h = 20 m m = 200 g = 0.2 k g Δ t = 0.001 s S = 1 c m 2 p m a x = 5 ∗ 1 0 6 d y n c m 2 = 50 N c m 2 h = 20m\newline
m =200g =0.2kg\newline
\Delta t = 0.001s\newline
S =1cm^2\newline
p_{max}= 5*10^6\frac{dyn}{cm^2}=50\frac{N}{cm^2} h = 20 m m = 200 g = 0.2 k g Δ t = 0.001 s S = 1 c m 2 p ma x = 5 ∗ 1 0 6 c m 2 d y n = 50 c m 2 N
g = 9.8 m s 2 g = 9.8\frac{m}{s^2} g = 9.8 s 2 m
Apple speed at the end of the fall: \text{Apple speed at the end of the fall:} Apple speed at the end of the fall:
v = 2 g h = 2 ∗ 9.8 ∗ 20 ≈ 19.8 m s v =\sqrt{2gh}=\sqrt{2*9.8*20}\approx19.8\frac{m}{s} v = 2 g h = 2 ∗ 9.8 ∗ 20 ≈ 19.8 s m
Apple’s impulse at the end of the fall: \text{Apple's impulse at the end of the fall:} Apple’s impulse at the end of the fall:
J = m v = 0.2 ∗ 19.8 = 3.96 k g m s J = mv =0.2*19.8=3.96\ kg\frac{m}{s} J = m v = 0.2 ∗ 19.8 = 3.96 k g s m
The force of the apple on the body: \text{The force of the apple on the body:} The force of the apple on the body:
J = F Δ t J= F\Delta t J = F Δ t
F = J Δ t = 3.96 0.001 = 3960 N F = \frac{J}{\Delta t}= \frac{3.96}{0.001}=3960N F = Δ t J = 0.001 3.96 = 3960 N
The pressure of the apple on the body during the impact: \text{The pressure of the apple on the body during the impact:} The pressure of the apple on the body during the impact:
p = F S = 3960 1 = 3960 N c m 2 p = \frac{F}{S}=\frac{3960}{1}=3960\frac{N}{cm^2} p = S F = 1 3960 = 3960 c m 2 N
p > p m a x hence p>p_{max}\text{ hence } p > p ma x hence
impact force will cause human trauma \text{impact force will cause
human trauma} impact force will cause human trauma
Answer: \text{Answer:} Answer:
a ) F = 3960 N a)F =3960N a ) F = 3960 N
b ) impact force will cause human trauma b)\text{impact force will cause human trauma} b ) impact force will cause human trauma
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