Two identical beads each of mass 'm' can slide freely on a ring of mass M lying on smooth horizontal surface. The masses are brought to diametrically opposite positions and both are imparted with same tangential speed v in same direction. Find the kinetic energy of the ring when beads are about to collide.

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the System
We have two identical beads each of mass \( m \) on a ring of mass \( M \). The beads are initially at diametrically opposite positions on the ring and are given the same tangential speed \( v \) in the same direction.

### Step 2: Analyze Initial Momentum
Initially, the total momentum of the system (both beads and the ring) is given by:
\[ P_{\text{initial}} = mv + mv = 2mv \]
The ring is initially at rest, so its initial momentum is zero.

### Step 3: Apply Conservation of Momentum
Since there are no external forces acting on the system, the total momentum of the system must be conserved. Let \( V_r \) be the final velocity of the ring when the beads are about to collide. The final momentum of the system can be expressed as:
\[ P_{\text{final}} = 2mV_b + MV_r \]
where \( V_b \) is the final velocity of the beads.

### Step 4: Determine the Final Velocities
When the beads are about to collide, they will have some velocity \( V_b \) in the direction of their initial motion. The momentum conservation equation becomes:
\[ 2mv = 2mV_b + MV_r \]

### Step 5: Relate the Velocities of Beads and Ring
Since the beads are moving on the ring, their motion will affect the ring's motion. The beads will exert a force on the ring, causing it to move. By symmetry and conservation of momentum, we can assume:
\[ V_b = V_r \]
Thus, we can rewrite the momentum conservation equation as:
\[ 2mv = 2mV_r + MV_r \]

### Step 6: Solve for the Ring's Velocity
Rearranging the momentum equation gives:
\[ 2mv = V_r(2m + M) \]
Solving for \( V_r \):
\[ V_r = \frac{2mv}{2m + M} \]

### Step 7: Calculate the Kinetic Energy of the Ring
The kinetic energy \( KE \) of the ring is given by:
\[ KE = \frac{1}{2}MV_r^2 \]
Substituting \( V_r \) from the previous step:
\[ KE = \frac{1}{2}M\left(\frac{2mv}{2m + M}\right)^2 \]
Calculating this gives:
\[ KE = \frac{1}{2}M \cdot \frac{4m^2v^2}{(2m + M)^2} \]
\[ KE = \frac{2Mm^2v^2}{(2m + M)^2} \]

### Final Answer
The kinetic energy of the ring when the beads are about to collide is:
\[ KE = \frac{2Mm^2v^2}{(2m + M)^2} \]

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