Question

A 0.450-kg hammer is moving horizontally at 7.00 m/s when it strikes a nail and comes to rest after driving the nail 1.00 cm into a board. Assume constant acceleration of the hammer-nail pair. Calculate the duration of the impact. What was the average force exerted on the nail?

Answer 1

Answer:a)0.00286 s

b)1103 N

Explanation:

Given

mass of hammer $m=0.45 kg$

Initial speed $v=7 m/s$

hammer drives nail $x=1 cm$ into a board

Using $v^2-u^2=2 as$

where u=initial velocity

v=final velocity

a=acceleration

s=distance moved

$0-7^2=2\cdot a\cdot x$

$-49=2\cdot a\cdot 0.01$

$a=-2.45\times 10^3 m/s^2$

Time taken by it to stop completely

$v=u+at$

$0=7-2.45\times 10^3\cdot t$

$t=0.00286 s$

(b)Average $Force\ F_{avg}$

$Change\ in\ momentum=Impulse=force\times time$

$F_{avg}=m\frac{\Delta v}{\Delta T}$

$F_{avg}=0.45\cdot \frac{7}{0.00286}$

$F_{avg}=1103 N$