Answer to Question #235548 in Mechanics | Relativity 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

$S_1= 2 \;cm \\ S_2 = 10 \;cm$

Total distance traveled $= S_1+S_2 = 2+10 = 12 \;cm = 0.12 \;m$

$u=420 \;m/s \\ v=280 \;m/s$

(a)

$v^2 =u^2 + 2as \\ 280^2 = 420^2 + 2a \times 0.12 \\ a= \frac{280^2 -420^2}{2 \times 0.12} \\ a = -40833.33 \;m/s^2$

(b)

$v=u+at \\ 280 = 420 + (-408333.33) \times t \\ t = \frac{280-420}{-408333.33} \\ t = 3.428 \times 10^{-4} \;s \\ t=0.3428 \;milliseconds$

(c) If v=0

$v^2 =u^2 + 2as \\ 0 = 420^2 + 2 \times (-408333.33) \times (0.02 + S_2) \\ (0.02 +S_2) = \frac{420^2}{2 \times 408333.33} \\ 0.02 + S_2 = 0.216 \\ S_2 = 0.196 \;m \\ S-2 = 19.6 \;cm$