A ball of mass m moving horizontally at a speed v collides with the bob of a simple pendulum at rest. The mass of the bob is also m. If the collision is perfectly elastic and both balls sticks, the height to which the two balls rise after the collision will be given by:

Answer

To solve the problem step by step, we will analyze the situation using the principles of conservation of momentum and the work-energy theorem.

### Step 1: Understand the scenario
We have a ball of mass \( m \) moving horizontally with a speed \( v \) that collides with a pendulum bob of mass \( m \) which is initially at rest. The collision is perfectly inelastic, meaning both masses stick together after the collision.

### Step 2: Apply conservation of momentum
In a perfectly inelastic collision, the total momentum before the collision is equal to the total momentum after the collision.

Before the collision:
- Momentum of the moving ball = \( mv \)
- Momentum of the pendulum bob = \( 0 \) (since it is at rest)

Total momentum before collision:
\[ p_{\text{initial}} = mv + 0 = mv \]

After the collision, let the combined mass (the ball and the bob) move with a velocity \( u \). The total mass after the collision is \( 2m \).

Using conservation of momentum:
\[ mv = (2m)u \]

### Step 3: Solve for the velocity \( u \)
From the equation \( mv = 2mu \), we can simplify to find \( u \):
\[ u = \frac{v}{2} \]

### Step 4: Use the work-energy theorem
After the collision, the kinetic energy of the combined mass will be converted into potential energy as it rises to a height \( h \).

The kinetic energy (KE) of the two masses just after the collision is:
\[ \text{KE} = \frac{1}{2} (2m) u^2 = m u^2 \]

Substituting \( u = \frac{v}{2} \):
\[ \text{KE} = m \left(\frac{v}{2}\right)^2 = m \frac{v^2}{4} = \frac{mv^2}{4} \]

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

### Step 5: Set kinetic energy equal to potential energy
According to the work-energy theorem:
\[ \text{KE} = \text{PE} \]
\[ \frac{mv^2}{4} = 2mgh \]

### Step 6: Solve for height \( h \)
Cancel \( m \) from both sides:
\[ \frac{v^2}{4} = 2gh \]

Now, solve for \( h \):
\[ h = \frac{v^2}{8g} \]

### Final Answer
The height to which the two balls rise after the collision is:
\[ h = \frac{v^2}{8g} \]