A `30.0 kg` hammer, moving with speed `20.0ms^(-1)` strikes a steel spike `2.30 cm` in diameter. The hammer rebounds with speed `10.0ms^(-1)` after `0.110s.` What is the average strain in the spike during the impact.?
Text Solution
Verified by Experts
The force acting on the hammer changes its momentum according to `mv_(i)+vecF(/_\t)=mv_(f)`, so `|vecF|=(m|v_(f)-v_(i)|)/(/_\t)` Hence `|vecF|=(30.0|-10.0-20.0|)/(0.11s)=8.18xx10^(3)N` By Newton's third law, that is alos the magnitude of the average force exerted on the spike by the hammer during the blow. Thus, the stress in the spike is Stress`=F/A=(8.18xx10^(3)N)/(pi(0.0230m)^(2)//4)=1.97xx10^(7)N//m^(2)` and the strain in Strain`= "Stress"/Y=(1.97xx10^(7)N//m^(2))/(20.0xx10^(10)N//m^(2))=9.85xx10^(-5)`
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CENGAGE PHYSICS-PROPERTIES OF SOLIDS AND FLUIDS-INTEGER_TYPE
