A bar weighing 400 g, moving on a smooth horizontal surface with a speed ʋ = 10 m / s, hits the same, but stationary

A bar weighing 400 g, moving on a smooth horizontal surface with a speed ʋ = 10 m / s, hits the same, but stationary bar and loses half of its speed. Find the amount of heat released when the bars collide. The movement of the bars is considered to be translational.

m = 400 g = 0.4 kg.

V10 = 10 m / s.

V20 = 0 m / s.

V1 = V10 / 2.

Q -?

When the bars hit, part of the mechanical energy is converted into internal energy.

The amount of thermal energy Q, which is released upon impact, is equal to the change in the kinetic energy of the bars upon impact ΔEk: Q = ΔEk.

Ek1 = m * V1 ^ 2/2.

Ek2 = m * V1 ^ 2/2 + m * V2 ^ 2/2.

m * V10 = m * V1 + m * V2.

V2 = V10 – V1 = V10 / 2.

after the collision, the bars will move at the same speed.

Q = Ek1 – Ek2 = m * V01 ^ 2/2 – m * V1 ^ 2/2 – m * V2 ^ 2/2 = m * V01 ^ 2/2 – m * V01 ^ 2/8 – m * V01 ^ 2/8 = m * V0 ^ 12/4.

Q = 0.4 kg * (10 m / s) ^ 2/4 = 10 J.

Answer: when the bars hit, Q = 10 J of heat was released.