Find the loss of KE of the two particles of mass `m_(1)=3 kg & m_(2)=6 kg`, moving towards each other with speed `u_(1)=5 m//s, u_(2)=10 m//s` respectively. The coefficient of restitution of collision of the particles is e = 0.5.

Answer

To find the loss of kinetic energy during the collision of two particles, we will follow these steps:

### Step 1: Identify the given values
- Mass of particle 1, \( m_1 = 3 \, \text{kg} \)
- Mass of particle 2, \( m_2 = 6 \, \text{kg} \)
- Initial velocity of particle 1, \( u_1 = 5 \, \text{m/s} \)
- Initial velocity of particle 2, \( u_2 = 10 \, \text{m/s} \)
- Coefficient of restitution, \( e = 0.5 \)

### Step 2: Calculate the initial kinetic energy (KE_initial)
The initial kinetic energy of the two particles can be calculated using the formula:
\[ KE_{\text{initial}} = \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 \]
Substituting the values:
\[ KE_{\text{initial}} = \frac{1}{2} \times 3 \times (5)^2 + \frac{1}{2} \times 6 \times (10)^2 \]
Calculating each term:
\[ = \frac{1}{2} \times 3 \times 25 + \frac{1}{2} \times 6 \times 100 \]
\[ = \frac{75}{2} + \frac{600}{2} \]
\[ = 37.5 + 300 = 337.5 \, \text{J} \]

### Step 3: Apply the conservation of momentum
Using the conservation of momentum before and after the collision:
\[ m_1 u_1 + m_2 (-u_2) = m_1 v_1 + m_2 v_2 \]
Substituting the values:
\[ 3 \times 5 - 6 \times 10 = 3 v_1 + 6 v_2 \]
\[ 15 - 60 = 3 v_1 + 6 v_2 \]
\[ -45 = 3 v_1 + 6 v_2 \quad \text{(Equation 1)} \]

### Step 4: Use the coefficient of restitution
The coefficient of restitution is defined as:
\[ e = \frac{v_2 - v_1}{u_1 + u_2} \]
Substituting the known values:
\[ 0.5 = \frac{v_2 - v_1}{5 + 10} \]
\[ 0.5 = \frac{v_2 - v_1}{15} \]
Multiplying both sides by 15:
\[ 7.5 = v_2 - v_1 \quad \text{(Equation 2)} \]

### Step 5: Solve the equations
From Equation 2:
\[ v_2 = v_1 + 7.5 \]
Substituting \( v_2 \) in Equation 1:
\[ -45 = 3 v_1 + 6(v_1 + 7.5) \]
Expanding:
\[ -45 = 3 v_1 + 6 v_1 + 45 \]
Combining like terms:
\[ -45 = 9 v_1 + 45 \]
Subtracting 45 from both sides:
\[ -90 = 9 v_1 \]
Dividing by 9:
\[ v_1 = -10 \, \text{m/s} \]
Now substituting back to find \( v_2 \):
\[ v_2 = -10 + 7.5 = -2.5 \, \text{m/s} \]

### Step 6: Calculate the final kinetic energy (KE_final)
Now, we can calculate the final kinetic energy:
\[ KE_{\text{final}} = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]
Substituting the values:
\[ KE_{\text{final}} = \frac{1}{2} \times 3 \times (-10)^2 + \frac{1}{2} \times 6 \times (-2.5)^2 \]
Calculating each term:
\[ = \frac{1}{2} \times 3 \times 100 + \frac{1}{2} \times 6 \times 6.25 \]
\[ = 150 + 18.75 = 168.75 \, \text{J} \]

### Step 7: Calculate the loss of kinetic energy
The loss of kinetic energy is given by:
\[ \Delta KE = KE_{\text{initial}} - KE_{\text{final}} \]
Substituting the values:
\[ \Delta KE = 337.5 - 168.75 = 168.75 \, \text{J} \]

### Final Answer
The loss of kinetic energy during the collision is **168.75 J**.

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