Answer to Question #235044 in Classical Mechanics for Ali

Question

a bullet 2 cm long is fired at 420 m/s and passes straight through a 10 cm thick wooden-board, penetrate at 280 m/s . (a) what is the average acceleration of the bullet through the board? (b) what is the total time that the bullet is in contact with the board? (c) what minimum thickness could the board have if it was supposed the bullet?

Answer 1

Given:

$v_1=420\:\rm m/s$

$v_2=280\:\rm m/s$

$d=10\:\rm cm=0.1\; m$

(a) the average acceleration of the bullet through the board

$$a=\frac{v_2^2-v_1^2}{2d}=\frac{420^2-280^2}{2*0.1}=-490000\:\rm m/s^2$$

(b) the total time that the bullet is in contact with the board

$$t=\frac{2d}{v_1+v_2}=\frac{2*0.1}{420+280}=2.86*10^{-4}\;\rm s$$