An indestructible bullet 2.00 cm long is fired straight through a board that is

Question

An indestructible bullet 2.00 cm long is fired straight through a board that is 10.0 cm thick. The bullet strikes the board with a speed of 480 m/s and emerges with a speed of 285 m/s. (To simplify, assume that the bullet accelerates only while the front tip is in contact with the wood.)

a) What is the average acceleration of the bullet through the board?

b) what is the total time the bulllet is in contact with the board? (enter the total time for the bullet to completely emerge form the board.

c) What thickness of board (calculated to .1 cm) would it take to stop the bullet, assuming the acceleration through all boards is the same?

Explanation / Answer

Assuming constant acceleration we can use the kinematic eqn v^2 = v0^2 + 2*a*delta x to solve for a

a = (v^2 -v0^2)/(2*delta x) = (285^2 - 480^2)/(2*0.10) = -7.46x10^5m/s^2

b) For the time use v = v0 + a*t or t = (v - v0)/a = (285-480)/(-7.46x10^5) = 2.61x10^-4 s = 0.261ms

c) To stop the bullet use v^2 = v0^2 +2*a*x and solve for x (Note here v =0)

So x = -v0^2/(2*a) = -(480^2)/(2*(-7.46x10^5)) = 0.1544 m = 15.44cm

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