An object is released from rest. The time it takes to fall through as distance h and the speed of the object as it falls through this distance are measured with a pendulum clock. The entire apparatus is taken on the moon and the experiment is repeated

Options

A. the measured times are same
B. the measured speeds are same
C. the actual times in the fall are equal
D. the actual speeds are equal

Answer

To solve this problem, we need to analyze the motion of an object falling under the influence of gravity on both Earth and the Moon. The key points to consider are the equations of motion, the effects of gravity, and the measurements taken.

### Step-by-Step Solution:

1. **Understanding Initial Conditions**:
- The object is released from rest, which means its initial velocity \( u = 0 \).

2. **Using the Equation of Motion**:
- The equation of motion that relates the final velocity \( v \), initial velocity \( u \), acceleration \( g \), and distance \( h \) is given by:
\[ v^2 = u^2 + 2gh \]
- Since \( u = 0 \), the equation simplifies to:
\[ v^2 = 2gh \]
- Therefore, the final velocity \( v \) as the object falls through distance \( h \) is:
\[ v = \sqrt{2gh} \]

3. **Calculating Time of Fall**:
- The time \( t \) taken to fall through distance \( h \) can be derived from the second equation of motion:
\[ h = ut + \frac{1}{2}gt^2 \]
- Again, since \( u = 0 \), we have:
\[ h = \frac{1}{2}gt^2 \]
- Rearranging gives:
\[ t^2 = \frac{2h}{g} \quad \Rightarrow \quad t = \sqrt{\frac{2h}{g}} \]

4. **Comparing Earth and Moon**:
- On Earth, the acceleration due to gravity \( g \) is approximately \( 9.81 \, \text{m/s}^2 \).
- On the Moon, the acceleration due to gravity \( g' \) is approximately \( \frac{1}{6}g \) (or about \( 1.62 \, \text{m/s}^2 \)).
- Thus, the velocity on the Moon becomes:
\[ v' = \sqrt{2g'h} = \sqrt{2 \cdot \frac{1}{6}g \cdot h} = \sqrt{\frac{1}{3} \cdot 2gh} = \frac{1}{\sqrt{3}} \sqrt{2gh} \]
- The time taken to fall through distance \( h \) on the Moon is:
\[ t' = \sqrt{\frac{2h}{g'}} = \sqrt{\frac{2h}{\frac{1}{6}g}} = \sqrt{12 \frac{h}{g}} = 2\sqrt{3} \sqrt{\frac{h}{g}} \]

5. **Conclusion**:
- The measured times and speeds will be different on the Moon compared to Earth due to the difference in gravitational acceleration.
- Therefore, the correct options are:
- The measured times are the same (as they are measured with the same clock).
- The measured speeds are not the same.
- The actual times in the fall are not equal.
- The actual speeds are not equal.

### Final Answer:
- The correct options are:
- The measured times are the same.
- The measured speeds are not the same.