A ball of mass 4 kg moving on a smooth horizontal surface makes an elastic collision with another ball of mass m at rest in the line of motion of first ball. If after collision first ball moves in the same direction with one fourth of its velocity before collision, then mass of second ball is

Options

A. 4 kg
B. 4.4 kg
C. 2.4 kg
D. 2 kg

Answer

To solve the problem step by step, we will use the principles of conservation of momentum and the characteristics of elastic collisions.

### Step 1: Understand the Problem
We have two balls:
- Ball 1 (mass \( m_1 = 4 \, \text{kg} \)) is moving with an initial velocity \( V \).
- Ball 2 (mass \( m_2 = m \)) is at rest.

After the collision, Ball 1 moves with a velocity of \( \frac{V}{4} \).

### Step 2: Use the Conservation of Momentum
The law of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision.

**Before the collision:**
- Momentum of Ball 1 = \( 4V \)
- Momentum of Ball 2 = \( 0 \)

Total initial momentum:
\[ P_{initial} = 4V + 0 = 4V \]

**After the collision:**
- Momentum of Ball 1 = \( 4 \times \frac{V}{4} = V \)
- Momentum of Ball 2 = \( m \cdot u \) (where \( u \) is the final velocity of Ball 2)

Total final momentum:
\[ P_{final} = V + mu \]

Setting the initial momentum equal to the final momentum:
\[ 4V = V + mu \]

### Step 3: Solve for \( mu \)
Rearranging the equation gives:
\[ 4V - V = mu \implies 3V = mu \]
\[ u = \frac{3V}{m} \]

### Step 4: Use the Elastic Collision Condition
In an elastic collision, the coefficient of restitution \( e \) is defined as:
\[ e = \frac{\text{velocity of separation}}{\text{velocity of approach}} = 1 \]

**Velocity of separation:**
- For Ball 1: \( \frac{V}{4} \)
- For Ball 2: \( u \)

The velocity of separation is:
\[ u - \frac{V}{4} \]

**Velocity of approach:**
- For Ball 1: \( V \)
- For Ball 2: \( 0 \)

The velocity of approach is:
\[ V - 0 = V \]

Setting the coefficient of restitution to 1:
\[ 1 = \frac{u - \frac{V}{4}}{V} \]

### Step 5: Solve for \( u \)
Rearranging gives:
\[ u - \frac{V}{4} = V \]
\[ u = V + \frac{V}{4} = \frac{5V}{4} \]

### Step 6: Substitute \( u \) into the Momentum Equation
Now, substitute \( u \) back into the momentum equation:
\[ 3V = m \cdot \frac{5V}{4} \]

### Step 7: Solve for \( m \)
Cancelling \( V \) (assuming \( V \neq 0 \)):
\[ 3 = m \cdot \frac{5}{4} \]
\[ m = \frac{3 \cdot 4}{5} = \frac{12}{5} = 2.4 \, \text{kg} \]

### Conclusion
The mass of the second ball is \( 2.4 \, \text{kg} \).

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