Question

2. Consider the ball example in the introduction when a ball is dropped from 3 meters. After the ball bounces, it raises to a height of 2 meters. The mass of the ball is 0.5 kg

a. What is the speed of the ball right before the bounce?





b. How much energy was converted into heat after the ball bounced off the ground. (Hint: Thermal Energy (TE) will now need to be included in your conservation of energy equation and you will now need to know the mass of the ball)




c. What is the speed of the ball immediately after the ball bounces off the ground?

Answer 1

.5 m v^2 = m g h
so
v = sqrt (2 g h)
v = sqrt (2*9.81*3) = 7.67 m/s

b)
loss of energy = m g (3-2)
= .5 * 9.81 * 1 = 4.9 Joules

c) v = sqrt (2 g h) but h is 2 instead of 3 now