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
[tex]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[/tex]
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
