A bar weighing 1.2 kg lies on a horizontal surface. It is hit by a bullet weighing 20 g, flying horizontally

A bar weighing 1.2 kg lies on a horizontal surface. It is hit by a bullet weighing 20 g, flying horizontally at a speed υ0, and gets stuck in it. With the coefficient of sliding friction equal to 0.3, the bar will travel 4 m to a complete stop. What is the bullet velocity υ0?

mb = 1.2 kg.

mp = 20 g = 0.02 kg.

g = 9.8 m / s2.

μ = 0.3.

S = 4 m.

Vп -?

We will assume that the bar was moving with constant acceleration a until it came to a complete stop.

a = Vb ^ 2/2 * S, where the speed of the bar after being hit by a bullet.

Acceleration of a bar with a bullet and during braking we express by 2 Newton’s law: (mb + mp) * a = Ftr.

Ftr = μ * N = μ * (mb + mp) * g.

(mb + mp) * a = μ * (mb + mp) * g.

a = μ * g.

Vb ^ 2/2 * S = μ * g.

Vb = √ (μ * g * 2 * S).

Vb = √ (0.3 * 9.8 m / s2 * 2 * 4 m) = 4.85 m / s.

Let’s write down the law of conservation of momentum: Vp * mp = (mb + mp) * Vb.

Vp = (mb + mp) * Vb / mp.

Vp = (1.2 kg + 0.02 kg) * 4.85 m / s / 0.02 kg = 296 m / s.

Answer: before hitting the bar, the bullet moved at a speed of Vp = 296 m / s.