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Conservation of Momentum and gravity

  1. Jan 16, 2009 #1
    1. The problem statement, all variables and given/known data
    A student performs a ballistic pendulum experiment using an apparatus similar to that shown in the figure. Initially the bullet is fired at the block while the block is at rest (at its lowest swing point). After the bullet hits the block, the block rises to its highest position, see dashed block in the figure, and continues swinging back and forth. The following data is obtained: the maximum height the pendulum rises is 3 cm, at the maximum height the pendulum subtends an angle of 36.9◦, the mass of the bullet is 97 g, and the mass of the pendulum bob is 788 g. The acceleration of gravity is 9.8 m/s2 .


    2. Relevant equations

    I thought this was the equation I needed to use: (1/2)(m1+m2)v^2 (final) =( m1 +m2)gh
    (to solve for v(final))

    Along with: (m1+m2)V (initial)= (m1 +m2) v(final)

    3. The attempt at a solution

    I continually get around 2.277, yet I know that's wrong.

    Also, I don't really know where the trig is applied....
     
  2. jcsd
  3. Jan 17, 2009 #2
    The question doesn't say what you have to compute and I only know that you "get around 2.277 and that it's wrong" (what quantity? units?) so I'll just outline the general approach.

    m - mass of the bullet
    M - mass of the block
    v - speed of the bullet
    V - speed of the block+bullet system
    h - distance the system rises
    K - initial kinetic energy of the system after impact
    U - potential energy of the system at the highest position

    To get the speed of the system after the impact just apply the conservation of momentum:

    mv = (m + M)V

    Then conservation of energy:

    K = U

    K = 1/2 * (m + M)V^2

    U = (m + M)gh

    The rest is just rearranging and substituting.
     
  4. Jan 17, 2009 #3
    This looks to be the problem, since only the bullet is moving initially, the block is at rest.
     
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