To find the loss of kinetic energy (KE) of two particles during a collision, we will follow these steps:
### Step 1: Calculate Initial Kinetic Energy
The initial kinetic energy (KE_initial) of the two particles can be calculated using the formula:
\[
KE = \frac{1}{2} m u^2
\]
where \(m\) is the mass and \(u\) is the initial velocity.
For particle 1 (mass \(m_1 = 3 \, \text{kg}\), velocity \(u_1 = 5 \, \text{m/s}\)):
\[
KE_1 = \frac{1}{2} \times 3 \times (5)^2 = \frac{1}{2} \times 3 \times 25 = 37.5 \, \text{J}
\]
For particle 2 (mass \(m_2 = 6 \, \text{kg}\), velocity \(u_2 = 10 \, \text{m/s}\)):
\[
KE_2 = \frac{1}{2} \times 6 \times (10)^2 = \frac{1}{2} \times 6 \times 100 = 300 \, \text{J}
\]
Thus, the total initial kinetic energy is:
\[
KE_{\text{initial}} = KE_1 + KE_2 = 37.5 + 300 = 337.5 \, \text{J}
\]
### Step 2: Apply Conservation of Momentum
The total momentum before and after the collision must be conserved. The initial momentum \(p_{\text{initial}}\) is given by:
\[
p_{\text{initial}} = m_1 u_1 + m_2 (-u_2) = 3 \times 5 + 6 \times (-10) = 15 - 60 = -45 \, \text{kg m/s}
\]
Let \(V_1\) and \(V_2\) be the final velocities of masses \(m_1\) and \(m_2\) respectively. The final momentum \(p_{\text{final}}\) is:
\[
p_{\text{final}} = m_1 V_1 + m_2 V_2
\]
Setting \(p_{\text{initial}} = p_{\text{final}}\):
\[
3 V_1 + 6 V_2 = -45 \quad \text{(Equation 1)}
\]
### Step 3: Use Coefficient of Restitution
The coefficient of restitution \(e\) relates the velocities before and after the collision:
\[
e = \frac{V_2 - V_1}{u_1 + u_2}
\]
Given \(e = 0.5\), \(u_1 = 5\), and \(u_2 = 10\):
\[
0.5 = \frac{V_2 - V_1}{5 + 10} = \frac{V_2 - V_1}{15}
\]
Thus,
\[
V_2 - V_1 = 0.5 \times 15 = 7.5 \quad \text{(Equation 2)}
\]
### Step 4: Solve the Equations
Now we have two equations:
1. \(3 V_1 + 6 V_2 = -45\)
2. \(V_2 - V_1 = 7.5\)
From Equation 2, we can express \(V_2\) in terms of \(V_1\):
\[
V_2 = V_1 + 7.5
\]
Substituting into Equation 1:
\[
3 V_1 + 6(V_1 + 7.5) = -45
\]
\[
3 V_1 + 6 V_1 + 45 = -45
\]
\[
9 V_1 + 45 = -45
\]
\[
9 V_1 = -90 \quad \Rightarrow \quad V_1 = -10 \, \text{m/s}
\]
Substituting back to find \(V_2\):
\[
V_2 = -10 + 7.5 = -2.5 \, \text{m/s}
\]
### Step 5: Calculate Final Kinetic Energy
Now we can calculate the final kinetic energy (KE_final):
For particle 1:
\[
KE_{1,\text{final}} = \frac{1}{2} \times 3 \times (-10)^2 = \frac{1}{2} \times 3 \times 100 = 150 \, \text{J}
\]
For particle 2:
\[
KE_{2,\text{final}} = \frac{1}{2} \times 6 \times (-2.5)^2 = \frac{1}{2} \times 6 \times 6.25 = 18.75 \, \text{J}
\]
Thus, the total final kinetic energy is:
\[
KE_{\text{final}} = KE_{1,\text{final}} + KE_{2,\text{final}} = 150 + 18.75 = 168.75 \, \text{J}
\]
### Step 6: Calculate Loss of Kinetic Energy
The loss of kinetic energy is given by:
\[
\Delta KE = KE_{\text{initial}} - KE_{\text{final}} = 337.5 - 168.75 = 168.75 \, \text{J}
\]
### Final Answer
The loss of kinetic energy during the collision is approximately:
\[
\Delta KE \approx 169 \, \text{J}
\]
