A bullet of mass A and velocity B is fired into a wooden block of mass C. If the bullet geta embedded in the wooden block.then velocity of wooden block after collision

Options

A. ` (A+B)/(AC)`
B. ` (A+C)/(B+C)`
C. ` (AC)/(B+C)`
D. ` (AB)/(A+C)`

Answer

To solve the problem of finding the velocity of the wooden block after the bullet embeds into it, we can use the principle of conservation of momentum. Here’s a step-by-step solution:

### Step 1: Understand the System
We have:
- A bullet of mass \( A \) and velocity \( B \).
- A wooden block of mass \( C \) initially at rest.

### Step 2: Write the Initial Momentum
Before the collision, the momentum of the system consists of the momentum of the bullet and the momentum of the wooden block. The wooden block is at rest, so its initial momentum is zero.

\[ \text{Initial Momentum} = \text{Momentum of Bullet} + \text{Momentum of Block} = A \times B + C \times 0 = A \times B \]

### Step 3: Write the Final Momentum
After the bullet embeds into the wooden block, the total mass of the system becomes \( A + C \) and the final velocity of the combined system is \( V_f \).

\[ \text{Final Momentum} = (\text{Mass of Bullet} + \text{Mass of Block}) \times \text{Final Velocity} = (A + C) \times V_f \]

### Step 4: Apply Conservation of Momentum
According to the law of conservation of momentum, the initial momentum of the system must equal the final momentum of the system:

\[ A \times B = (A + C) \times V_f \]

### Step 5: Solve for Final Velocity \( V_f \)
Rearranging the equation to solve for \( V_f \):

\[ V_f = \frac{A \times B}{A + C} \]

### Final Answer
The velocity of the wooden block after the collision is:

\[ V_f = \frac{A \times B}{A + C} \]

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