The bullet flew at a speed of 400 m / s (V zero), pierced a wooden block and flew out of it at a speed of 300 m / s

The bullet flew at a speed of 400 m / s (V zero), pierced a wooden block and flew out of it at a speed of 300 m / s (V first). Find the temperature at which the bullet is heated if its specific heat capacity is 140 J / kg * C. Disregard heat loss.

Initial kinetic energy of a bullet:

E0 = M V0 ^ 2/2;

kinetic energy of the bullet that pierced the bar:

E1 = M V1 ^ 2/2;

where M is the mass of the bullet; V0, V1 – initial and final bullet velocity, respectively.

The decrease in the kinetic energy of the bullet is equal to the work of the resistance force A:

A = E0 – E1 = M (V0 ^ 2 – V1 ^ 2) / 2.

Ignoring heat loss:

А = с М ΔТ,

where ΔТ is the increase in the temperature of the bullet; s = 140 J / (kg × oC).

ΔТ = (V0 ^ 2 – V1 ^ 2) / (2s) = (400 ^ 2 – 300 ^ 2) / (2 × 140) = 250 оС.

Answer: ΔТ = 250 оС.