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 500 m/s and emerges with a speed of 290 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?
m/s2

(b) What is the total time that the bullet is in contact with the board? (Enter the total time for the bullet to completely emerge from the board.)
s

(c) What thickness of board (calculated to 0.1 cm) would it take to stop the bullet, assuming that the acceleration through all boards is the same?
cm

Explanation / Answer

Thickness of the board  S = 10 cm                                       = 0.1 m Initial velocity of the bullet just before hit the board u = 400 m / s Velocity of the bullet just after emerging from the board v = 295 m / s From the relation v 2 - u 2 = 2aS Accleration of the bullet a = (v 2 - u 2 ) / 2S                                         = -364875 m / s 2 (b). let the total time that the bullet is in contact with the board be t Then from the relation v = u + at                                   t = ( v - u ) / a                                     = 2.877 x 10 -4 s (c). Final velocity V = 0 From the relation V 2 - u 2 = 2aS '                                    S ' = [V 2 - u 2] / 2a                                         = 0.219                                         = 21.9 cm The minimum thickness of the board to stop the bullet = 21.9 cm (c). Final velocity V = 0 From the relation V 2 - u 2 = 2aS '                                    S ' = [V 2 - u 2] / 2a                                         = 0.219                                         = 21.9 cm The minimum thickness of the board to stop the bullet = 21.9 cm

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