A marble and a cube are placed at the top of a ramp. Starting from rest at the same height, the marble rolls and the cube slides (no kinetic friction) down the ramp. Determine the ratio of the center-of-mass speed of the cube to the center-of-mass speed of the marble at the bottom of the ramp.

Options

A. 1.14
B. `1.20`
C. 0.707
D. 1.18

Answer

To solve the problem, we need to analyze the motion of both the marble and the cube as they descend the ramp. We will use the principles of energy conservation to find the center-of-mass speeds of both objects at the bottom of the ramp.

### Step 1: Understand the System
Both the marble and the cube start from rest at the same height \( h \) on the ramp. The marble rolls down while the cube slides down without friction.

### Step 2: Apply Conservation of Energy for the Cube
For the cube, which slides down without any rotation, the potential energy at the top is converted entirely into translational kinetic energy at the bottom.

- **Potential Energy at the top**:
\[ PE_{\text{cube}} = mgh \]
- **Kinetic Energy at the bottom**:
\[ KE_{\text{cube}} = \frac{1}{2} mv_c^2 \]
- Setting potential energy equal to kinetic energy:
\[ mgh = \frac{1}{2} mv_c^2 \]
- Simplifying, we find:
\[ gh = \frac{1}{2} v_c^2 \implies v_c^2 = 2gh \implies v_c = \sqrt{2gh} \]

### Step 3: Apply Conservation of Energy for the Marble
For the marble, which rolls down the ramp, both translational and rotational kinetic energy need to be considered.

- **Potential Energy at the top**:
\[ PE_{\text{marble}} = mgh \]
- **Kinetic Energy at the bottom**:
\[ KE_{\text{marble}} = \frac{1}{2} mv_m^2 + \frac{1}{2} I \omega^2 \]
- For a solid sphere, the moment of inertia \( I \) is given by \( I = \frac{2}{5} m r^2 \) and the relationship between linear velocity \( v_m \) and angular velocity \( \omega \) is \( v_m = r\omega \). Thus, \( \omega = \frac{v_m}{r} \).

- Substituting \( I \) and \( \omega \) into the kinetic energy equation:
\[ KE_{\text{marble}} = \frac{1}{2} mv_m^2 + \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v_m}{r}\right)^2 \]
\[ = \frac{1}{2} mv_m^2 + \frac{1}{5} mv_m^2 = \frac{1}{2} mv_m^2 + \frac{2}{10} mv_m^2 = \frac{7}{10} mv_m^2 \]

- Setting potential energy equal to total kinetic energy:
\[ mgh = \frac{7}{10} mv_m^2 \]
- Simplifying, we find:
\[ gh = \frac{7}{10} v_m^2 \implies v_m^2 = \frac{10}{7} gh \implies v_m = \sqrt{\frac{10}{7} gh} \]

### Step 4: Calculate the Ratio of Speeds
Now we can find the ratio of the center-of-mass speed of the cube to that of the marble:
\[ \frac{v_c}{v_m} = \frac{\sqrt{2gh}}{\sqrt{\frac{10}{7} gh}} = \sqrt{\frac{2gh}{\frac{10}{7} gh}} = \sqrt{\frac{2 \cdot 7}{10}} = \sqrt{\frac{14}{10}} = \sqrt{1.4} \]
Calculating this gives:
\[ \sqrt{1.4} \approx 1.18 \]

### Final Answer
The ratio of the center-of-mass speed of the cube to that of the marble is approximately \( 1.18 \).