What is the average strain in a steel spike when struck by a 30kg sledge hammer?

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The average strain in a steel spike, when struck by a 30kg sledge hammer moving at 20.9 cm/s and rebounding at 10.0 cm/s over a duration of 0.110 seconds, can be calculated using the formula for force derived from change in momentum. The relevant equations include F = (mvf - mvi)/t and S = (F/A)/(deltaL/initialL), with Young's modulus for steel being 2 x 10^11 Pa. The correct approach involves using the change in momentum rather than initial and final velocities directly.

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A 30kg sledge hammer strikes a steel spike of 2.30 cm diameter while moving at 20.9 cm/s. The hammer rebounds with a speed of 10.0 cm/s after 0.110 s. What is the average strain in the spike during impact?



2. F = (mvf-mvi)/t
S = (F/A)/(deltaL/initialL)
Young's modulus for steel = 2 x 10^11




3. I used (vf-vi)/t to find the force but apparently I'm supposed to use change in momentum/time. I don't understand why.
 
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nevermind, i got it.
 

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