Question

A bullet 2cm log is fired at 420m/s and passes straight a 10cm thick board exiting at 280m/s

a) what is the average acceleration of the bullet through the board?
b)what is the total time the bullet is in contact with the board?
c)what minimum thickness could the board have if it was supposed to bring the bullet to a stop?

Answer 1

Solving for the acceleration of the bullet

acceleration = (vf^2 – vi^2) / 2d

acceleration = ((280 m/s)^2 – (420 m/s)^2) / (2 * 0.12 m)

acceleration = (78400 - 176400) / 0.24 m

acceleration = -98000 / 0.24

acceleration = -408333 m/s^2

Solving for contact time with board

t^2 = 2d/a

t^2 = 2 * 0.12 m / 408333 m/s^2

t^2 = 0.24 m / 408333 m/s^2

t^2 = 5.8775558 x 10^-7

t = 0.0007666 s or 767 microseconds


(I was only able to do A and B)

Answer 2

Step-by-step explanation:

(a)Solving for the acceleration of the bullet

acceleration = (vf^2 – vi^2) / 2d

acceleration = ((280 m/s)^2 – (420 m/s)^2) / (2 * 0.12 m)

acceleration = (78400 - 176400) / 0.24 m

acceleration = -98000 / 0.24

acceleration = -408333 m/s^2

(a)Solving for contact time with board

t^2 = 2d/a

t^2 = 2 * 0.12 m / 408333 m/s^2

t^2 = 0.24 m / 408333 m/s^2

t^2 = 5.8775558 x 10^-7

t = 0.0007666 s or 767 microseconds