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A ball is dropped on a floor from a height of `2.0m`. After the collision it rises up to a height of `1.5m`. Assume that `40%` of the mechanical energy lost goes as thermal energy into the ball.Calculate the rise in the temperature of the ball in the collision.Heat capacity of the ball is `800J K^(-1)` |
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Answer» Correct Answer - A::B::C `0.4 [mgDeltah] = msDeltatheta ` `=Deltatheta = (0.4gDeltah)/s = (0.4 xx 9.8 xx 0.5)/800` `=2.5 xx (10^-3)^@C`. |
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