Answer
(a) 0.2
(b) $v_1 = 6.3 ~m/s$
$v_2 = 5.6 ~m/s$
(c) The energy went into thermal energy in the ball and in the floor.
Work Step by Step
(a) Before the bounce:
$E_1 = PE_1 = mgy_1 = 2.0(mg)$
After the bounce:
$E_2 = PE_2 = mgy_2 = 1.6(mg)$
We can calculate the fraction of energy lost during the bounce.
$\frac{E_1-E_2}{E_1} = \frac{2.0(mg)-1.6(mg)}{2.0(mg)} = \frac{0.4}{2.0} = 0.2$
(b) We can use conservation of energy to solve this part of the question.
Before the bounce:
$KE_1 = PE_1$
$\frac{1}{2}mv_1^2 = mgy_1 $
$\frac{1}{2}mv_1^2 = 2.0(mg)$
$v_1 = \sqrt{4.0~g} = \sqrt{(4.0 ~m)(9.80 ~m/s^2)} = 6.3 ~m/s$
After the bounce:
$KE_2 = PE_2$
$\frac{1}{2}mv_2^2 = mgy_2 $
$\frac{1}{2}mv_2^2 = 1.6(mg)$
$v_2 = \sqrt{3.2~g} = \sqrt{(3.2 ~m)(9.80 ~m/s^2)} = 5.6 ~m/s$
(c) The energy went into thermal energy in the ball and in the floor.
