A 0.43 kg hammer is moving horizontally at 5.2 m/s when it strikes a nail and comes to rest after driving the nail 0.014 m into a board. What is the duration (in seconds) of the impact?

5.385×10⁻³

Explanation:

First we find the acceleration

Using the equation of motion,

v² = u²+2as Equation 1

Where v = Final velocity, u = initial velocity, a = acceleration, s = distance.

make a the subject of the equation,

a = (v²-u²)/2s Equation 2

Given: v = 0 m/s (comes to rest), u = 5.2 m/s, s = 0.014 m

Substitute into equation 2

a = (0²-5.2²)/(2×0.014)

a = -27.04/0.028

a = -965.71 m/s²

Finally Using

a = (v-u)/t

where t = Duration of impact

make t the subject of the equation

t = (v-u)/a Equation 3

Given: v = 0 m/s, u = 5.2 m/s, a = -965.71 m/s²

Substitute into equation 3

t = (0-5.2)/-965.71

t = -5.2/-965.71

t = 5.385×10⁻³ s.

Hence the duration of impact =  5.385×10⁻³

compass can be used day and night, cloudy or sunny