Kinetic energy of particles of mass 10 g and 40 g is same, the ratio of their linear momentum is

Options

A. `1/4`
B. `1/2`
C. `1/sqrt2`
D. `sqrt2/1`

Explanation

To solve the problem, we need to find the ratio of linear momentum of two particles with the same kinetic energy but different masses.

### Step-by-Step Solution:

1. **Understanding Kinetic Energy**:
The formula for kinetic energy (KE) is given by:
\[ KE = \frac{1}{2} mv^2 \]
where \( m \) is the mass and \( v \) is the velocity of the particle.

2. **Setting Up the Equation**:
Let the mass of the first particle be \( m_1 = 10 \, \text{g} \) and the mass of the second particle be \( m_2 = 40 \, \text{g} \). Since the kinetic energies of both particles are the same, we can write:
\[ KE_1 = KE_2 \]
This implies:
\[ \frac{1}{2} m_1 v_1^2 = \frac{1}{2} m_2 v_2^2 \]

3. **Cancelling Out Common Factors**:
We can cancel \(\frac{1}{2}\) from both sides:
\[ m_1 v_1^2 = m_2 v_2^2 \]

4. **Expressing Linear Momentum**:
The linear momentum \( P \) is defined as:
\[ P = mv \]
Therefore, for the two particles, we have:
\[ P_1 = m_1 v_1 \quad \text{and} \quad P_2 = m_2 v_2 \]

5. **Relating Momentum to Kinetic Energy**:
From the earlier equation \( m_1 v_1^2 = m_2 v_2^2 \), we can express \( v_1 \) and \( v_2 \) in terms of momentum:
\[ v_1 = \frac{P_1}{m_1} \quad \text{and} \quad v_2 = \frac{P_2}{m_2} \]

6. **Substituting Back**:
Substitute \( v_1 \) and \( v_2 \) back into the kinetic energy equation:
\[ m_1 \left(\frac{P_1}{m_1}\right)^2 = m_2 \left(\frac{P_2}{m_2}\right)^2 \]
Simplifying this gives:
\[ \frac{P_1^2}{m_1} = \frac{P_2^2}{m_2} \]

7. **Finding the Ratio of Momenta**:
Rearranging gives:
\[ \frac{P_1^2}{P_2^2} = \frac{m_1}{m_2} \]
Taking the square root of both sides results in:
\[ \frac{P_1}{P_2} = \sqrt{\frac{m_1}{m_2}} = \sqrt{\frac{10}{40}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]

8. **Final Result**:
Thus, the ratio of their linear momentum is:
\[ \frac{P_1}{P_2} = \frac{1}{2} \]