A pendulum bob of mass `10^-2kg` is raised to a height `5xx10^-2m` and then released. At the bottom of its swing, it picks up a mass `10^-3kg`. To what height will the combined mass rise?

Explanation

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

### Step 1: Calculate the velocity of the pendulum bob at the lowest point of its swing.

The pendulum bob is raised to a height \( h_1 = 5 \times 10^{-2} \, \text{m} \). When it is released and reaches the lowest point, all the potential energy is converted into kinetic energy.

Using the formula for velocity at the lowest point:
\[ v_1 = \sqrt{2gh_1} \]
where \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity).

Substituting the values:
\[ v_1 = \sqrt{2 \times 10 \times 5 \times 10^{-2}} = \sqrt{1} = 1 \, \text{m/s} \]

### Step 2: Apply conservation of momentum when the pendulum bob picks up the mass.

Let:
- \( m_1 = 10^{-2} \, \text{kg} \) (mass of the pendulum bob)
- \( m_2 = 10^{-3} \, \text{kg} \) (mass picked up)
- \( v_2 = 0 \, \text{m/s} \) (initial velocity of the mass \( m_2 \))

According to the conservation of momentum:
\[ m_1 v_1 + m_2 v_2 = (m_1 + m_2) v \]

Substituting the known values:
\[ (10^{-2} \, \text{kg})(1 \, \text{m/s}) + (10^{-3} \, \text{kg})(0) = (10^{-2} + 10^{-3}) v \]

This simplifies to:
\[ 10^{-2} = (1.1 \times 10^{-2}) v \]

Solving for \( v \):
\[ v = \frac{10^{-2}}{1.1 \times 10^{-2}} = \frac{10}{11} \, \text{m/s} \]

### Step 3: Calculate the height the combined mass will rise to.

Using conservation of energy, the kinetic energy at the lowest point will equal the potential energy at the highest point:
\[ \frac{1}{2} (m_1 + m_2) v^2 = (m_1 + m_2) g h_2 \]

Canceling \( (m_1 + m_2) \) from both sides:
\[ \frac{1}{2} v^2 = g h_2 \]

Substituting \( v = \frac{10}{11} \, \text{m/s} \):
\[ \frac{1}{2} \left(\frac{10}{11}\right)^2 = 10 h_2 \]

Calculating \( h_2 \):
\[ h_2 = \frac{\frac{1}{2} \left(\frac{100}{121}\right)}{10} = \frac{50}{1210} = \frac{5}{121} \approx 0.0413 \, \text{m} \]

### Final Answer:
The combined mass will rise to a height of approximately \( 0.0413 \, \text{m} \) or \( 4.13 \, \text{cm} \).

---