A spherical ball of mass `1 kg` moving with a uniform speed of `1 m//s` collides symmetrically with two identical spherical balls of mass `1 kg` each at rest touching each other. If the two balls move with `0.5 m//s` in two directions at the same angle of `60^(@)` with the direction of the first ball, the loss of kinetic energy on account of the collision is :-

Options

A. `0.125 J`
B. `0.5 J`
C. `1.0 J`
D. `0.75 J`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the Initial Kinetic Energy (KE_initial)
The initial kinetic energy of the system can be calculated using the formula:
\[ KE_{\text{initial}} = \frac{1}{2} m u^2 \]
where:
- \(m = 1 \, \text{kg}\) (mass of the moving ball)
- \(u = 1 \, \text{m/s}\) (initial speed of the moving ball)

Substituting the values:
\[ KE_{\text{initial}} = \frac{1}{2} \times 1 \times (1)^2 = \frac{1}{2} \times 1 = 0.5 \, \text{J} \]

### Step 2: Calculate the Final Kinetic Energy (KE_final)
After the collision, the two balls move with a speed of \(0.5 \, \text{m/s}\) at an angle of \(60^\circ\) to the direction of the first ball. We need to calculate the final kinetic energy of all three balls.

1. **Kinetic Energy of the First Ball (after collision)**:
- Let’s assume the first ball continues to move with a velocity \(v\) (we will find this later).
- The kinetic energy of the first ball:
\[ KE_{\text{first}} = \frac{1}{2} m v^2 \]

2. **Kinetic Energy of the Two Balls**:
- Each of the two balls has a velocity of \(0.5 \, \text{m/s}\).
- The total kinetic energy for the two balls:
\[ KE_{\text{two}} = 2 \times \left(\frac{1}{2} m (0.5)^2\right) = 2 \times \left(\frac{1}{2} \times 1 \times (0.5)^2\right) = 2 \times \left(\frac{1}{2} \times 1 \times 0.25\right) = 2 \times 0.125 = 0.25 \, \text{J} \]

3. **Total Final Kinetic Energy**:
- Now we need to find the velocity \(v\) of the first ball after the collision. By conservation of momentum in the x-direction:
\[ m u = m v + 2 \times m \left(0.5 \cos(60^\circ)\right) \]
Since \( \cos(60^\circ) = \frac{1}{2} \):
\[ 1 = v + 2 \times 1 \times 0.5 \times \frac{1}{2} \]
\[ 1 = v + 0.5 \]
\[ v = 0.5 \, \text{m/s} \]

4. **Kinetic Energy of the First Ball**:
\[ KE_{\text{first}} = \frac{1}{2} \times 1 \times (0.5)^2 = \frac{1}{2} \times 1 \times 0.25 = 0.125 \, \text{J} \]

5. **Total Final Kinetic Energy**:
\[ KE_{\text{final}} = KE_{\text{first}} + KE_{\text{two}} = 0.125 + 0.25 = 0.375 \, \text{J} \]

### Step 3: Calculate the Loss of Kinetic Energy
The loss of kinetic energy due to the collision is given by:
\[ \text{Loss in KE} = KE_{\text{initial}} - KE_{\text{final}} \]
Substituting the values:
\[ \text{Loss in KE} = 0.5 \, \text{J} - 0.375 \, \text{J} = 0.125 \, \text{J} \]

### Final Answer:
The loss of kinetic energy on account of the collision is \(0.125 \, \text{J}\).

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