017 #4

    PeroK

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    In the question, you are given the impact speed of the bullet as ##v##, but then you use ##v_1## as the speed of the bullet and ##v_2## as the speed of the bob. And you are using ##v## as the speed of the bob at the top of the swing?

    The first step is to find the velocity of the bob at the top of the circle. You have sort of worked that out to be ##\sqrt{gl}##. Now you need to work out the initial speed of the bob using potential energy.
     
  6. Apr 1, 2017 #5

    haruspex

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    At what point?
    It's an impact. Work will not be conserved.
    I do not understand the mv22 term. m is the mass of the bullet, but you defined v2 as a speed of the bob.
    M has gained PE and still has KE. The last term above appears deficient.
     
  7. Apr 3, 2017 #6
    isn't this an elastic collision meaning work will be conserved?
     
  8. Apr 3, 2017 #7
     
  9. Apr 3, 2017 #8

    PeroK

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    No, it's not an elastic collison. In fact, once you have solved it, you could if you like calculate how much kinetic energy was lost in the collision.
     
  10. Apr 3, 2017 #9
    Oh, see now what I got confused to thank you! and for future reference how do you know it's not and elastic/inelastic collision
     
  11. Apr 3, 2017 #10

    haruspex

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    It is inelastic; it is not elastic.
    In general, you should not assume it is perfectly elastic without good cause. If one object becomes embedded in or passes through the other you can be sure it is completely inelastic. If the impact is very sudden, implying very high accelerations, it is probably at least partly inelastic.
    If the conditions for momentum conservation are met, and that gives you enough information to solve the problem, you should not assume work is conserved.
     
  12. Apr 4, 2017 #11
    Consider how much energy was passed from the bullet to the pendulum, in terms of the variables, and also consider, based on the action of the pendulum, another way to represent this energy in terms of the variables, particularly those that can be measured easily
    This should allow you to come up with an equation that incorporates the variables that can be easily measured
     
  13. Apr 4, 2017 #12
    In an elastic collision the total kinetic energy of the closed system remains the same. For example, if there are two objects making up the closed system, and they collide elastically, the total K.E. of the two objects will remain the same, even though the individual K.E.'s may change.

    In an inelastic collision total K.E. is not conserved. The reason total K.E. is not conserved is because some of the K.E. is transformed to another form of energy. Of course total energy is always conserved in a closed system.

    For practical purposes, if the loss of K.E. is negligible, then you can ignore it and treat the collision as elastic. In fact, although we use the collision of two billiard balls as an example of an elastic collision, in reality it is not perfectly elastic. But we tend to ignore the loss of total K.E. because it is negligible.

    Speaking of practical applications, this problem is related to the ballistic pendulum. This is an old method for measuring the speed of a bullet. It does not require direct measurement of the transit time.
     
    Last edited: Apr 4, 2017
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