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Homework Help: Law to preserve motion quantity help ( picture included to help describe )

  1. Sep 23, 2008 #1
    Speed of a bullet shot into a pendulum ,picture included (dl attch. if pic wont show)

    1. The problem statement, all variables and given/known data

    You're supposed to be able to define the speed of a bullet with the help of a pendulum.
    The bullet is shot into a big tree piece ( The pendulum ) with a known mass (weight). The tree piece's center of gravity is lifted upwards when hit by the bullet. What is the speed of the bullet if the bullet's mass (weight) is 7.5g ( 0.0075 kg ), the tree piece's mass (weight) is 4.3 kg and the tree piece is lifted upwards in height of 8.8 cm ( 0.088 m )?

    bullet: m1 = 7.5 g = 0.0075 kg :: v (speed) = ? ( This is the question )

    tree piece: m2 = 4.3 kg ( it is the pendulum, see picture below )

    height: h = 8.8 cm = 0.088 m ( the length that the tree piece is lifted upwards ).


    2. Relevant equations

    E(sum) p (before) = E(sum) p (after)

    m1v1 + m2v2 = m1u1 + m2u2,

    - where m1v1 and m2v2 are 'motion quantity' before the collision whereas m1u1 and m2u2 are the 'motion quantity' after the collision.

    I think this is what they want me to use.

    3. The attempt at a solution

    Ok I've been trying at this for a while now but I can't figure out what to DO with the HEIGHT that is given ( tree piece lifted 8.8 cm ). I can't find a way to define the length the tree piece moves right on the x-plane.

    Basically all solutions I've come up with have all been wrong. I tried to use the "Law to preserve energy", the one with E(potential) = E(kinetic) But it didn't work out either.


    Any TIPS on how to solve this? Thanks in advance.

    Attached Files:

    Last edited: Sep 23, 2008
  2. jcsd
  3. Sep 23, 2008 #2


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    Homework Helper

    Re: Speed of a bullet shot into a pendulum ,picture included (dl attch. if pic wont s

    Remember you need to consider that there is a conservation of energy.

    In this case the Kinetic Energy m*v2/2 will be translated into m*g*h which is the change in height of the combined masses.
  4. Sep 23, 2008 #3
    I get this:

    m1v1 = m1u1+m2u2
    v1 = u1+(m2u2/m1)

    And also this:

    v1 = sqr(2gh) (= 1.3139... )

    sqr(2gh) = u1+m2u2/m1

    Now I can't do anything with this because u1 and u2 are still undefined. You need to know the lengths of the ropes or something. Grrr. Wait. When exactly does the Kinetic energy convert? Is it when it HITS the box or when the box is at height 8.8 cm? Should be when it's at 8.8 cm and at the peak of it's swing on the top right, right?

    But then again this doesn't help at all. We've got no angles or other Forces except (m1+m2)g at that point. Wth goddamnit. I need to define u1 or u2 before I can know what v1 is, right? But it's impssible, there's too little info. What am I missing??
  5. Sep 23, 2008 #4
    Wait a sec let's see...

    m1v1 + m2v2 = m1u1 + m2u2 : v2=0
    m1v1 = m1u1 + m2u2 : Is it safe to assume u1 and u2 are the same since they become, like, 1?

    then - m1v1 = (m1+m2)u
    v1 = (m1+m2)u/m1

    Now is it that the Kinetic energy appears at the HIT, and turns into Potential energy when it stops moving. So in face the Kinetic energy of the bullet itself merges with the pendulum and becomes another form of Kinetic energy, thus: (1/2)(m1+m2)u2 = (m1+m2)g*h

    so we finally get u = sqr(2gh) which we put in v1 = (m1+m2)u/m1 to get what v is!

    Goddamn, seems energy just becomes whenever, quite simple in fact, maybe too simple? With forces you always have to figure out the geometrics but all you needed to figure out here was jus a bit how energy works i guess...
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