Someone shoots a gun while holding their arm stiffly outstretched in front of th

Question

Someone shoots a gun while holding their arm stiffly outstretched in front of them and with the back of their shoulder braced against a wall. The bullet leaving the gun has a mass of 12 g and is traveling at 490 m/s (.44 magnum). It reaches this velocity (from rest) in 0.0001 s. Assuming that all of the recoil force is directed along the long axis of the bones of the forearm, what is the stress in the arm bone? Think of the arm as being one long bone, and the cross-sectional area of this bone is 3 cm2.

Explanation / Answer

Mass of the bullet m = 12 g = 0.012 kg

Initial velocity u = 0

Final velocity v = 490 m/s

Change in momentum = m( v - u )

= mv = 0.012 x490 = 5.88 kgm/s

Time t = 0.0001 s

Force F = change in momentum / time

= 5.88 / 0.0001

= 58800 N

= 58800 x10 5 dyne

Cross-sectional area A = 3 cm 2

The stress in the arm bone = F / A

= (58800 x10 5 dyne ) /(3 cm 2)

= 1.96 x10 9 dyne/cm 2

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