Answer to Question #271795 in Classical Mechanics for Tasnim

Question

In a car collision, the bag expands suddenly and cushions

the impact of the passenger. Assume the impact velocity is

105 km/h. For a 70-kg person with a 30-cm allowed

stopping distance, (a) calculate the stopping force per cm²

applied to the person, if the average applied force is

uniformly distributed over a 1000-cm² area of the

passenger's body. (b) calculate the duration of the collision

between the passenger and the inflated bag of the collision

protection device. (c) If the rupture strength for body tissue

is about 5 X 10⁶dyn/cm², will the impact force lead to

serious injury for the passenger?

Answer 1

$p_{max}= 5*10^6\frac{dyn}{cm^2}= 50\frac{N}{cm^2}$

$v= 105\frac{km}{h}=29.17\frac{m}{s}$

$l = 30cm = 0.3m$

$m=70kg$

$S = 1000cm^2$

$a)$

$Fl = EK$

$EK = \frac{mv^2}{2}=\frac{70*29.17^2}{2}=29781.11N$

$F =\frac{EK}{l}=\frac{29781.11}{0.3}=99270.37N$

$p = \frac{F}{S}=99.27\frac{N}{cm^2}$

$b)\newline J = F\Delta t$

$J =mv = 70*29.17=2041.9kg*\frac{m}{s}$

$\Delta t= \frac{J}{F}=\frac{2041.9}{99270.37}\approx0.021s$

$c)$

$p =99.27\frac{N}{cm^2}$

$p>p_{max}$

$\text{high likelihood of injury}$

$\text{Answer:}$

$a)99.27\frac{N}{cm^2}$

$b)0.021s$

$c) \text{high likelihood of injury}$