A body of mass `3 kg` moving with a velocity of `4 m//s` towards left collides head on with a body of mass `4 kg` moving in opposite direction with a velocity of `3 m//s`. After collision the two bodies stick together and move with a common velocity which is

Options

A. zero
B. `12 m//s` towards left
C. `12 m//s` towards right
D. `12/7` m//s towards left

Explanation

To solve the problem, we will use the principle of conservation of momentum. Let's break down the steps:

### Step 1: Identify the masses and velocities
- Mass of the first body (m1) = 3 kg
- Velocity of the first body (u1) = -4 m/s (towards left, we take left as negative)
- Mass of the second body (m2) = 4 kg
- Velocity of the second body (u2) = 3 m/s (towards right, we take right as positive)

### Step 2: Write the conservation of momentum equation
According to the conservation of momentum:
\[ m_1 u_1 + m_2 u_2 = (m_1 + m_2) v \]
where \(v\) is the common velocity after the collision.

### Step 3: Substitute the values into the equation
Substituting the known values into the equation:
\[ 3 \cdot (-4) + 4 \cdot 3 = (3 + 4) v \]

### Step 4: Calculate the left-hand side
Calculating the left-hand side:
\[ -12 + 12 = 0 \]

### Step 5: Set the equation to find the common velocity
Now, we have:
\[ 0 = 7v \]

### Step 6: Solve for v
Dividing both sides by 7:
\[ v = 0 \]

### Conclusion
The common velocity after the collision is \(0 \, \text{m/s}\).

### Final Answer
The two bodies stick together and move with a common velocity of \(0 \, \text{m/s}\).

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