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A 0.22LR caliber bullet has a mass of 1.90 g and a muzzle velocity of 500 m/s. The bullet is fired into a door made of a single thickness of pine boards, with a thickness of 0.75 in. The average stopping force exerted by the wood is 960 N. How fast (in m/s) would the bullet be traveling after it penetrated through the door

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Numenius

Answer:

The final speed of the bullet is 480.4 m/s.

Explanation:

mass of bullet, m = 1.9 g

initial speed, u = 500 m/s

thickness, d = 0.75 inch = 0.01905 m

Force, F = 960 N

Let the final speed is v.

According to the work energy theorem,

Work = change in kinetic energy

[tex]W = F d = 0.5 m{\left (v^2 - u^2 \right )}[/tex]

-960 x 0.01905 = 0.5 x 0.0019 x (v^2 - 500 x 500)

-18.288 = 0.00095 (v^2 - 250000)

v = 480.4 m/s

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