A bullet weighing 9 g, moving at a speed of 800 m / s, pierced a board 2.5 cm thick and, when exiting the board

A bullet weighing 9 g, moving at a speed of 800 m / s, pierced a board 2.5 cm thick and, when exiting the board, had a speed of 200 m / s. Determine the average drag force for the bullet in the board.

Given:

m = 9 grams = 0.009 kilograms – bullet mass;

v = 800 m / s – bullet speed in front of the board;

v1 = 200 m / s – bullet velocity after the board;

l = 2.5 cm = 0.025 meters – the thickness of the board.

It is required to find the resistance force F (Newton).

Let’s find the time it took the bullet to pass through the board:

t = (v – v1) / a.

l = v * t – a * t ^ 2/2 = v * (v – v1) / a – a * (v – v1) ^ 2/2 * a ^ 2 =

= v * (v – v1) / a – (v – v1) ^ 2/2 * a = (v – v1) * (v + v1) / 2 * a, hence:

a = (v – v1) * (v + v1) / 2 * l = (800 – 200) * (800 + 200) / 2 * 0.025 = (600 * 1000) / 0.05 =

600,000 / 0.05 = 12,000,000 m / s ^ 2.

F = m * a = 0.009 * 12,000,000 = 108,000 Newton = 108 kN.

Answer: the resistance force is 108 kN.