Simple harmonic motion - answer slightly off

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SUMMARY

The discussion focuses on a physics problem involving simple harmonic motion (SHM) where a 4.00 kg block is suspended from a spring with a spring constant of 500 N/m. A 50.0 g bullet, traveling at 125 m/s, embeds itself in the block. The calculation of the percentage of kinetic energy transferred to mechanical energy reveals a discrepancy between the calculated value of 1.23% and the book's value of 1.17%. The consensus is that the book's answer is incorrect due to premature rounding in calculations.

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  • Understanding of simple harmonic motion (SHM)
  • Knowledge of conservation of momentum principles
  • Familiarity with kinetic energy equations
  • Ability to perform calculations with significant figures
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  • Study the principles of conservation of momentum in inelastic collisions
  • Learn about the effects of rounding in physics calculations
  • Explore the derivation of kinetic energy formulas in different contexts
  • Investigate the relationship between amplitude and energy in SHM
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Students studying physics, particularly those focusing on mechanics and simple harmonic motion, as well as educators looking for clarification on energy transfer in inelastic collisions.

Lord Anoobis
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Homework Statement


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?

Homework Equations


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

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?
 
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Book answer is simply wrong -- due to too early rounding, as you indicate. Well spotted !
 
BvU said:
Book answer is simply wrong -- due to too early rounding, as you indicate. Well spotted !
Thanks a lot. Always good to know where the error lies.
 

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