# A projectile with a speed of 400 m / s penetrates 6 cm of armor and flies out of it at a speed of 200 m / s.

A projectile with a speed of 400 m / s penetrates 6 cm of armor and flies out of it at a speed of 200 m / s. Determine the smallest thickness of armor not penetrated by such a projectile.

Given: V1 (projectile velocity before armor penetration) = 400 m / s; s (armor thickness, distance traveled) = 6 cm (0.06 m); V2 (speed after departure from the armor) = 200 m / s.

1) Determine the friction force acting on the projectile: ΔEk = A; m * V1 ^ 2/2 – m * V2 ^ 2/2 = Ftr. * s, whence Ftr. = 0.5 * m * (V1 ^ 2 – V2 ^ 2) / s = 0.5 * m * (400 ^ 2 – 200 ^ 2) / 0.06 = 106 * m.

2) Calculate the smallest armor thickness: ΔEk = A; m * V1 ^ 2/2 – 0 = Ftr. * s1, whence s1 = m * V1 ^ 2 / (2 * Ftr.) = m * 400 ^ 2 / (2 * 10 ^ 6 * m) = 0.08 m (8 cm). One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.