- #1

rohanprabhu

- 414

- 2

I know it's old already.. but this came up when I was lecturing my mom on the evil effects of urban myths and stuff.. so here it goes..

I'm throwing a 5 rupee coin [Weight: 0.009 kg], from the Empire state building's roof [Height: 381m]. The coin hits a passerby on his head, as in.. vertically i.e. the coin's circumference hits him:

http://img248.imageshack.us/img248/3748/howithitsmo3.jpg

So, the GPE of the coin is: [itex]\textrm{U} = \textrm{mgh} = (0.009 \times 9.8 \times 381) ~J = 33.7~J[/itex]. Assuming that the coin hits the head and stops is 0.1s (which I assume is more than what would practically happen) it means that when the coin hits the person, it hits with a velocity:

[tex]

v = \sqrt{\frac{2\textrm{U}}{\textrm{m}}}

[/tex]

[tex]

v = 86.54~\textrm{m/s}

[/tex]

meaning, it has a momentum of:

[tex]

p = 0.778 ~\textrm{kg} \cdot \textrm{m/s}

[/tex]

which was imparted in 0.1s, meaning a total force of:

[tex]

F = \frac{0.778}{0.1} = 7.78~\textrm{kg} \cdot \textrm{m} / \textrm{s}^2

[/tex]

I stained the coins circumference with black ink, and placed it on paper, to get the area that the coin would hit. I got a rough rectangle of 0.2cm x 0.5 cm, meaning a total area of [itex]10^{-5}~\textrm{m}^2[/itex].

So, the pressure on the skull at that point would be:

[tex]

P = \frac{7.78}{10^{-5}}~\textrm{Pa} = 7.78 \times 10^{5}~\textrm{Pa}

[/tex]

which is like, around 8 times the atmospheric pressure.

So, the question is.. could this kill a person? And, if the time within which the coin stops is like, 10 times less, the pressure is 10 times more. Since, I'm not sure about the interval.. if the coin hits in like 0.01s, the pressure would be 80 times the atmospheric pressure. Could that kill a person?

I'm throwing a 5 rupee coin [Weight: 0.009 kg], from the Empire state building's roof [Height: 381m]. The coin hits a passerby on his head, as in.. vertically i.e. the coin's circumference hits him:

http://img248.imageshack.us/img248/3748/howithitsmo3.jpg

So, the GPE of the coin is: [itex]\textrm{U} = \textrm{mgh} = (0.009 \times 9.8 \times 381) ~J = 33.7~J[/itex]. Assuming that the coin hits the head and stops is 0.1s (which I assume is more than what would practically happen) it means that when the coin hits the person, it hits with a velocity:

[tex]

v = \sqrt{\frac{2\textrm{U}}{\textrm{m}}}

[/tex]

[tex]

v = 86.54~\textrm{m/s}

[/tex]

meaning, it has a momentum of:

[tex]

p = 0.778 ~\textrm{kg} \cdot \textrm{m/s}

[/tex]

which was imparted in 0.1s, meaning a total force of:

[tex]

F = \frac{0.778}{0.1} = 7.78~\textrm{kg} \cdot \textrm{m} / \textrm{s}^2

[/tex]

I stained the coins circumference with black ink, and placed it on paper, to get the area that the coin would hit. I got a rough rectangle of 0.2cm x 0.5 cm, meaning a total area of [itex]10^{-5}~\textrm{m}^2[/itex].

So, the pressure on the skull at that point would be:

[tex]

P = \frac{7.78}{10^{-5}}~\textrm{Pa} = 7.78 \times 10^{5}~\textrm{Pa}

[/tex]

which is like, around 8 times the atmospheric pressure.

So, the question is.. could this kill a person? And, if the time within which the coin stops is like, 10 times less, the pressure is 10 times more. Since, I'm not sure about the interval.. if the coin hits in like 0.01s, the pressure would be 80 times the atmospheric pressure. Could that kill a person?

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