A bullet weighing 0.2 kg acquires a speed of 40 m / s leaving the barrel of a rifle 50 cm long.

A bullet weighing 0.2 kg acquires a speed of 40 m / s leaving the barrel of a rifle 50 cm long.Find a) the acceleration of the bullet b) what force acted on the bullet

m = 0.2 kg.

V0 = 0 m / s.

V = 40 m / s.

L = 50cm = 0.5m.

a -?

F -?

We will assume that the bullet in the barrel moves uniformly, so its acceleration will be determined by the formula: a = (V ^ 2 – V0 ^ 2) / 2 * L, where V, V0 are the final and initial velocity of the bullet in the barrel, L is the path of the bullet in the trunk.

Since the bullet begins its movement from a state of rest V0 = 0 m / s, the formula will take the form: a = V ^ 2/2 * L.

a = (40 m / s) ^ 2/2 * 0.5 m = 1600 m / s2.

According to 2 Newton’s laws: F = m * a.

F = 0.2 kg * 1600 m / s2 = 320 N.

Answer: a = 1600 m / s2, F = 320 N.