A shere of mass `m` , moving with velocity `V`, enters a hanging bag of sand and stop. If the mass of the bag is `M` and it is reised by height `h`, then the velocity of the sphere will be

Options

A. `(M+m)/m sqrt(2gh)`
B. `M/msqrt(2gh)`
C. `m/(M+m) sqrt(2gh)`
D. `m/Msqrt(2gh)`

Explanation

To solve the problem step by step, we will use the principles of conservation of linear momentum and conservation of energy.

### Step 1: Understand the problem
We have a sphere of mass \( m \) moving with velocity \( V \) that enters a hanging bag of sand with mass \( M \). After the sphere enters the bag, the combined system rises to a height \( h \). We need to find the velocity of the sphere before it enters the bag.

### Step 2: Apply the conservation of linear momentum
According to the conservation of linear momentum, the initial momentum of the system must equal the final momentum of the system.

- **Initial momentum**: The momentum of the moving sphere is given by:
\[ p_{\text{initial}} = m \cdot V \]

- **Final momentum**: After the sphere enters the bag, the total mass of the system becomes \( m + M \). Let \( v \) be the velocity of the system after the collision. The final momentum is:
\[ p_{\text{final}} = (m + M) \cdot v \]

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

From this, we can express \( v \):
\[ v = \frac{m \cdot V}{m + M} \]

### Step 3: Apply the conservation of energy
Next, we apply the conservation of energy principle. The initial kinetic energy of the system must equal the potential energy at height \( h \).

- **Initial kinetic energy**: The kinetic energy of the system after the collision is:
\[ KE_{\text{initial}} = \frac{1}{2} (m + M) v^2 \]

- **Final potential energy**: The potential energy when the system is raised to height \( h \) is:
\[ PE_{\text{final}} = (m + M) \cdot g \cdot h \]

Setting the initial kinetic energy equal to the final potential energy:
\[ \frac{1}{2} (m + M) v^2 = (m + M) \cdot g \cdot h \]

### Step 4: Simplify the equation
We can cancel \( (m + M) \) from both sides (assuming \( m + M \neq 0 \)):
\[ \frac{1}{2} v^2 = g \cdot h \]

Multiplying both sides by 2 gives:
\[ v^2 = 2gh \]

### Step 5: Substitute \( v \) from momentum equation
Now we substitute the expression for \( v \) from the momentum conservation into the energy equation:
\[ \left(\frac{m \cdot V}{m + M}\right)^2 = 2gh \]

### Step 6: Solve for \( V \)
Squaring both sides gives:
\[ \frac{m^2 \cdot V^2}{(m + M)^2} = 2gh \]

Rearranging to find \( V^2 \):
\[ V^2 = \frac{(m + M)^2 \cdot 2gh}{m^2} \]

Taking the square root:
\[ V = \frac{(m + M)}{m} \sqrt{2gh} \]

### Final Answer
Thus, the velocity of the sphere is:
\[ V = \frac{(m + M)}{m} \sqrt{2gh} \]