A particle with linear momentum of magnitude P is subjected to a force `F = Kt ( K gt 0 )` which is directed along the direction of initial momentum. The time after which its linear momentum changes to 3P is

Options

A. `sqrt((2P)/(K))`
B. `2sqrt((P)/(K))`
C. `sqrt((2K)/(P))`
D. `2sqrt((2K)/(P))`

Answer

To solve the problem, we will use the relationship between force, momentum, and time. The force acting on the particle is given as \( F = Kt \), where \( K > 0 \). We need to find the time \( t \) when the linear momentum changes from \( P \) to \( 3P \).

### Step-by-step Solution:

1. **Use Newton's Second Law**:
According to Newton's second law, the force \( F \) is equal to the rate of change of momentum \( \frac{dp}{dt} \):
\[ F = \frac{dp}{dt} \]

2. **Substitute the Given Force**:
Substitute \( F = Kt \) into the equation:
\[ Kt = \frac{dp}{dt} \]

3. **Rearranging the Equation**:
Rearranging gives us:
\[ dp = Kt \, dt \]

4. **Integrate Both Sides**:
We will integrate both sides. The left side will be integrated from the initial momentum \( P \) to the final momentum \( 3P \), and the right side will be integrated from \( t = 0 \) to \( t \):
\[ \int_{P}^{3P} dp = \int_{0}^{t} Kt \, dt \]

5. **Calculate the Left Side**:
The left side integrates to:
\[ [p]_{P}^{3P} = 3P - P = 2P \]

6. **Calculate the Right Side**:
The right side integrates as follows:
\[ \int_{0}^{t} Kt \, dt = K \int_{0}^{t} t \, dt = K \left[ \frac{t^2}{2} \right]_{0}^{t} = K \frac{t^2}{2} \]

7. **Set the Integrals Equal**:
Now we set the results of the integrals equal to each other:
\[ 2P = K \frac{t^2}{2} \]

8. **Solve for \( t^2 \)**:
Rearranging gives:
\[ t^2 = \frac{4P}{K} \]

9. **Take the Square Root**:
Taking the square root to find \( t \):
\[ t = \sqrt{\frac{4P}{K}} = \frac{2\sqrt{P}}{\sqrt{K}} \]

### Final Answer:
The time after which the linear momentum changes to \( 3P \) is:
\[ t = \frac{2\sqrt{P}}{\sqrt{K}} \]