Answer
$(a)\space 17.92\space cm$
$(b)\space 1.59\space m/s$
Work Step by Step
Please see the attached image first.
Here we use the conservation of mechanical energy.
Kinetic energy + Potential energy = Constant
So we can write,
$Mgh+0=\frac{1}{2}MV^{2}+\frac{1}{2}I\omega^{2}-(1)$
We know that the rotational inertia of the marble, $I=\frac{2}{5}MR^{2},\space \omega=\frac{V}{R}$
From $(1)=\gt$
$Mgh=\frac{1}{2}MV^{2}+\frac{1}{2}(\frac{2}{5}MR^{2})(\frac{V^{2}}{R^{2}})$
$gh=\frac{V^{2}}{2}(\frac{7}{5})$
$V= \sqrt {\frac{10gh}{7}}-(2)$
(a) At halfway down; V = 1.12 m/s
So from $(2)=\gt\space 1.12\space m/s=\sqrt {\frac{10\times9.8\space m/s^{2}\times h/2}{7}}$
$2\times0.0896\space m=h$
$h=17.92\space cm$ (Starting height)
(b) At the bottom of the slope, h = 17.92 cm
So, $(1)=\gt$
$V=\sqrt {\frac{10\times9.8\space m/s^{2}\times 0.18\space m}{7}}=1.59\space m/s$
Speed at the bottom = 1.59 m/s
