A massive ball moving with speed v collieds with a tiny ball which is intially at rest having a mass very much smaller than the mass of the first ball. The collision is elastic, then immediately after the impact, the second ball will move with a speed approximately equal to :-

Options

A. v
B. 2v
C. v/2
D. `oo`

Answer

To solve the problem of a massive ball colliding elastically with a tiny ball initially at rest, we will use the principles of conservation of momentum and the properties of elastic collisions. Here’s a step-by-step solution:

### Step 1: Define the masses and velocities
Let:
- \( M \) = mass of the massive ball (initially moving)
- \( m \) = mass of the tiny ball (initially at rest)
- \( v \) = initial speed of the massive ball
- \( V_1 \) = final speed of the massive ball after the collision
- \( V_2 \) = final speed of the tiny ball after the collision

### Step 2: Write the conservation of momentum equation
Since momentum is conserved in the collision, we can write:
\[ M \cdot v + m \cdot 0 = M \cdot V_1 + m \cdot V_2 \]
This simplifies to:
\[ M \cdot v = M \cdot V_1 + m \cdot V_2 \quad \text{(1)} \]

### Step 3: Write the equation for elastic collision
For an elastic collision, the relative velocity of separation is equal to the relative velocity of approach. This can be expressed as:
\[ V_2 - V_1 = v \quad \text{(2)} \]

### Step 4: Substitute \( V_1 \) from equation (2) into equation (1)
From equation (2), we can express \( V_1 \) in terms of \( V_2 \):
\[ V_1 = V_2 - v \]
Substituting this into equation (1):
\[ M \cdot v = M \cdot (V_2 - v) + m \cdot V_2 \]
Expanding this gives:
\[ M \cdot v = M \cdot V_2 - M \cdot v + m \cdot V_2 \]
Rearranging the terms:
\[ M \cdot v + M \cdot v = (M + m) \cdot V_2 \]
This simplifies to:
\[ 2M \cdot v = (M + m) \cdot V_2 \quad \text{(3)} \]

### Step 5: Solve for \( V_2 \)
Now we can solve for \( V_2 \):
\[ V_2 = \frac{2M \cdot v}{M + m} \]

### Step 6: Simplify \( V_2 \) using the condition \( m \ll M \)
Since \( m \) is very much smaller than \( M \) (i.e., \( m \ll M \)), we can neglect \( m \) in the denominator:
\[ V_2 \approx \frac{2M \cdot v}{M} = 2v \]

### Conclusion
Thus, the speed of the tiny ball after the collision is approximately:
\[ V_2 \approx 2v \]

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

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