#### Answer

(a) The speed of the bike at the bottom of the hill is $~19.4~m/s$
(b) With a less massive rider, the speed at the bottom would be slower.

#### Work Step by Step

(a) Let $M$ be the mass of the bike. Let $m$ be the mass of each wheel. We can find the total kinetic energy of the bike when it is moving at a speed $v$:
$KE = KE_{tr}+KE_{rot}$
$KE = \frac{1}{2}Mv^2+2\times \frac{1}{2}I~\omega^2$
$KE = \frac{1}{2}Mv^2+(mR^2)~(\frac{v}{R})^2$
$KE = \frac{1}{2}Mv^2+mv^2$
$KE = (\frac{M}{2}+m)~v^2$
We can use conservation of energy to find the speed of the bike at the bottom of the hill:
$KE_f = U_0$
$(\frac{M}{2}+m)~v^2 = Mgh$
$v^2 = \frac{Mgh}{\frac{M}{2}+m}$
$v = \sqrt{\frac{Mgh}{\frac{M}{2}+m}}$
$v = \sqrt{\frac{(80.0~kg)(9.80~m/s^2)(20.0~m)}{\frac{80.0~kg}{2}+1.5~kg}}$
$v = 19.4~m/s$
The speed of the bike at the bottom of the hill is $~19.4~m/s$
(b) If the rider is less massive, then a smaller fraction of the kinetic energy would be in the form of translational kinetic energy. Therefore, with a less massive rider, the speed at the bottom would be slower.