A bag of mass `M` hangs by a long massless rope. A bullet of mass in, moving horizontally with velocity `u`, is caught in the bag. Then for the combined (bag `+` bullet) system, just after collision

Options

A. momentum is `"mu"M//(M+m)`
B. kinetic energy is `"mu"^2//2`
C. momentum is `"mu"(M+m)//M`
D. kinetic energy is `m^(2)u^(2)//2(M+m)`

Explanation

To solve the problem step by step, we will use the principle of conservation of momentum and the formula for kinetic energy.

### Step 1: Identify the masses and initial velocities
- Let the mass of the bag be \( M \).
- Let the mass of the bullet be \( m \).
- The initial velocity of the bullet is \( u \) (horizontal).
- The initial velocity of the bag is \( 0 \) (since it is hanging and at rest).

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

The total initial momentum (before the collision) is:
\[ \text{Initial Momentum} = \text{Momentum of bullet} + \text{Momentum of bag} = m \cdot u + M \cdot 0 = m \cdot u \]

Let \( V \) be the velocity of the combined system (bag + bullet) just after the collision. The total mass after the collision is \( M + m \).

The total momentum after the collision is:
\[ \text{Final Momentum} = (M + m) \cdot V \]

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

### Step 3: Solve for the velocity \( V \)
Rearranging the equation to solve for \( V \):
\[ V = \frac{m \cdot u}{M + m} \]

### Step 4: Calculate the kinetic energy of the system after the collision
The kinetic energy \( KE \) of the combined system after the collision can be calculated using the formula:
\[ KE = \frac{1}{2} \cdot \text{Total Mass} \cdot V^2 \]
Substituting the total mass and the expression for \( V \):
\[ KE = \frac{1}{2} \cdot (M + m) \cdot \left(\frac{m \cdot u}{M + m}\right)^2 \]

### Step 5: Simplify the kinetic energy expression
Now, simplifying the expression:
\[ KE = \frac{1}{2} \cdot (M + m) \cdot \frac{m^2 \cdot u^2}{(M + m)^2} \]
\[ KE = \frac{1}{2} \cdot \frac{m^2 \cdot u^2}{M + m} \]

### Conclusion
Thus, the kinetic energy of the system just after the collision is:
\[ KE = \frac{m^2 \cdot u^2}{2(M + m)} \]