Two bodies A and B having masses in the ratio of 3 : 1 possess the same kinetic energy. The ratio of their linear momenta is then

Options

A. `3:1`
B. `9:1`
C. `1:1`
D. `sqrt(3):1`

Answer

To solve the problem, we need to find the ratio of the linear momenta of two bodies A and B, given that their masses are in the ratio of 3:1 and they possess the same kinetic energy.

### Step-by-Step Solution:

1. **Understanding Kinetic Energy and Momentum**:
- The kinetic energy (KE) of an object is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
- The momentum (p) of an object is given by the formula:
\[ p = mv \]

2. **Setting Up the Problem**:
- Let the mass of body A be \( m_A = 3m \) and the mass of body B be \( m_B = m \) (where \( m \) is a common factor).
- Since both bodies have the same kinetic energy, we can denote their kinetic energies as \( KE_A \) and \( KE_B \).

3. **Expressing Kinetic Energies**:
- For body A:
\[ KE_A = \frac{1}{2} m_A v_A^2 = \frac{1}{2} (3m) v_A^2 \]
- For body B:
\[ KE_B = \frac{1}{2} m_B v_B^2 = \frac{1}{2} m v_B^2 \]
- Since \( KE_A = KE_B \), we can set the equations equal to each other:
\[ \frac{1}{2} (3m) v_A^2 = \frac{1}{2} m v_B^2 \]

4. **Simplifying the Equation**:
- Cancel \( \frac{1}{2} \) and \( m \) (assuming \( m \neq 0 \)):
\[ 3 v_A^2 = v_B^2 \]

5. **Finding the Relationship Between Velocities**:
- Taking the square root of both sides:
\[ v_B = \sqrt{3} v_A \]

6. **Calculating the Linear Momentum**:
- The momentum of body A:
\[ p_A = m_A v_A = (3m) v_A \]
- The momentum of body B:
\[ p_B = m_B v_B = m (\sqrt{3} v_A) = m \sqrt{3} v_A \]

7. **Finding the Ratio of Linear Momenta**:
- The ratio of the linear momenta \( \frac{p_A}{p_B} \):
\[ \frac{p_A}{p_B} = \frac{(3m) v_A}{m \sqrt{3} v_A} \]
- Simplifying this gives:
\[ \frac{p_A}{p_B} = \frac{3}{\sqrt{3}} = \sqrt{3} \]

8. **Final Ratio**:
- Thus, the ratio of their linear momenta is:
\[ p_A : p_B = \sqrt{3} : 1 \]

### Conclusion:
The ratio of the linear momenta of bodies A and B is \( \sqrt{3} : 1 \).