A ball of mass 'm' moving with a horizontal velocity 'v' strikes the bob of mass 'm' of a pendulum at rest. During this collision, the ball sticks with the bob of the pendulum. The height to which the combined mass raises is (g = acceleration due to gravity).

Options

A. `v^(2)/(4g)`
B. `v^(2)/(8g)`
C. `v^(2)/g`
D. `v^(2)/(2g)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the scenario
We have a ball of mass 'm' moving with a horizontal velocity 'v' that strikes a pendulum bob of mass 'm' at rest. After the collision, the ball sticks to the bob, and we need to find the height 'h' to which the combined mass rises.

### Step 2: Apply the conservation of momentum
Before the collision, the momentum of the system is given by the moving ball:
\[ \text{Initial momentum} = m \cdot v + m \cdot 0 = mv \]

After the collision, the two masses stick together, so the total mass is \( m + m = 2m \). Let \( v' \) be the velocity of the combined mass after the collision. By the conservation of momentum:
\[ mv = (m + m) v' \]
\[ mv = 2m v' \]

### Step 3: Solve for the velocity after the collision
Dividing both sides by \( 2m \):
\[ v' = \frac{v}{2} \]

### Step 4: Use energy conservation to find the height
After the collision, the kinetic energy of the combined mass is converted into potential energy at the maximum height 'h'. The kinetic energy just after the collision is:
\[ KE = \frac{1}{2} (2m) (v')^2 = m (v')^2 = m \left(\frac{v}{2}\right)^2 = m \cdot \frac{v^2}{4} \]

The potential energy at height 'h' is given by:
\[ PE = (2m)gh = 2mgh \]

Setting the kinetic energy equal to the potential energy:
\[ m \cdot \frac{v^2}{4} = 2mgh \]

### Step 5: Simplify and solve for 'h'
Dividing both sides by \( m \):
\[ \frac{v^2}{4} = 2gh \]

Now, divide both sides by 2g:
\[ h = \frac{v^2}{8g} \]

### Final Answer
The height to which the combined mass rises is:
\[ h = \frac{v^2}{8g} \]