Question

A 30.0 kg hammer, moving with speed speed 20.0 $m{s}^{-1}$, strikes a steel Spike 2.30 cm in diameter. The hammer rebounds with speed 10.0 $m{s}^{-1}$ after 0.110 s .What is the average Strain in the Spike during the impact?

Answer 1

The force acting on the hammer changes its momentum according to $m{v}_i+\underset{F}{\to }\left(\Delta t\right)=m{v}_f,so|\underset{F}{\to }|=\frac{m|v_f-v_i|}{\Delta t}$
Hence,$|\underset{F}{\to }|=\frac{30.01-10.0-20.01}{0.110s}=8.18\times {10}^{3}N$
By Newtons third law, that is also the magnitude of the average force exerted on the spike by the hammer during the blow. Thus, the stress in the spike is
$Stress=\frac{F}{A}=\frac{8.18\times {10}^{3}N}{\pi (0.0230m{)}^2/4}=1.97\times {10}^{7}N/m^2$
and the strain is
$Strain=\frac{Stress}{Y}=\frac{1.97\times {10}^{7}N/m^2}{20.0\times {10}^{1}0N/m^2}=9.85\times {10}^{-5}$