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Wave Motion Equations

  1. Jul 10, 2017 #1
    1. The problem statement, all variables and given/known data
    A mass of 120 g rolls down a frictionless hill, reaching a speed of 4.2 m/s. This mass collides with another mass of 300 g that is at rest and attached to a spring with constant 30 N/m. The two masses stick together and enter into periodic motion. What is the equation for the motion?
    A) x(t)=0.266 sin(8.45t)
    B) x(t)=0.420 cos(10.0t)
    C) x(t)=0.0706 sin(8.45t)
    D) x(t)=0.497 cos(15.8t)

    2. Relevant equations
    x(t) = A cos (wt+ phi) equation for wave motion

    3. The attempt at a solution
    I am able to narrow it down to A or C, by using the equation w = (k/m)^1/2, which gives me 8.45. I am unsure how to find A. I have a feeling you need to implement the velocity, but I have tried that, and have also tried using momentum equation to find velocity after impact.
     
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  3. Jul 10, 2017 #2

    Orodruin

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    Please show your work.
     
  4. Jul 10, 2017 #3

    scottdave

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    Since they stick together, it is an inelastic collision, so you cannot look at kinetic energy, to find velocity of the stuck-together masses. You should be able to use momentum. The momentum of the two masses just before collision is the same as momentum of the 2 masses immediately after collision. Once you have initial velocity, you can find the kinetic energy of this combined mass, then use that with the spring energy formula, to find how far it will deflect the spring. That distance will be your Amplitude.
     
  5. Jul 10, 2017 #4
    0 + 0.12kg x 4.2 m/s = 0.420 kg x velocity. solve for velocity to get 1.2m/s initially. Now, this is where I get lost. KE = 1/2mv^2 so filling that in I get
    KE = (1/2)(0.42kg)(1.2m/s)^2 = 0.302J. Which spring energy formula are you referring to? The only one that comes to mind is F=kx and that is not correct if you use KE as F and solve for x.
     
  6. Jul 10, 2017 #5

    scottdave

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    Energy in a compressed (or stretched) spring = (1/2)*k*x2, where x is the distance from the rest position.
     
  7. Jul 10, 2017 #6
    Using your equation, 0.302J = (1/2)*30*x2 , x = 0.142 m from rest position. I know that the amplitude is actually 0.266m. What am I missing?
     
  8. Jul 10, 2017 #7

    scottdave

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    Unfortunately, I now think that the momentum is not the way to go about it (I cannot get any of your choices, using that). Perhaps they consider that kinetic energy is completely transferred. Because you can get one of the choices, using that.
     
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