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. 0.447
B. 1.14
C. 0.707
D. 1.18

Answer

To solve the problem of determining 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, we will apply the principles of energy conservation. Here’s a step-by-step solution:

### Step 1: Analyze the Cube
1. **Initial Conditions**: The cube starts from rest at height \( h \).
2. **Energy Conservation**: The potential energy at the top converts to kinetic energy at the bottom.
\[ \text{Potential Energy at top} = \text{Kinetic Energy at bottom} \]
\[ mgh = \frac{1}{2} mv_c^2 \]
where \( v_c \) is the speed of the cube at the bottom.

3. **Simplifying**: Cancel \( m \) from both sides (since \( m \neq 0 \)):
\[ gh = \frac{1}{2} v_c^2 \]

4. **Solving for \( v_c \)**:
\[ v_c^2 = 2gh \implies v_c = \sqrt{2gh} \]

### Step 2: Analyze the Marble
1. **Initial Conditions**: The marble also starts from rest at height \( h \).
2. **Energy Conservation**: The marble has both translational and rotational kinetic energy when it reaches the bottom.
\[ mgh = \frac{1}{2} mv_m^2 + \frac{1}{2} I \omega^2 \]
where \( v_m \) is the speed of the marble at the bottom and \( I \) is the moment of inertia of the marble.

3. **Moment of Inertia**: For a solid sphere, \( I = \frac{2}{5} m r^2 \) and \( \omega = \frac{v_m}{r} \).
\[ \omega^2 = \left(\frac{v_m}{r}\right)^2 = \frac{v_m^2}{r^2} \]

4. **Substituting \( I \) and \( \omega \)**:
\[ mgh = \frac{1}{2} mv_m^2 + \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v_m^2}{r^2}\right) \]
\[ mgh = \frac{1}{2} mv_m^2 + \frac{1}{5} mv_m^2 \]
\[ mgh = \left(\frac{1}{2} + \frac{1}{5}\right) mv_m^2 \]

5. **Finding a common denominator**:
\[ \frac{1}{2} + \frac{1}{5} = \frac{5}{10} + \frac{2}{10} = \frac{7}{10} \]
Thus,
\[ mgh = \frac{7}{10} mv_m^2 \]

6. **Solving for \( v_m \)**:
\[ gh = \frac{7}{10} v_m^2 \implies v_m^2 = \frac{10gh}{7} \implies v_m = \sqrt{\frac{10gh}{7}} \]

### Step 3: Calculate the Ratio
1. **Finding the ratio \( \frac{v_c}{v_m} \)**:
\[ \frac{v_c}{v_m} = \frac{\sqrt{2gh}}{\sqrt{\frac{10gh}{7}}} \]
\[ = \sqrt{\frac{2gh \cdot 7}{10gh}} = \sqrt{\frac{14}{10}} = \sqrt{1.4} \]

2. **Calculating the numerical value**:
\[ \sqrt{1.4} \approx 1.183 \]

### Final Answer
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 is approximately \( 1.18 \).

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