# The steel ball falls from a certain height onto the steel plate, reaching a speed of 15 m / s before the collision.

**The steel ball falls from a certain height onto the steel plate, reaching a speed of 15 m / s before the collision. After hitting the plate, the ball bounced to a height of 5 cm from the plate. Determine how many degrees the ball has heated up after hitting the plate. Neglect heat loss to the environment. The heating of the stove is also neglected. Specific heat 0.46 kJ / kg coefficient g = 10 N / kg**

V = 15 m / s.

h = 5 cm = 0.05 m.

g = 10 m / s2.

C = 0.46 kJ / kg * ° C = 460 J / kg * ° C.

Δt -?

The amount of heat that goes into heating the ball is expressed by the formula: Q = C * m * Δt.

We express the same amount of heat from the law of conservation of energy: Q = Ek – En.

We express the kinetic Ek and the potential energy En by the formulas: Ek = m * V ^ 2/2, En = m * g * h.

Q = m * V ^ 2/2 – m * g * h = m * (V ^ 2/2 – g * h).

m * (V ^ 2/2 – g * h) = C * m * Δt.

The change in the temperature of the ball Δt is expressed by the formula: Δt = (V ^ 2/2 – g * h) / C.

Δt = ((15 m / s) ^ 2/2 – 10 m / s2 * 0.05 m) / 460 J / kg * ° C = 0.24 ° C.

Answer: The temperature of the steel ball has increased by Δt = 0.24 ° C.