1. **State the problem:** A hollow cylinder rolls up an incline with initial velocity $4.20$ m/s and reaches a vertical height of $1.00$ m. A hollow sphere with the same mass and radius is given the same initial velocity. We need to find the vertical height the hollow sphere reaches. 2. **Relevant formulas and concepts:** - Rolling without slipping means kinetic energy is split between translational and rotational energy. - Total kinetic energy: $$K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$ - For rolling, $\omega = \frac{v}{R}$. - Moment of inertia for hollow cylinder: $$I_{cyl} = mR^2$$ - Moment of inertia for hollow sphere: $$I_{sphere} = \frac{2}{3}mR^2$$ - Conservation of energy: initial kinetic energy converts to potential energy at max height $h$: $$\frac{1}{2}mv^2 + \frac{1}{2}I\left(\frac{v}{R}\right)^2 = mgh$$ 3. **Calculate height for hollow cylinder (given):** $$\frac{1}{2}mv^2 + \frac{1}{2}mR^2\left(\frac{v}{R}\right)^2 = mgh_{cyl}$$ Simplify rotational term: $$\frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mgh_{cyl}$$ $$mv^2 = mgh_{cyl}$$ Cancel $m$: $$v^2 = gh_{cyl}$$ Given $h_{cyl} = 1.00$ m, check consistency: $$v^2 = g \times 1.00$$ $$v = \sqrt{9.8 \times 1.00} = 3.13 \text{ m/s}$$ But given $v=4.20$ m/s, so the problem likely assumes the given height is correct for the hollow cylinder. 4. **Calculate height for hollow sphere:** $$\frac{1}{2}mv^2 + \frac{1}{2} \times \frac{2}{3}mR^2 \left(\frac{v}{R}\right)^2 = mgh_{sphere}$$ Simplify rotational term: $$\frac{1}{2}mv^2 + \frac{1}{3}mv^2 = mgh_{sphere}$$ $$\frac{1}{2}mv^2 + \frac{1}{3}mv^2 = mgh_{sphere}$$ $$\left(\frac{1}{2} + \frac{1}{3}\right) mv^2 = mgh_{sphere}$$ $$\frac{3}{6} + \frac{2}{6} = \frac{5}{6}$$ $$\frac{5}{6} mv^2 = mgh_{sphere}$$ Cancel $m$: $$\frac{5}{6} v^2 = gh_{sphere}$$ Solve for $h_{sphere}$: $$h_{sphere} = \frac{5}{6} \frac{v^2}{g}$$ 5. **Express $h_{sphere}$ in terms of $h_{cyl}$:** From step 3, for hollow cylinder: $$v^2 = gh_{cyl}$$ Substitute into $h_{sphere}$: $$h_{sphere} = \frac{5}{6} \frac{gh_{cyl}}{g} = \frac{5}{6} h_{cyl}$$ 6. **Calculate numerical value:** $$h_{sphere} = \frac{5}{6} \times 1.00 = 0.833 \text{ m}$$ **Final answer:** The hollow sphere rolls up to a vertical height of approximately **0.833 m**.