A bag P (mass M) hangs by a long thread and bullet (mass m) comes horizontally with velocity v and gets caught in the bag. Then for the combined (bag + bullet) system

Options

A. momentum is `(m v M)/(M +m)`
B. kinetic energy is `(mv^(2))/(2)`
C. momentum is `(mv(M + m))/(M)`
D. Kinetic energy is `(m^(2)v^(2))/(2(M + m))`

Answer

To solve the problem, we will use the principles of conservation of momentum and kinetic energy. Let's break down the solution step by step.

### Step 1: Understand the System
We have a bag (mass \( M \)) hanging by a thread, and a bullet (mass \( m \)) is moving horizontally with velocity \( v \). When the bullet gets caught in the bag, we need to analyze the combined system of the bag and the bullet.

### Step 2: Apply Conservation of Momentum
According to the law of conservation of momentum, the total momentum before the collision must equal the total momentum after the collision.

**Initial Momentum:**
- The bullet has momentum \( p_{\text{bullet}} = m \cdot v \).
- The bag is at rest, so its momentum \( p_{\text{bag}} = 0 \).

Thus, the total initial momentum \( p_{\text{initial}} \) is:
\[ p_{\text{initial}} = m \cdot v + 0 = m \cdot v \]

**Final Momentum:**
After the bullet is caught in the bag, the total mass becomes \( M + m \), and let \( V \) be the velocity of the combined system. Therefore, the final momentum \( p_{\text{final}} \) is:
\[ p_{\text{final}} = (M + m) \cdot V \]

Setting the initial momentum equal to the final momentum:
\[ m \cdot v = (M + m) \cdot V \]

### Step 3: Solve for Final Velocity \( V \)
Rearranging the equation to find \( V \):
\[ V = \frac{m \cdot v}{M + m} \]

### Step 4: Calculate Kinetic Energy of the Combined System
The kinetic energy of the combined system after the collision can be calculated using the formula:
\[ KE = \frac{1}{2} \cdot (M + m) \cdot V^2 \]

Substituting \( V \) from the previous step:
\[ KE = \frac{1}{2} \cdot (M + m) \cdot \left(\frac{m \cdot v}{M + m}\right)^2 \]

### Step 5: Simplify the Kinetic Energy Expression
\[ KE = \frac{1}{2} \cdot (M + m) \cdot \frac{m^2 \cdot v^2}{(M + m)^2} \]
\[ KE = \frac{1}{2} \cdot \frac{m^2 \cdot v^2}{M + m} \]

### Final Results
1. The final velocity \( V \) of the combined system is:
\[ V = \frac{m \cdot v}{M + m} \]

2. The kinetic energy \( KE \) of the combined system is:
\[ KE = \frac{1}{2} \cdot \frac{m^2 \cdot v^2}{M + m} \]