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Spring-Gun and Bullet Momentum and Energy

  1. Nov 6, 2012 #1
    1. The problem statement, all variables and given/known data
    A ball of mass m which is projected with speed vi into the barrel of a spring-gun of mass M initially at rest on a frictionless surface, as shown in the attached file below. The ball sticks in the barrel at the point of maximum compression of the spring. No energy is lost in friction.

    A) In terms of the given masses and the kinetic energy, what energy is stored in the spring at its maximum compression?

    B) If the mass of the ball and the gun are equal and the spring constant is given as k, determine the maximum compression of the spring in terms of the initial kinetic energy and the spring constant k.


    2. Relevant equations
    m1v1 + m2v2 = m1v1' + m2v2'
    Kinetic Energy=(1/2)mv2
    Spring Energy=(1/2)kx2


    3. The attempt at a solution
    A) At maximum compression, the speed of the ball will equal that of the gun since the ball is stuck in the barrel. Therefore I used the conservation of momentum:
    mvi = (m+M)vf
    vf = (mvi)/(M+m)
    At this point, I wasn't sure about my work... I equated the energy stored in the spring to the difference between the initial KE and the final KE:
    Espring=(1/2)(m)(vi)2 - (1/2)(M+m)(vf)2
    If I substitute the fraction (mvi)/(M+m) for vf, would I have the right answer?

    B) I would set (1/2)(m)(vi)2 - (1/2)(M+m)(vf)2 equal to: (1/2)(k)(x2) where M=m. Then I would solve for x... would that be correct?

    Thanks in advance!
     

    Attached Files:

  2. jcsd
  3. Nov 6, 2012 #2

    haruspex

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    Minor quibble: yes, at maximum compression, the speed of the ball will equal that of the gun, but that's because it is at maximum compression, and has nothing to do with the ball's becoming stuck.
    Other than that, your method looks sound.
     
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