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Simple harmonic motion - answer slightly off

  1. Oct 13, 2015 #1
    1. The problem statement, all variables and given/known data
    A 4.00kg block is suspended from a spring with k = 500N/m. A 50.0g bullet is fired into the block from directly below with a speed of 125m/s and becomes embedded in the block.

    a) Find the amplitude of the resulting SHM.
    b) What percentage of the original kinetic energy of the bullet is transferred to mechanical energy of the oscillator?

    2. Relevant equations
    Conservation of momentum, ##p_f = p_i##
    ##K = \frac{1}{2}mv^2##

    3. The attempt at a solution
    The problem is solved but there is something about the given answer or part (b) that is bugging me, so let's ignore (a). For (b) I arrived at

    ##percentage = 100(\frac{K_m}{K_b})##

    ##percentage = 100(\frac{m(v_b)^2}{(m + M)(v_m)^2})##

    where ##K_b## is the initial kinetic energy of the bullet and ##K_m## is the kinetic energy of system after the bullet becomes embedded in the block. The velocity of the block in this state is given by

    ##v_m = \frac{m v_b}{m + M}##

    The trouble is that the preferred method of working with symbols until the very end gives an answer of 1.23% whereas the book answer is 1.17%. However, I noticed that if you use ##v_m = 1.5m/s## the correct answer pops up.

    I've noticed this sort of thing on a number of occasions where using a rounded figure in the final answer turns out to be more correct, so to speak, which goes against the usual method. Did I blow it somewhere or is it really a case of early rounding?
     
  2. jcsd
  3. Oct 13, 2015 #2

    BvU

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    Book answer is simply wrong -- due to too early rounding, as you indicate. Well spotted !
     
  4. Oct 13, 2015 #3
    Thanks a lot. Always good to know where the error lies.
     
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