A rebound and kinetic energy problem with a little specific heat

1. The problem statement, all variables and given/known data
A ball is thrown vertically downward strikes a surface with a speed of 15m/s. It then bounces, and reaches a maximum height of 5 meters. Neglect air resistance on the ball.

a) What is the speed of the ball immediately after it rebounds from the surface?
b)What fraction of the ball's initial kinetic energy is apparently lost during the bounce?
c)If the specific heat of the ball is 1,800 j/kg*C, and if all of the lost energy is absorbed by the molecules of the ball, by how much does the temperature of the ball increase?

2. Relevant equations
KE=1/2mv^2
Q=mL(delta "T")

3. The attempt at a solution
my answer for a was crazy so i am pretty much stuck with the rest
 
Use law of conservation of energy - at any level (when there are no energy loss) potential energy+kinetic energy = constant. When ball is hitting a surface at 15 m/s, there is only a kinetic energy. When it reaches maximum height 5 m - there is only potential energy. Calculate what height would it be if all of kinetic energy at start where converted to potential. From difference you will find loss of energy during contact with surface ;] hope this will help. By the way - potential energy equation near Earth's surface: EP = mgh
 
but how do i find the mass ....i found the the joules lost to be 63(mass) joules
 
Well i don't know ;] so maybe express your answer as a function of mass, deltaT=(some number)/m.
 

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