A `2 kg` ball , moving at `10 m//s` , collides head - on with a `3 kg` ball moving in the opposite direction at `20 m//s`. If the coefficient of restitution is `1//3` , then the energy lost in the collision is

Options

A. `120 J`
B. `240 J`
C. `360 J`
D. `480 J`

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Understand the given data
We have:
- Mass of ball 1 (m1) = 2 kg
- Initial velocity of ball 1 (u1) = 10 m/s (to the right)
- Mass of ball 2 (m2) = 3 kg
- Initial velocity of ball 2 (u2) = -20 m/s (to the left, hence negative)

### Step 2: Apply the conservation of momentum
The total momentum before the collision must equal the total momentum after the collision. Therefore, we can write the equation:

\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]

Substituting the values:

\[ 2 \times 10 + 3 \times (-20) = 2 v_1 + 3 v_2 \]

This simplifies to:

\[ 20 - 60 = 2 v_1 + 3 v_2 \]

\[ -40 = 2 v_1 + 3 v_2 \quad \text{(Equation 1)} \]

### Step 3: Use the coefficient of restitution
The coefficient of restitution (e) is given as \( \frac{1}{3} \). The formula for the coefficient of restitution is:

\[ e = \frac{v_2 - v_1}{u_1 - u_2} \]

Substituting the known values:

\[ \frac{1}{3} = \frac{v_2 - v_1}{10 - (-20)} \]

This simplifies to:

\[ \frac{1}{3} = \frac{v_2 - v_1}{30} \]

Cross-multiplying gives:

\[ v_2 - v_1 = 10 \quad \text{(Equation 2)} \]

### Step 4: Solve the equations simultaneously
Now we have two equations:
1. \( -40 = 2 v_1 + 3 v_2 \)
2. \( v_2 - v_1 = 10 \)

From Equation 2, we can express \( v_2 \) in terms of \( v_1 \):

\[ v_2 = v_1 + 10 \]

Substituting this into Equation 1:

\[ -40 = 2 v_1 + 3(v_1 + 10) \]

Expanding this gives:

\[ -40 = 2 v_1 + 3 v_1 + 30 \]

Combining like terms:

\[ -40 = 5 v_1 + 30 \]

Rearranging gives:

\[ 5 v_1 = -70 \implies v_1 = -14 \, \text{m/s} \]

Now substituting \( v_1 \) back into Equation 2 to find \( v_2 \):

\[ v_2 = -14 + 10 = -4 \, \text{m/s} \]

### Step 5: Calculate initial and final kinetic energies
1. **Initial Kinetic Energy (KE_initial)**:

\[ KE_{initial} = \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 \]

\[ = \frac{1}{2} \times 2 \times 10^2 + \frac{1}{2} \times 3 \times (-20)^2 \]

\[ = 1 \times 100 + 1.5 \times 400 = 100 + 600 = 700 \, \text{J} \]

2. **Final Kinetic Energy (KE_final)**:

\[ KE_{final} = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]

\[ = \frac{1}{2} \times 2 \times (-14)^2 + \frac{1}{2} \times 3 \times (-4)^2 \]

\[ = 1 \times 196 + 1.5 \times 16 = 196 + 24 = 220 \, \text{J} \]

### Step 6: Calculate energy lost
The energy lost in the collision is given by:

\[ \text{Energy lost} = KE_{initial} - KE_{final} \]

\[ = 700 - 220 = 480 \, \text{J} \]

### Final Answer
The energy lost in the collision is **480 Joules**.

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