Two blocks of msses 10 kg and 30 kg are placed along a vertical line. The first block is raised through a height of 7 cm. By what distance should the second mass be moved to raise the centre of mass by 1 cm?

Options

A. 2 cm upward
B. 1 cm upward
C. 2cm downward
D. 1 cm downward

Explanation

To solve the problem step by step, let's denote the masses and their movements clearly:

1. **Identify the masses and their initial positions**:
- Let \( m_1 = 10 \, \text{kg} \) (first block)
- Let \( m_2 = 30 \, \text{kg} \) (second block)
- The first block is raised through a height of \( h_1 = 7 \, \text{cm} \).

2. **Determine the initial position of the center of mass (CM)**:
- The initial position of the center of mass can be calculated using the formula:
\[ y_{CM} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2} \]
- Assume the initial position of \( m_1 \) (10 kg) is at \( y_1 = 0 \) cm and \( m_2 \) (30 kg) is at \( y_2 = h \) cm (we can assume \( h \) is the height of the second block).

3. **Calculate the initial center of mass**:
- The initial center of mass is:
\[ y_{CM, initial} = \frac{10 \cdot 0 + 30 \cdot h}{10 + 30} = \frac{30h}{40} = \frac{3h}{4} \]

4. **Calculate the new position of the first block after being raised**:
- After raising the first block by 7 cm, its new position becomes \( y_1' = 7 \, \text{cm} \).

5. **Calculate the new center of mass**:
- The new center of mass after raising the first block is:
\[ y_{CM, new} = \frac{10 \cdot 7 + 30 \cdot y_2'}{10 + 30} = \frac{70 + 30y_2'}{40} \]

6. **Set the change in center of mass to 1 cm**:
- The problem states that the center of mass is raised by 1 cm:
\[ y_{CM, new} - y_{CM, initial} = 1 \, \text{cm} \]
- Substituting the expressions we derived:
\[ \frac{70 + 30y_2'}{40} - \frac{3h}{4} = 1 \]

7. **Simplify and solve for \( y_2' \)**:
- Multiply through by 40 to eliminate the denominator:
\[ 70 + 30y_2' - 30h = 40 \]
- Rearranging gives:
\[ 30y_2' = 40 - 70 + 30h \]
\[ 30y_2' = 30h - 30 \]
\[ y_2' = h - 1 \]

8. **Determine the distance the second mass should be moved**:
- Since we need to find the distance \( d \) that the second mass should be moved, we can express it as:
\[ d = h - y_2 \]
- If the second mass is initially at height \( h \) and needs to be moved to height \( h - 1 \), then:
\[ d = -1 \, \text{cm} \]
- This indicates that the second mass must be moved downward by 1 cm.

**Final Answer**: The second mass should be moved downward by 1 cm.