Two pendulum bobs of masses m and `2m` collide head on elastically at the lowest point in their motion. If both the balls are released from a height H above the lowest point, to what heights do they rise for the first time after collision?

Explanation

To solve the problem of two pendulum bobs of masses \( m \) and \( 2m \) colliding elastically at the lowest point in their motion, we will follow these steps:

### Step 1: Calculate the initial velocities of the pendulum bobs just before the collision.

Using the conservation of energy, we can find the velocities of both bobs just before the collision.

For the first bob (mass \( m \)):
- It is released from height \( H \).
- The potential energy at height \( H \) converts to kinetic energy at the lowest point.

\[ PE = KE \implies mgh = \frac{1}{2} mv_1^2 \]

Solving for \( v_1 \):

\[ v_1^2 = 2gH \implies v_1 = \sqrt{2gH} \]

For the second bob (mass \( 2m \)):
- It is also released from height \( H \).

\[ PE = KE \implies 2mgH = \frac{1}{2} (2m)v_2^2 \]

Solving for \( v_2 \):

\[ v_2^2 = 2gH \implies v_2 = \sqrt{2gH} \]

### Step 2: Apply conservation of momentum and kinetic energy for the elastic collision.

Let \( u_1 \) and \( u_2 \) be the velocities of the bobs after the collision.

Using the conservation of momentum:

\[ m v_1 + 2m v_2 = m u_1 + 2m u_2 \]

Substituting \( v_1 \) and \( v_2 \):

\[ m \sqrt{2gH} + 2m \sqrt{2gH} = m u_1 + 2m u_2 \]

Simplifying:

\[ 3m \sqrt{2gH} = m u_1 + 2m u_2 \]

Dividing through by \( m \):

\[ 3\sqrt{2gH} = u_1 + 2u_2 \quad \text{(1)} \]

Using conservation of kinetic energy:

\[ \frac{1}{2} m v_1^2 + \frac{1}{2} (2m) v_2^2 = \frac{1}{2} m u_1^2 + \frac{1}{2} (2m) u_2^2 \]

Substituting the values:

\[ \frac{1}{2} m (2gH) + \frac{1}{2} (2m)(2gH) = \frac{1}{2} m u_1^2 + \frac{1}{2} (2m) u_2^2 \]

Simplifying:

\[ m gH + 2mgH = \frac{1}{2} m u_1^2 + mg u_2^2 \]

\[ 3mgH = \frac{1}{2} m u_1^2 + mg u_2^2 \]

Dividing through by \( m \):

\[ 3gH = \frac{1}{2} u_1^2 + g u_2^2 \quad \text{(2)} \]

### Step 3: Solve the equations (1) and (2).

From equation (1):

\[ u_1 = 3\sqrt{2gH} - 2u_2 \]

Substituting \( u_1 \) into equation (2):

\[ 3gH = \frac{1}{2} (3\sqrt{2gH} - 2u_2)^2 + g u_2^2 \]

This will yield a quadratic equation in terms of \( u_2 \). Solving this will give us the values of \( u_1 \) and \( u_2 \).

### Step 4: Calculate the heights after the collision.

Using the conservation of energy again for both bobs after the collision:

For bob \( m \):

\[ \frac{1}{2} m u_1^2 = m g h_1 \implies h_1 = \frac{u_1^2}{2g} \]

For bob \( 2m \):

\[ \frac{1}{2} (2m) u_2^2 = (2m) g h_2 \implies h_2 = \frac{u_2^2}{2g} \]

### Final Heights:

After calculating \( u_1 \) and \( u_2 \) from the previous steps, we can find \( h_1 \) and \( h_2 \).