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Conservation of Linear Momentum (spring)

  1. Nov 18, 2013 #1
    1. The problem statement, all variables and given/known data

    A mass 3M moving horizontally with velocity v[itex]_{0}[/itex] on a frictionless surface, strikes head-on and sticks to a horizontal spring system of natural length l and spring constant k with masses M at each end. The spring has negligible mass.

    Determine the maximum compression of the spring.

    2. Relevant equations

    Conservation of Linear Momentum:
    Initial Linear Momentum = Final Linear Momentum
    Conservation of Energy:
    Initial Kinetic Energy + Initial Elastic Potential Energy = Final Kinetic Energy + Final Elastic Potential Energy

    3. The attempt at a solution

    3Mv[itex]_{0}[/itex] = (3M + M)Vfinal
    Vfinal = [itex]\frac{3}{4}[/itex]v[itex]_{0}[/itex]

    At Maximum compression: all the Kinetic Energy would have been converted to Elastic Potential Energy; let x be the maximum compression.

    [itex]\frac{1}{2}[/itex](4M)([itex]\frac{3}{4}[/itex]v[itex]_{0}[/itex])[itex]^{2}[/itex] = [itex]\frac{1}{2}[/itex](k)(x)[itex]^{2}[/itex]

    [itex]\frac{9}{8}[/itex]Mv[itex]_{0}[/itex][itex]^{2}[/itex] = [itex]\frac{kx^{2}}{2}[/itex]

    x = √([itex]\frac{9Mv_{0}^{2}}{4k}[/itex])

    According to the answersheet I was given, the final answer should be x = √([itex]\frac{9Mv_{0}^{2}}{20k}[/itex]) instead.

    Would really appreciate it if anyone could help me out. Thanks!
     
  2. jcsd
  3. Nov 18, 2013 #2
    Your assumption that at max compression KE = 0 has no basis. In fact, it is not compatible with conservation of momentum.
     
  4. Nov 18, 2013 #3
    I see, would it be correct to write

    [itex]\frac{1}{2}[/itex](4M)([itex]\frac{3}{4}[/itex]v[itex]_{0}[/itex])[itex]^{2}[/itex] = [itex]\frac{1}{2}[/itex](k)(x)[itex]^{2}[/itex] + [itex]\frac{1}{2}[/itex](5M)(V1[itex]^{2}[/itex])

    If it is then what method should I use to find v1?
     
  5. Nov 18, 2013 #4
    The latter equation is meaningless, because 5M is the mass of the entire system, but V1, even if it was the velocity of the center of mass, is not the only contribution to kinetic energy.

    Perhaps the easiest approach here would to use the center of mass frame. Find its velocity. Find the kinetic energy of the center of mass. Then, the internal energy in the center of mass frame is the original energy minus the center of mass energy. In the center of mass frame, the system undergoes simple harmonic motion, whose energy is known.
     
  6. Nov 18, 2013 #5
    I think I got it. Thanks!
     
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