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

Step-by-step 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* 10^3 m/s^2`

Time taken by it to stop completely

`v=u+at`

`0=7-2.45* 10^3\cdot t`

`t=0.00286 s`

(b)Average
`Force\ F_(avg)`

`Change\ in\ momentum=Impulse=force* time`

`F_(avg)=m(\Delta v)/(\Delta T)`

`F_(avg)=0.45\cdot (7)/(0.00286)`

`F_(avg)=1103 N`