A `6 kg` mass collides with a body at rest. After the collision, they travel together with a velocity one third the velocity of `6 kg` mass. The mass of the second body is

Options

A. 6 kg
B. 3 kg
C. 12 kg
D. 18 kg

Explanation

To solve the problem of a `6 kg` mass colliding with a body at rest and then moving together with a velocity that is one third of the original mass's velocity, we can use the principle of conservation of momentum. Here’s the step-by-step solution:

### Step 1: Define the variables
Let:
- Mass of the first body (m1) = 6 kg
- Mass of the second body (m2) = M (unknown)
- Initial velocity of the first body (u1) = V (unknown)
- Initial velocity of the second body (u2) = 0 (since it is at rest)
- Final velocity of both bodies after the collision (v) = V/3

### Step 2: Write the momentum conservation equation
According to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.

Initial momentum (before collision):
\[ P_{initial} = m_1 \cdot u_1 + m_2 \cdot u_2 \]
\[ P_{initial} = 6 \cdot V + M \cdot 0 = 6V \]

Final momentum (after collision):
\[ P_{final} = (m_1 + m_2) \cdot v \]
\[ P_{final} = (6 + M) \cdot \left(\frac{V}{3}\right) \]

### Step 3: Set the initial momentum equal to the final momentum
\[ 6V = (6 + M) \cdot \left(\frac{V}{3}\right) \]

### Step 4: Simplify the equation
To eliminate V from both sides (assuming V ≠ 0):
\[ 6 = \frac{(6 + M)}{3} \]

### Step 5: Multiply both sides by 3
\[ 18 = 6 + M \]

### Step 6: Solve for M
Subtract 6 from both sides:
\[ M = 18 - 6 \]
\[ M = 12 \]

### Conclusion
The mass of the second body is **12 kg**.

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