A0.650 kg hammer is moving horizontally at 4.00 m/s when it strikes a nail and comes to rest after driving it 1.00 cm into a board. (a) calculate the duration of the impact.
(b) what was the average force exerted on the nail?

a) 0.005s

b) F = -520 N , this is taking the direction into account.

Explanation:

let d be the distance covered by the hummer and nail, a be the acceleration of the hummer and nail

a)   m = 0.650 kg

     Vi = 4.00 m/s

      d = 0.01 m

(Vf)^2 = (Vi)^2 + 2(a)(d)

      0 = (4)^2 + 2(a)(0.01)

      a = -800 m/s^2

then:

Vf = Vi + at

0 = 4 + (-800)t

 t = 0.005s

Therefore, the impact lasted for 0.005s.

b)  the average force is given by:

F = ma = (0.650)(-800) =  - 520 N

Therefore, the magnitude of the average force is 520 N.

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