A metal ball with a mass of 2 kg fell on a lead plate with a mass of 1 kg and stopped. In this case

| June 20, 2021 | education

A metal ball with a mass of 2 kg fell on a lead plate with a mass of 1 kg and stopped. In this case, the plate warmed up by 3.2 degrees. From what height did the ball fall if 80% of the amount of heat released during the impact went into heating the plate?

Data: m1 (mass of the dropped ball) = 2 kg; m2 (mass of the plate on which the ball fell) = 1 kg; Δt (change in plate temperature) = 3.2 ºС; η (percentage of heat from the impact of the ball, which went to heating the plate) = 80% (0.8).

Constants: g (acceleration of gravity) = 9.81 m / s2; С (specific heat capacity of lead) = 140 J / (kg * ºС).

The height of the fall of the ball is determined from the equality: η * En = Q; η * m1 * g * h = C * m2 * Δt and h = C * m2 * Δt / (η * m1 * g).

Calculation: h = 140 * 1 * 3.2 / (0.8 * 2 * 9.81) = 28.54 m.



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