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The centre of mass of two masses m & m' ...

The centre of mass of two masses m & m' moves by distance `(x)/(5)` when mass m is moved by distance x and m' is kept fixed. The ratio `(m')/(m)` is

A

2

B

4

C

`1//4`

D

None of these

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To solve the problem of finding the ratio \( \frac{m'}{m} \) given that the center of mass of two masses \( m \) and \( m' \) moves by a distance \( \frac{x}{5} \) when mass \( m \) is moved by a distance \( x \) and mass \( m' \) is kept fixed, we can follow these steps: ### Step-by-Step Solution: 1. **Understanding the Center of Mass Formula:** The center of mass \( x_{cm} \) for two masses \( m \) and \( m' \) located at positions \( x_1 \) and \( x_2 \) respectively is given by: \[ x_{cm} = \frac{m \cdot x_1 + m' \cdot x_2}{m + m'} \] 2. **Setting Up the Problem:** Let's assume: - Mass \( m \) is initially at position \( x_1 \) (let's say \( x_1 = 0 \) for simplicity). - Mass \( m' \) is at position \( x_2 \) (we can assume \( x_2 = d \), where \( d \) is some distance). Thus, the center of mass will initially be: \[ x_{cm} = \frac{m \cdot 0 + m' \cdot d}{m + m'} = \frac{m' \cdot d}{m + m'} \] 3. **Moving Mass \( m \):** When mass \( m \) is moved by a distance \( x \), its new position becomes \( x_1' = x \). The position of mass \( m' \) remains the same at \( x_2 = d \). The new center of mass \( x_{cm}' \) is: \[ x_{cm}' = \frac{m \cdot x + m' \cdot d}{m + m'} \] 4. **Change in Center of Mass:** According to the problem, the center of mass moves by a distance \( \frac{x}{5} \): \[ x_{cm}' - x_{cm} = \frac{x}{5} \] 5. **Substituting the Expressions:** Substituting the expressions for \( x_{cm}' \) and \( x_{cm} \): \[ \frac{m \cdot x + m' \cdot d}{m + m'} - \frac{m' \cdot d}{m + m'} = \frac{x}{5} \] This simplifies to: \[ \frac{m \cdot x}{m + m'} = \frac{x}{5} \] 6. **Cross Multiplying:** Cross multiplying gives: \[ 5m \cdot x = (m + m') \cdot x \] Dividing both sides by \( x \) (assuming \( x \neq 0 \)): \[ 5m = m + m' \] 7. **Rearranging the Equation:** Rearranging gives: \[ 5m - m = m' \implies 4m = m' \] 8. **Finding the Ratio:** Thus, the ratio \( \frac{m'}{m} \) is: \[ \frac{m'}{m} = \frac{4m}{m} = 4 \] ### Final Answer: The ratio \( \frac{m'}{m} \) is \( 4 \).

To solve the problem of finding the ratio \( \frac{m'}{m} \) given that the center of mass of two masses \( m \) and \( m' \) moves by a distance \( \frac{x}{5} \) when mass \( m \) is moved by a distance \( x \) and mass \( m' \) is kept fixed, we can follow these steps: ### Step-by-Step Solution: 1. **Understanding the Center of Mass Formula:** The center of mass \( x_{cm} \) for two masses \( m \) and \( m' \) located at positions \( x_1 \) and \( x_2 \) respectively is given by: \[ x_{cm} = \frac{m \cdot x_1 + m' \cdot x_2}{m + m'} ...
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Knowledge Check

  • The centre of mass of a system of two particle of masses m_1 and m_2 is at a distance d_1 from m_1 and at a distance d_2 from mass m_2 such that.

    A
    `(d_1)/(d_2) = (m_2)/(m_1)`
    B
    `(d_1)/(d_2) = (m_1)/(m_2)`
    C
    `(d_1)/(d_2) = (m_1)/(m_1 + m_2)`
    D
    `(d_1)/(d_2) = (m_2)/(m_1 + m_2)`
  • Where will be the centre of mass on combining two masses m and M(M gt m) ?

    A
    Towards m
    B
    Towards M
    C
    Between m and M
    D
    Anywhere
  • Where will be the centre of mass on combining two masses m and M (M gt m)

    A
    Towards m
    B
    Towards M
    C
    Between m and M
    D
    Anywhere
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