A heavy ball moving with speed `v` collides with a tiny ball. The collision is elastic, then immediately after the impact, the second ball will move with a speed approximately equal to

Options

A. (a) `v`
B. (b) `2v`
C. (c) `v/2`
D. (d) `v/3`

Explanation

To solve the problem of a heavy ball colliding elastically with a tiny ball, we can use the principles of conservation of momentum and conservation of kinetic energy. Here's a step-by-step solution:

### Step 1: Define the masses and initial velocities
Let:
- Mass of the heavy ball = \( M \)
- Mass of the tiny ball = \( m \)
- Initial velocity of the heavy ball = \( v \)
- Initial velocity of the tiny ball = \( 0 \) (since it is at rest)

### Step 2: Apply conservation of momentum
The total momentum before the collision must equal the total momentum after the collision. Therefore, we can write:
\[ M \cdot v + m \cdot 0 = M \cdot v' + m \cdot u \]
where \( v' \) is the final velocity of the heavy ball and \( u \) is the final velocity of the tiny ball.

### Step 3: Apply conservation of kinetic energy
Since the collision is elastic, the total kinetic energy before the collision must equal the total kinetic energy after the collision:
\[ \frac{1}{2} M v^2 + \frac{1}{2} m \cdot 0^2 = \frac{1}{2} M (v')^2 + \frac{1}{2} m u^2 \]
This simplifies to:
\[ \frac{1}{2} M v^2 = \frac{1}{2} M (v')^2 + \frac{1}{2} m u^2 \]

### Step 4: Solve the equations
From the momentum equation:
\[ M v = M v' + m u \quad \text{(1)} \]
From the kinetic energy equation:
\[ M v^2 = M (v')^2 + m u^2 \quad \text{(2)} \]

### Step 5: Analyze the scenario
Given that the mass of the heavy ball \( M \) is much larger than the mass of the tiny ball \( m \), we can make an approximation. After the collision, the heavy ball will continue moving with a velocity close to its initial velocity \( v \), and the tiny ball will gain a significant amount of speed.

### Step 6: Approximate the final speed of the tiny ball
In a perfectly elastic collision where one object is much heavier, the tiny ball will move away with a speed approximately equal to twice the speed of the heavy ball before the collision:
\[ u \approx 2v \]

### Conclusion
Thus, the speed of the tiny ball immediately after the impact will be approximately equal to \( 2v \).

### Final Answer
The second ball will move with a speed approximately equal to \( 2v \).

---