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

`[2.5xx10^(-3) .^(@)C]`

Explanation

Mechanical energy lost `=mg(h_(2)-h_(1))` 40% of mech. Energy lost = heat gained by ball `(40)/(100)xxmg(h_(2)-h_(1))=800m xx Delta T`
`(2)/(5)xx10(2-1.5)=800xxDelta T`
`Delta T=(2)/(800)=2.5xx10^(-3) .^(@)C`