Two masses `M_1 & M_2` have same kinetic energy. If `V_2 = 2V_1` , then find ratio of their momentum

Options

A. `root(2)`
B. `(root(2))`
C. 2
D. `1/2`

Answer

To solve the problem, we need to find the ratio of the momenta of two masses \( M_1 \) and \( M_2 \) that have the same kinetic energy, given that \( V_2 = 2V_1 \).

### Step-by-Step Solution:

1. **Write the expression for kinetic energy**:
The kinetic energy \( KE \) of a mass is given by the formula:
\[ KE = \frac{1}{2} m v^2 \]
For mass \( M_1 \) with velocity \( V_1 \):
\[ KE_1 = \frac{1}{2} M_1 V_1^2 \]
For mass \( M_2 \) with velocity \( V_2 \):
\[ KE_2 = \frac{1}{2} M_2 V_2^2 \]

2. **Set the kinetic energies equal**:
Since both masses have the same kinetic energy:
\[ \frac{1}{2} M_1 V_1^2 = \frac{1}{2} M_2 V_2^2 \]
We can simplify this by canceling \( \frac{1}{2} \):
\[ M_1 V_1^2 = M_2 V_2^2 \]

3. **Substitute the relation between velocities**:
We know from the problem statement that \( V_2 = 2V_1 \). Substituting this into the equation gives:
\[ M_1 V_1^2 = M_2 (2V_1)^2 \]
Simplifying the right side:
\[ M_1 V_1^2 = M_2 \cdot 4V_1^2 \]

4. **Cancel \( V_1^2 \)**:
Since \( V_1^2 \) is common on both sides (assuming \( V_1 \neq 0 \)):
\[ M_1 = 4 M_2 \]

5. **Find the momentum of each mass**:
The momentum \( P \) of a mass is given by:
\[ P = m v \]
Therefore, the momentum of \( M_1 \) is:
\[ P_1 = M_1 V_1 \]
And the momentum of \( M_2 \) is:
\[ P_2 = M_2 V_2 = M_2 (2V_1) = 2 M_2 V_1 \]

6. **Calculate the ratio of momenta**:
Now we can find the ratio of their momenta:
\[ \frac{P_1}{P_2} = \frac{M_1 V_1}{2 M_2 V_1} \]
Cancelling \( V_1 \) from the numerator and denominator:
\[ \frac{P_1}{P_2} = \frac{M_1}{2 M_2} \]
Substituting \( M_1 = 4 M_2 \):
\[ \frac{P_1}{P_2} = \frac{4 M_2}{2 M_2} = \frac{4}{2} = 2 \]

### Final Answer:
The ratio of their momenta \( \frac{P_1}{P_2} \) is \( 2 \).