To solve the problem step by step, we will analyze the collision of the three identical balls and apply the principles of conservation of momentum and kinetic energy.
### Step 1: Understand the Initial Conditions
- We have three identical balls (A, B, and C), each with a mass \( m = 0.1 \, \text{kg} \).
- Ball A has an initial velocity of \( v_A = 10\sqrt{3} \, \text{m/s} \).
- Balls B and C are at rest and in contact with each other, positioned such that their centers are on a line perpendicular to the initial velocity of ball A.
### Step 2: Analyze the Collision Geometry
- When ball A collides with balls B and C, it is aimed directly at the contact point between B and C.
- The centers of the balls form an equilateral triangle at the moment of collision, with angles of \( 60^\circ \) between the lines connecting the centers.
- The angles \( \alpha \) at which balls B and C will move after the collision are \( 30^\circ \) each.
### Step 3: Apply Conservation of Momentum
- The initial momentum of the system is entirely due to ball A:
\[
p_{\text{initial}} = m v_A = 0.1 \times 10\sqrt{3} = 1\sqrt{3} \, \text{kg m/s}
\]
- After the collision, ball A comes to rest, so its momentum is \( 0 \).
- Let the velocities of balls B and C after the collision be \( v_1 \) (since they are identical). The momentum of balls B and C can be expressed as:
\[
p_{\text{final}} = m v_1 \cos(30^\circ) + m v_1 \cos(30^\circ = 2m v_1 \cos(30^\circ)
\]
- Setting the initial momentum equal to the final momentum:
\[
1\sqrt{3} = 2(0.1)v_1 \left(\frac{\sqrt{3}}{2}\right)
\]
### Step 4: Solve for \( v_1 \)
- Simplifying the equation:
\[
1\sqrt{3} = 0.1v_1\sqrt{3}
\]
- Dividing both sides by \( \sqrt{3} \):
\[
1 = 0.1v_1
\]
- Thus, we find:
\[
v_1 = \frac{1}{0.1} = 10 \, \text{m/s}
\]
### Step 5: Calculate Kinetic Energy Before and After Collision
- **Initial Kinetic Energy (KE_initial)**:
\[
KE_{\text{initial}} = \frac{1}{2} m v_A^2 = \frac{1}{2} \times 0.1 \times (10\sqrt{3})^2 = \frac{1}{2} \times 0.1 \times 300 = 15 \, \text{J}
\]
- **Final Kinetic Energy (KE_final)**:
\[
KE_{\text{final}} = \frac{1}{2} m v_1^2 + \frac{1}{2} m v_1^2 = 2 \times \frac{1}{2} m v_1^2 = m v_1^2 = 0.1 \times (10)^2 = 10 \, \text{J}
\]
### Step 6: Determine the Nature of the Collision
- The change in kinetic energy is:
\[
\Delta KE = KE_{\text{final}} - KE_{\text{initial}} = 10 \, \text{J} - 15 \, \text{J} = -5 \, \text{J}
\]
- Since the kinetic energy after the collision is less than before, the collision is inelastic.
### Conclusion
- The collision is inelastic, and the change in kinetic energy is \( -5 \, \text{J} \).
