A 750 g disk and a 760 g ring, both 15 cm in diameter, are rolling along a horizontal surface at 1.5 m/s when they encounter a 15° slope. How far up the slope does each travel before rolling back down?

The mass of the disc=750 gm = 0.75kg

The mass of the ring=760 gm = 0.76kg
Diameter of the disk and ring = 15 cm = 0.15 m
Velocity of both on horizontal surface= 1.5 m/sec
Angle of inclination = 15^o

From the law of energy conservation

Total energy on horizontal floor = Total energy on inclined plane
Kinetic energy of transnational motion + Kinetic energy of rotational motion = Potential energy on inclined plane

$\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2=mgh........................\left(1\right)$

Calculate for Disc
In equation (1) substitute $I=\frac{\left(M{R}^2\right)}{2}\mathrm{and}\omega =v/R$

$$\begin{gathered} \frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{1}{2}m{R}^2\right)=mgh \\ {\left(\frac{v}{R}\right)}^2=mgh \\ \frac{3}{4}{v}^2=gh \end{gathered}$$

$$\begin{gathered} substituteh=L\sin \theta ,weget \\ \frac{3}{4}{v}^2=gL\sin \theta \\ L=\frac{3v^2}{4g\sin \theta } \end{gathered}$$

Now substitute the given value

$$\begin{gathered} L=\frac{3\times {\left(1.5{\mathrm{ms}}^{-1}\right)}^2}{4\sin 15°\times 9.8m{s}^{-2}} \\ L=\frac{6.75}{10.145}m \\ L=0.66\mathrm{m} \end{gathered}$$

Calculate for Ring

Substitute value of $I=M{R}^2\mathrm{and}\omega =v/R$in equation (1)

$$\begin{gathered} \frac{1}{2}M{v}^2+\frac{1}{2}\left(M{R}^2\right) \\ {\left(\frac{v}{R}\right)}^2=Mgh \\ {v}^2=gh \end{gathered}$$

Substitute the value of $h=L\sin \theta$

$$\begin{gathered} v^2=g{L}^{\text{'}}\sin \theta \\ {L}^{\text{'}}=\frac{v^2}{4g\sin \theta } \end{gathered}$$

Substitute the given values, we get

$L=\frac{{\left.\left(1.5m{s}^{-1}\right)\right)}^2}{4(\sin 15°)\times (9.8m{s}^{-2})}=0.89m$