A solid ball and a hollow ball, each with a mass of \(1.00 \mathrm{~kg}\) and radius of \(0.100 \mathrm{~m}\), start from rest and roll down a ramp of length \(3.00 \mathrm{~m}\) at an incline of \(35.0^{\circ} .\) An ice cube of the same mass slides without friction down the same ramp. a) Which ball will reach the bottom first? Explain! b) Does the ice cube travel faster or slower than the solid ball at the base of the incline? Explain your reasoning. c) What is the speed of the solid ball at the bottom of the incline?

To find which ball will reach the bottom first, we need to determine the acceleration of each ball down the ramp. Since both balls have the same mass and radius, we can focus on their moment of inertia for rolling motion. The moment of inertia for a solid sphere is \(I = \frac{2}{5}mr^2\), and for a hollow sphere is \(I = \frac{2}{3}mr^2\). As the moment of inertia represents the resistance of an object to rotational acceleration, the solid ball with the smaller moment of inertia will have the greater acceleration down the ramp. Thus, the solid ball will reach the bottom first.

We can compare the ice cube's speed to the solid ball's speed by looking at their kinetic energy at the base of the incline. Since there is no friction acting on the ice cube, all of its gravitational potential energy is converted to translational kinetic energy at the base of the incline. In contrast, the solid ball has both translational and rotational kinetic energy. However, since the solid ball has a smaller moment of inertia and a greater acceleration down the incline, it will have a greater overall kinetic energy at the base. Therefore, the ice cube will travel slower than the solid ball at the base of the incline.

To find the speed of the solid ball at the bottom of the incline, we can use the conservation of mechanical energy. The ball's initial gravitational potential energy (\(mgh\)) is converted to both translational (\(\frac{1}{2}mv^2\)) and rotational (\(\frac{1}{2}I\omega^2\)) kinetic energy. Since \(v = r\omega\), we get: \(mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mr^2\right)(\frac{v}{r})^2\) Solving for \(v\): \(v^2 = \frac{10}{7}gh\) Using the given values for mass, radius, incline angle, and ramp length, we can calculate the height \(h\) as: \(h = \sin(35^\circ) \cdot 3.00 \mathrm{~m}\) Now, we can plug in the numbers to find the speed of the solid ball at the bottom of the incline: \(v = \sqrt{\frac{10}{7}gh}\) \(v = \sqrt{\frac{10}{7} \cdot 9.81 \mathrm{m/s^2} \cdot \sin(35^\circ) \cdot 3.00 \mathrm{~m}}\) Therefore, the speed of the solid ball at the bottom of the incline is approximately \(4.46 \mathrm{m/s}\).