1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Simple Harmonic Motion and Energy Problem

  1. Apr 30, 2016 #1
    1. The problem statement, all variables and given/known data
    In the figure, block 2 of mass 2.40 kg oscillates on the end of a spring in SHM with a period of 26.00 ms. The position of the block is given by x = (1.80 cm) cos(ωt + π/2). Block 1 of mass 4.80 kg slides toward block 2 with a velocity of magnitude 6.90 m/s, directed along the spring's length. The two blocks undergo a completely inelastic collision at time t = 6.50 ms. (The duration of the collision is much less than the period of motion.) What is the amplitude of the SHM after the collision?

    2. Relevant equations
    E=KE+PE=0.5k*a^2 where k is spring constant and a is amplitude
    KE=0.5m*v^2
    T=2π*sqrt(m/k)
    3. The attempt at a solution
    First rearrange the expression for T:
    k=(4*m*π^2)/T^2
    We also have initial mechanical energy:
    Ei=0.5k*a^2, where a can be read off from the given equation x = (1.80 cm) cos(ωt + π/2), that is 0.018m, and we have managed an expression for k.
    Thus we get Ei=45.42286025 J
    After the collision, block 1 transfers all its energy to block 2, thus the total mechanical energy increases by the the KE of block 1:
    KE1=0.5*M*v^2=0.5*4.8*6.9^2=114.264 J
    Now the new mechanical energy is
    Ef=Ei+KE1=159.6758603 J
    Finally we use the equation E=0.5k*a^2 to solve for the new amplitude and get
    a=0.0477 m
    This is not the correct answer and I did not use all the information given(t=6.5ms and ø=π/2).
    Please point out what I have missed thanks.
     

    Attached Files:

  2. jcsd
  3. Apr 30, 2016 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hi,

    With a a nickname like that you want to try and outsmart the composer of the exercise. Think what an inelastic collision entails and find a minimum path through the glob of info and relationships. One of your relationships contains all you need for the anser - and involves only a few variables.

    [edit] Sorry, misread the aim of the exercise -- when it's late it seems that it is possible to read 'frequency' instead of 'amplitude' :oldconfused: -- . Fortunately you haven't been distracted and still are on the right track in #11 thanks to haru and ehild
     
    Last edited: May 1, 2016
  4. Apr 30, 2016 #3
    Hi
    One day I shall prove to the world that I am so much smarter than the the composer of this problem, its just the composer might not live to witness this brilliant moment.

    Complete Inelastic collision means all the KE of block 1 is transferred to the new two-blocks system is that right? If so I think my equations follow this
     
  5. Apr 30, 2016 #4

    ehild

    User Avatar
    Homework Helper
    Gold Member

    No. Part of the KE of the whole system transforms to non-mechanical energy, like heat. What does complete inelastic collision mean?
     
  6. Apr 30, 2016 #5
    It means the greatest loss in KE, m1*Vi=(m1+m2)*Vf
     
  7. Apr 30, 2016 #6

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Yes, except that in this question both have an initial velocity.
     
  8. Apr 30, 2016 #7
    Does it mean that the actual velocity after collision would be Vf+(velocity of block 2)?
     
  9. Apr 30, 2016 #8

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    No, it means that you add up the initial momenta and equate that sum to the final momentum.
     
  10. May 1, 2016 #9

    ehild

    User Avatar
    Homework Helper
    Gold Member

    What are Vi and Vf? In general, both blocks have their velocities before and after the collision. How are the velocities related after the completely inelastic collision? What was the velocity of block 2 just before the collision?
     
  11. May 1, 2016 #10
    Right, so M1*V1+M2*V2=(M1+M2)*Vf
     
  12. May 1, 2016 #11
    M1*V1+M2*V2=(M1+M2)*Vf where V1 is given and V2 can be calculated from the equation of X(t). Now I am able to get Vf, and Vf=a*w right?
     
  13. May 1, 2016 #12

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Right
     
  14. May 1, 2016 #13

    ehild

    User Avatar
    Homework Helper
    Gold Member

    What is a? Are the two block stick together after the collision?
    You can calculate Vf, and then you can determine the whole energy of the system. For that, you need to know the elastic energy of the spring at the moment of collision.
     
  15. May 1, 2016 #14
    a is the amplitude.

    K=0.5 • (M1+M2) • Vf
    U=0.5k • [0.018cos(ωt + π/2)]^2
    K+U=E=0.5 • k • a^2
     
  16. May 1, 2016 #15

    ehild

    User Avatar
    Homework Helper
    Gold Member

    If the blocks move together the frequency of the SHM will change.
    Determine both V1 and V2 before the collision, and the common Vf after it. What is the energy of the spring at the moment of collision?
    What is the new ω when the blocks move together?
     
  17. May 1, 2016 #16
    The new w is
    w=sqrt(k/M1+M2)
    is that right?
     
  18. May 1, 2016 #17

    ehild

    User Avatar
    Homework Helper
    Gold Member

    if you add a pair of parentheses ...
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Simple Harmonic Motion and Energy Problem
  1. Oscillation problem (Replies: 4)

Loading...