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Homework Help: A few questions - I want to make sure I did it right

  1. Apr 13, 2006 #1
    I recently had a physics test and here are the questions that were given. I answered 3 out of 4. Would an expert please be able to tell me if I got them right or if not, please lead me in the right direction? I would wait to get my test back but that won't be for another 2 weeks and I am curious. Thanks in advance. :)

    A 0.20 kilogram mass is sliding on a horizontal, frictionless air track with a speed of 3.0 meters per second when it instantaneously hits and sticks to a 1.3 kilogram mass initially at rest on the track. The 1.3 kilogram mass is connected to one end of a massless spring, which has a spring constant of 100 Newtons per meter. The other end of the spring is fixed.

    A) Determine the following for the 0.20 kilogram mass immediately BEFORE the impact.

    1. Its linear momentum:

    = (0.20)(3)
    =0.6 kg*m/s

    2. It's kinetic energy:

    =0.9 J

    B) Determine the following for the COMBINED MASSES immediately AFTER the impact.

    1. Its linear momentum:

    = (0.20+1.3)(3)
    =4.5 kg*m/s

    2. It's kinetic energy:

    =6.75 J

    After the collision, the two masses undergo simple harmonic motion about their position at impact.

    C) Determine the amplitude of the harmonic motion.

    I had no idea what to use for this?

    D) Determine the period of the harmonic motion:


    =1/12.829 s

    Any help on this is much appreciated!
  2. jcsd
  3. Apr 13, 2006 #2
    Answers to A1 and A2 are correct. You have to recheck B1 and B2. The question asks for the momentum of the combined mass immediately after collision (ie just before the spring force starts to have an effect). That is determined by the initial momenta of the two masses. Having said this find the velocity. C asks for the maximum displacement of the masses on either side of equillibrium position. Use conservation of energy.
  4. Apr 13, 2006 #3
    So for B1:


    =0.4 m/s

    =0.6 kg*m/s



    v(final)=1.095 m/s

    =.891 J

    Is this the answer for B1 and B2?
    Last edited: Apr 13, 2006
  5. Apr 13, 2006 #4
    You're approach to the first one is right, but check you calculations.

    As for B2, it's just 0.5M(v(final)^2); M = m1+m2.
  6. Apr 13, 2006 #5
    B2 i see had the wrong calculations;


    It should come out to .899 J. But I double checked the calcuations for B1 - it still comes out to 0.4 - did I set it up incorrectly?
  7. Apr 13, 2006 #6
    Comes to .12 J with v=0.4m/s.

    I believe you had typed 0.46 or something before I posted my previous reply. You had missed m2 in the denominator.
  8. Apr 13, 2006 #7
    Oh yes - I had corrected the original post - question though - how come when I do v(final) using conservation of momentum and when i do it using conservation of mechanical energy - i come up with 2 different answers? You can see what I mean above - one answer gave me 0.4 m/s for vfinal (momentum) and the energy way gave me 1.095 m/s.
  9. Apr 13, 2006 #8
    That's becuase you've assumed that kinetic energy is conserved. It is not conserved since this is a completely inelastic colission.
  10. Apr 13, 2006 #9

    Arghh :yuck: I totally forgot!

    I am still unsure of questions C and D. Thanks for the help so far!
  11. Apr 13, 2006 #10
    On the right track for D, but calculation needs checking, again. Check my first post for a hint for C.
  12. Apr 13, 2006 #11
    D) Determine the period of the harmonic motion:


    1/2pi = 1.571
    k/m1+m2 = 100/1.5 = 66.7
    sqrt (66.7)= 8.167

    =1/12.829 s

    Where am i going wrong?
  13. Apr 13, 2006 #12
    You've stated that 1/2pi > 1... Please go through the whole thing once again slowly. It's all calculation errors and I can show you where you've gone wrong, but that's not going to help you at all. Also remember the direct relation between time period and angular frequency.
  14. Apr 13, 2006 #13
    3.14/2 = 1.57 - am i using the wrong value for pi?
  15. Apr 13, 2006 #14
    frequency = angular frequency/(2pi) ... f = w/(2pi)

    Time, T = 1/f = (2pi)/w
  16. Apr 26, 2006 #15

    =0.05 m

    Is this correct? Thanks!
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