A 25 g bullet with a muzzle velocity of 300 m/s is fired into a board 4.0 cm thick. For a soft board, the bullet goes through the board and emerges with a speed of 50 m/s. What was the average force exerted on the bullet by the board?

Question

Answer= 27344 N
Show Work

Answer 1

V^2 = Vo^2 + 2a*d,
a = (V^2-Vo^2)/2d = (50^2-300^2)/0.08 = -1,093,750 m/s^2.

F = M*a = 0.025 * (-1,093,750) =