Potential energy momentum

Question

A 0.950 kg hammer is moving horizontally at 4.50 m/s when it strikes a nail and comes to rest after driving it 1.00 cm into a board.

Calculate the duration of the impact in seconds. (Enter a number.)

What was the average force in newtons exerted downward on the nail? (Enter a number.)

Answer 1

From kinematics you know (v_fin² - v_in²)=2ad →a=(v_fin² - v_in²)/2d=(0-4.5²)/0.02 m=-1012.5 m/s²

 Also, v(t)= v(0) +at in this case v_fin=0 , v_in=4.5 m/s a,d a=-1012.5 m/s² →∆t=(4.5 m/s)/(1012.5 m/s²)=4.4 ms

Impulse is change of the object momentum F_avg∆t =(∆mv), and then

 F_avg=(∆mv)/∆t=0.95 kg)(0-4.5 m/s)/0.0044 s=-971.6 N

Negative means the direction of the force exerted by the nail on the hummer, as the reason of stopping.

Therefore, the force exerted by the hummer is 971.6 N, according to Newton's 3rd law.