A particle of mass `'m'` moving with a velocity `(3 hat(i)+2hat(j))m//s` collides with stationary mass `'M'` and finally `'m'` moves with a velocity `(-2hat(i)+hat(j))m//s` if `(m)/(M) = (1)/(13)` the velocity of the `M` after collision is?

Options

A. the impulse is `+- m (5 hat(i) + hat(j))`
B. the velocity of the M is `(1/13)(5hat(i) + hat(j))`
C. Both (a) and (b) are wrong
D. Both (a) and (b) are correct

Explanation

To solve the problem, we will use the principle of conservation of momentum. The momentum before the collision must equal the momentum after the collision.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Mass of particle `m` is moving with a velocity \( \vec{v}_1 = (3 \hat{i} + 2 \hat{j}) \, \text{m/s} \).
- Mass `M` is stationary, so its initial velocity \( \vec{v}_2 = 0 \).
- After the collision, the velocity of mass `m` is \( \vec{v}_1' = (-2 \hat{i} + \hat{j}) \, \text{m/s} \).
- The ratio of masses is given as \( \frac{m}{M} = \frac{1}{13} \), which implies \( M = 13m \).

2. **Calculate Initial Momentum**:
- The initial momentum of mass `m` is:
\[ \vec{p}_{\text{initial}} = m \vec{v}_1 + M \vec{v}_2 = m (3 \hat{i} + 2 \hat{j}) + 0 = m (3 \hat{i} + 2 \hat{j}) \]

3. **Calculate Final Momentum**:
- The final momentum of mass `m` after the collision is:
\[ \vec{p}_{\text{final, m}} = m \vec{v}_1' = m (-2 \hat{i} + \hat{j}) \]
- Let the final velocity of mass `M` be \( \vec{v}_2' \). Then the final momentum of mass `M` is:
\[ \vec{p}_{\text{final, M}} = M \vec{v}_2' = 13m \vec{v}_2' \]

4. **Apply Conservation of Momentum**:
- According to the conservation of momentum:
\[ \vec{p}_{\text{initial}} = \vec{p}_{\text{final, m}} + \vec{p}_{\text{final, M}} \]
- Substituting the values:
\[ m (3 \hat{i} + 2 \hat{j}) = m (-2 \hat{i} + \hat{j}) + 13m \vec{v}_2' \]

5. **Simplify the Equation**:
- Dividing the entire equation by `m` (assuming \( m \neq 0 \)):
\[ (3 \hat{i} + 2 \hat{j}) = (-2 \hat{i} + \hat{j}) + 13 \vec{v}_2' \]
- Rearranging gives:
\[ 13 \vec{v}_2' = (3 \hat{i} + 2 \hat{j}) - (-2 \hat{i} + \hat{j}) \]
\[ 13 \vec{v}_2' = (3 + 2) \hat{i} + (2 - 1) \hat{j} = 5 \hat{i} + \hat{j} \]

6. **Solve for \( \vec{v}_2' \)**:
- Now divide by 13:
\[ \vec{v}_2' = \frac{1}{13} (5 \hat{i} + \hat{j}) = \frac{5}{13} \hat{i} + \frac{1}{13} \hat{j} \]

### Final Answer:
The velocity of mass `M` after the collision is:
\[ \vec{v}_2' = \frac{5}{13} \hat{i} + \frac{1}{13} \hat{j} \, \text{m/s} \]