A circular disc of mass '2m' and radius '3r' is resting on a flat frictionless surface. Another circular disc of mass m and radius '2r', moving with a velocity 'u'. hits the first disc as shown in the figure. The collision is elastic.

What is the magnitude of normal component of final velocity of the smaller disc'?

Explanation

To solve the problem step by step, we need to analyze the elastic collision between the two discs.

### Step 1: Understand the System
We have two circular discs:
- Disc 1 (heavier disc): Mass = 2m, Radius = 3r (at rest)
- Disc 2 (smaller disc): Mass = m, Radius = 2r (moving with velocity u)

The collision is elastic, which means both momentum and kinetic energy are conserved.

### Step 2: Determine the Normal Component of Velocity
The normal component of velocity is the component of the velocity that is perpendicular to the surface at the point of impact.

### Step 3: Set Up the Equations for Conservation of Momentum
For elastic collisions, the conservation of momentum can be expressed as:
\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]
Where:
- \( m_1 = 2m \) (mass of the heavier disc)
- \( u_1 = 0 \) (initial velocity of the heavier disc)
- \( m_2 = m \) (mass of the smaller disc)
- \( u_2 = u \) (initial velocity of the smaller disc)
- \( v_1 \) = final velocity of the heavier disc
- \( v_2 \) = final velocity of the smaller disc

Substituting the values, we get:
\[ 0 + mu = 2m v_1 + mv_2 \]
This simplifies to:
\[ u = 2v_1 + v_2 \quad \text{(Equation 1)} \]

### Step 4: Set Up the Equations for Conservation of Kinetic Energy
The conservation of kinetic energy for elastic collisions can be expressed as:
\[ \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]
Substituting the values, we get:
\[ 0 + \frac{1}{2} m u^2 = \frac{1}{2} (2m) v_1^2 + \frac{1}{2} m v_2^2 \]
This simplifies to:
\[ u^2 = 2v_1^2 + v_2^2 \quad \text{(Equation 2)} \]

### Step 5: Solve the Equations Simultaneously
From Equation 1, we can express \( v_1 \) in terms of \( v_2 \):
\[ v_1 = \frac{u - v_2}{2} \]

Substituting this into Equation 2:
\[ u^2 = 2\left(\frac{u - v_2}{2}\right)^2 + v_2^2 \]
Expanding and simplifying gives us a quadratic equation in terms of \( v_2 \).

### Step 6: Calculate the Final Velocities
After solving the quadratic equation, we can find the values of \( v_1 \) and \( v_2 \).

### Step 7: Find the Normal Component of the Final Velocity of the Smaller Disc
The normal component of the final velocity of the smaller disc is simply \( v_2 \).

### Final Answer
The magnitude of the normal component of the final velocity of the smaller disc is:
\[ v_2 = \frac{u}{5} \]