Question

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)`

Answer 1

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`.