Question

a ball bounces to 80percent of its original height what fraction of its k.e. is lost in each bounce

Answer 1

Let a ball fall from a height h,
thenKinetic energy of ball at the time striking the ground= Potential energy of ball at height h

so, K.E=mgh

Kinetic energy of ball due to rebounce will be :

K.Er =P.E of ball at height h1= mgh1

Loss of kinetic energy due to re-bounce=K.E -K.Er

=mgh -mgh1
=mg(h-h1)
=mg(h-80h/100)
=mg(20h/100)
= 0.2 x mgh

Fractional loss in K.E in each re-bounce will be :
=K.E-K.Er/K.E
=0.2 x mgh/mgh
=0.2
=0.2 x100%
=20%

Fraction of its Mechanical energy lost in each rebound is 20%

Answer 2

Actual P.E = mgh

Decrease in P.E =
$\frac{80}{100} mgh = 0.8mgh$

decrease in each bounce =
mgh - 0.8 mgh = 0.2 mgh

therefore, fraction =>
$\frac{0.2mgh}{mgh} = 0.2 = \frac{2}{10}$

hope this helps.....