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Bullet and Block Circular Problem

  1. Jan 21, 2009 #1
    1. The problem statement, all variables and given/known data
    I am going through past Olympiad qualifying exams as practice for this year's test. I came upon the following problem in the 2008 F=ma and continue to come up with the wrong answer. I keep coming up with C and that is not the answer provided in the answer key.


    22. A bullet of mass m1 strikes a pendulum of mass m2 suspended from a pivot by a string of length
    L with a horizontal velocity v0. The collision is perfectly inelastic and the bullet sticks to the
    bob. Find the minimum velocity v0 such that the bob (with the bullet inside) completes a circular
    vertical loop.
    (a) 2√Lg
    (b) √5Lg
    (c) (m1 + m2)2√Lg/m1
    (d) (m1 − m2)√Lg/m2
    (e) (m1 + m2)√5Lg/m1




    2. Relevant equations

    m1v1= (m1+m2)v2
    1/2 mv ^2 = mgh

    3. The attempt at a solution

    m1v1 = (m1+m2)v2
    v2 = (m1v1)/(m1+m2)

    So the kinetic energy of the block should be:
    1/2 (m1+m2)v2^2
    Then I set the initial kinetic energy of the block equal to the potential energy it would have at height 2L.

    1/2 (m1+m2)v2^2 = 2L(m1+m2)g

    Since v2 = (m1v1)/(m1+m2):

    1/2(m1+m2)*((m1v1)/(m1+m2))^2 = 2L(m1+m2)g

    m1+m2 should cancel:

    ((m1v1)/(m1+m2))^2 = 4Lg

    (m1v1)^2 / (m1+m2) ^2 = 4Lg

    m1 ^2 * v1 ^ 2 = 4Lg / ((m1+m2) ^2)

    Take the Square Root of the whole thing:

    m1 * v1 = 2sqrt(Lg)/(m1+m2)

    SO my answer is (c) (m1 + m2)2√Lg/m1, which is apparently incorrect.

    If someone could help me correct this that would be great.

    Thanks.
     
  2. jcsd
  3. Jan 21, 2009 #2

    rl.bhat

    User Avatar
    Homework Helper

    Then I set the initial kinetic energy of the block equal to the potential energy it would have at height 2L.
    At the hight 2L the block has kinetic energy as well as the potential velocity.
    To find the velocity at the top for minimum tension( = 0), equate the centripetal force to the weight of the block.
     
    Last edited: Jan 22, 2009
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