Question

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?

Answer 1

Answer:

The duration of the impact is 0.005384 seconds

Explanation:

Given

m = 0.43 kg

v = 5.2 m/s

x = 0.014 m

Knowing the formulas

$v^{2}_{f} = v^{2}_{i} + 2ax\\0 = 5.2^{2} + 2a*0.014\\a = - 965.71 m/s^{2} \\\\vf = vi + at\\0 = 5.2 + (965.71)t\\t = 0.005384 s$

Answer 2

Answer:

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