At what speed did a bullet weighing 7 g fly if, under the action of a resistance force of 2 kN, it made a hole 15 cm deep?

To calculate the speed of the bullet before making the hole, we will use the equality: ΔEk = A (bullet work) = Fc * S; m * V2 / 2 = Fc * S and V = √ (Fc * S * 2 / m).

Values of variables: Fc – acting resistance force (Fc = 2 kN = 2 * 10 ^ 3 N); S – hole depth, bullet path (S = 15 cm = 0.15 m); m is the mass of the bullet (m = 7 g = 7 * 10 ^ -3 kg).

Let’s perform the calculation: V = √ (Fc * S * 2 / m) = √ (2 * 10 ^ 3 * 0.15 * 2 / (7 * 10 ^ -3)) = 292.77 m / s.

Answer: The bullet had a velocity of 292.77 m / s before making the hole.