A particle of mass `m` moving with a velocity `(3hati+2hatj)ms^-1` collides with another body of mass M and finally moves with velocity `(-2hati+hatj)ms^-1`. Then during the collision

Explanation

To solve the problem, we will calculate the impulse experienced by the particle during the collision. The impulse can be determined by finding the change in momentum of the particle.

### Step-by-Step Solution:

1. **Identify Initial and Final Velocities:**
- The initial velocity of the particle is given as \( \vec{v_i} = 3 \hat{i} + 2 \hat{j} \, \text{m/s} \).
- The final velocity of the particle after the collision is given as \( \vec{v_f} = -2 \hat{i} + \hat{j} \, \text{m/s} \).

2. **Calculate Initial Momentum:**
- The momentum of the particle before the collision can be calculated using the formula:
\[ \vec{p_i} = m \vec{v_i} = m (3 \hat{i} + 2 \hat{j}) = 3m \hat{i} + 2m \hat{j} \]

3. **Calculate Final Momentum:**
- The momentum of the particle after the collision is:
\[ \vec{p_f} = m \vec{v_f} = m (-2 \hat{i} + \hat{j}) = -2m \hat{i} + m \hat{j} \]

4. **Calculate Change in Momentum:**
- The change in momentum \( \Delta \vec{p} \) is given by:
\[ \Delta \vec{p} = \vec{p_f} - \vec{p_i} \]
- Substituting the values:
\[ \Delta \vec{p} = (-2m \hat{i} + m \hat{j}) - (3m \hat{i} + 2m \hat{j}) \]
- Simplifying this:
\[ \Delta \vec{p} = (-2m - 3m) \hat{i} + (m - 2m) \hat{j} = -5m \hat{i} - m \hat{j} \]

5. **Calculate Impulse:**
- The impulse \( \vec{J} \) experienced by the particle is equal to the change in momentum:
\[ \vec{J} = \Delta \vec{p} = -5m \hat{i} - m \hat{j} \]

6. **Final Answer:**
- Therefore, the impulse during the collision is:
\[ \vec{J} = -5m \hat{i} - m \hat{j} \]