A particle of mass `2 kg` moving with a velocity of `3 m//s` is acted upon by a force which changes its direction of motion by an angle of `90^(@)` without changing its speed. What is the magnitude of impulse experienced by the particle?

Options

A. `6Ns`
B. `2Ns`
C. `3sqrt(2)Ns`
D. `6sqrt(2)Ns`

Explanation

To solve the problem step by step, we will calculate the impulse experienced by the particle as it changes its direction of motion by 90 degrees without changing its speed.

### Step 1: Understand the Initial and Final Velocities
The particle has an initial mass \( m = 2 \, \text{kg} \) and is moving with an initial velocity \( v_i = 3 \, \text{m/s} \) in the positive x-direction. After the force acts on it, the particle's direction changes by \( 90^\circ \), and it moves in the positive y-direction with the same speed.

### Step 2: Calculate the Initial Momentum
The initial momentum \( p_i \) of the particle can be calculated using the formula:
\[ p_i = m \cdot v_i \]
Substituting the given values:
\[ p_i = 2 \, \text{kg} \cdot 3 \, \text{m/s} = 6 \, \text{kg m/s} \]

### Step 3: Calculate the Final Momentum
The final velocity \( v_f \) is also \( 3 \, \text{m/s} \) but in the y-direction. Therefore, the final momentum \( p_f \) is:
\[ p_f = m \cdot v_f = 2 \, \text{kg} \cdot 3 \, \text{m/s} = 6 \, \text{kg m/s} \]
However, since the direction has changed, we need to represent this as a vector:
\[ p_f = 0 \, \hat{i} + 6 \, \hat{j} \, \text{kg m/s} \]

### Step 4: Calculate the Change in Momentum
The change in momentum \( \Delta p \) is given by:
\[ \Delta p = p_f - p_i \]
Substituting the initial and final momentum:
\[ \Delta p = (0 \, \hat{i} + 6 \, \hat{j}) - (6 \, \hat{i} + 0 \, \hat{j}) = -6 \, \hat{i} + 6 \, \hat{j} \, \text{kg m/s} \]

### Step 5: Calculate the Magnitude of the Change in Momentum
The magnitude of the change in momentum (which is equal to the impulse \( I \)) can be calculated using:
\[ |I| = |\Delta p| = \sqrt{(-6)^2 + (6)^2} \]
Calculating this:
\[ |I| = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \, \text{kg m/s} \]

### Conclusion
The magnitude of the impulse experienced by the particle is:
\[ |I| = 6\sqrt{2} \, \text{N s} \]