Two particles of masses `m_(1)=6 kg & m_(2)=4 kg` move towards each other with `u_(1)=3 m//sec & u_(2)=6 m//sec` respectively. If they collide elastically, find the impulse of the particles.

Answer

To solve the problem step by step, we will follow the principles of conservation of momentum and the coefficient of restitution for elastic collisions.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of particle 1, \( m_1 = 6 \, \text{kg} \)
- Mass of particle 2, \( m_2 = 4 \, \text{kg} \)
- Initial velocity of particle 1, \( u_1 = 3 \, \text{m/s} \) (to the right)
- Initial velocity of particle 2, \( u_2 = -6 \, \text{m/s} \) (to the left, hence negative)

2. **Use Conservation of Momentum:**
The total momentum before the collision must equal the total momentum after the collision. Therefore, we can write:
\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]
Substituting the known values:
\[ 6 \times 3 + 4 \times (-6) = 6 v_1 + 4 v_2 \]
Simplifying this gives:
\[ 18 - 24 = 6 v_1 + 4 v_2 \]
\[ -6 = 6 v_1 + 4 v_2 \quad \text{(Equation 1)} \]

3. **Use the Coefficient of Restitution:**
For an elastic collision, the coefficient of restitution \( e = 1 \). The formula for the coefficient of restitution is:
\[ e = \frac{\text{Relative velocity after collision}}{\text{Relative velocity before collision}} \]
Thus,
\[ 1 = \frac{v_2 - v_1}{u_1 - u_2} \]
Substituting the values:
\[ 1 = \frac{v_2 - v_1}{3 - (-6)} \]
\[ 1 = \frac{v_2 - v_1}{9} \]
Therefore,
\[ v_2 - v_1 = 9 \quad \text{(Equation 2)} \]

4. **Solve the System of Equations:**
We now have two equations:
- Equation 1: \( 6 v_1 + 4 v_2 = -6 \)
- Equation 2: \( v_2 - v_1 = 9 \)

From Equation 2, we can express \( v_2 \) in terms of \( v_1 \):
\[ v_2 = v_1 + 9 \]
Substitute \( v_2 \) into Equation 1:
\[ 6 v_1 + 4(v_1 + 9) = -6 \]
Simplifying this:
\[ 6 v_1 + 4 v_1 + 36 = -6 \]
\[ 10 v_1 + 36 = -6 \]
\[ 10 v_1 = -42 \]
\[ v_1 = -4.2 \, \text{m/s} \]

Now substituting \( v_1 \) back to find \( v_2 \):
\[ v_2 = -4.2 + 9 = 4.8 \, \text{m/s} \]

5. **Calculate the Impulse:**
The impulse \( J \) experienced by each particle can be calculated using the impulse-momentum theorem:
\[ J = \Delta p = m(v - u) \]
For particle 1:
\[ J_1 = m_1(v_1 - u_1) = 6(-4.2 - 3) = 6(-7.2) = -43.2 \, \text{Ns} \]
The magnitude of the impulse is:
\[ |J_1| = 43.2 \, \text{Ns} \]

For particle 2:
\[ J_2 = m_2(v_2 - u_2) = 4(4.8 - (-6)) = 4(4.8 + 6) = 4(10.8) = 43.2 \, \text{Ns} \]

### Final Answer:
The impulse of both particles is \( 43.2 \, \text{Ns} \).