Monday, October 28, 2013

Cell phones as bullet proof vests?

In Florida this morning around 4:30 AM, a robber shot at a gas station clerk who was unable to open the safe.  Miraculously, the clerk walked away from the shooting with only mild chest pain.  In fact, the clerk didn't even realize he was hit until he took out his cell phone and found the bullet lodged in it.


I thought it was pretty amazing that the cell phone was able to stop the bullet.  I was curious about how much force the cell phone exerted on the bullet and the deceleration experienced by the bullet.  

In order to determine this, I made several assumptions.  I found the average mass of a bullet to be 200grains or .01kg.  I also determined that the average muzzle velocity is 850ft/s or 259m/s.  The article stated that the robber shot at the clerk on his way out of the store, so I assumed that the bullet traveled 5m before hitting the clerk.  A bullet traveling at 259m/s would travel 5m in .02s.  

So, we have the following information:
m = .01kg
v0 = 259m/s
vF = 0m/s
t = .02s

We now have enough information to calculate the net force exerted on the bullet by the cell phone.  

Force = change in momentum / time

change in momentum of bullet = mFvF - m0v0
                                                 =(.01kg)(0m/s) - (.01kg)(259m/s)
                                                 =-2.59kg m/s
Fcell phone = (-2.59kg m/s) / .02s
             = -129.5N 

Now that have calculated the net force, we can figure out the deceleration experienced by the bullet using:

F = ma
-129.5N = (.01kg)a
           a = -12950m/s2

That's incredible!  


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