A mass of 10 gm moving with a velocity of 100 cm / s strikes a pendulum bob of mass 10 gm . The two masses stick together. The maximum height reached by the system now is `(g=10m//s^(2))`

Options

A. 7.5 cm
B. 5cm
C. 2.5 cm
D. 1.25cm

Explanation

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

### Step 1: Understanding the Problem
We have two masses:
- Mass 1 (m1) = 10 gm (0.01 kg)
- Mass 2 (m2) = 10 gm (0.01 kg)

Mass 1 is moving with a velocity of 100 cm/s (1 m/s) and strikes mass 2, which is at rest. After the collision, the two masses stick together.

### Step 2: Conservation of Momentum
Using the conservation of momentum before and after the collision:

\[ m_1 \cdot v_1 + m_2 \cdot v_2 = (m_1 + m_2) \cdot v_f \]

Where:
- \(v_1 = 100 \, \text{cm/s} = 1 \, \text{m/s}\)
- \(v_2 = 0 \, \text{cm/s}\) (mass 2 is at rest)
- \(v_f\) is the final velocity after collision.

Substituting the values:

\[ (0.01 \, \text{kg}) \cdot (1 \, \text{m/s}) + (0.01 \, \text{kg}) \cdot (0) = (0.01 \, \text{kg} + 0.01 \, \text{kg}) \cdot v_f \]

\[ 0.01 = 0.02 \cdot v_f \]

Solving for \(v_f\):

\[ v_f = \frac{0.01}{0.02} = 0.5 \, \text{m/s} = 50 \, \text{cm/s} \]

### Step 3: Conservation of Energy
After the collision, the kinetic energy of the combined mass will be converted into potential energy at the maximum height.

Kinetic Energy (KE) just after the collision:

\[ KE = \frac{1}{2} (m_1 + m_2) v_f^2 \]

\[ KE = \frac{1}{2} (0.01 + 0.01) (0.5)^2 = \frac{1}{2} (0.02) (0.25) = 0.0025 \, \text{J} \]

Potential Energy (PE) at maximum height \(h\):

\[ PE = (m_1 + m_2) g h \]

Setting \(KE = PE\):

\[ 0.0025 = (0.02) (10) h \]

\[ 0.0025 = 0.2h \]

Solving for \(h\):

\[ h = \frac{0.0025}{0.2} = 0.0125 \, \text{m} = 1.25 \, \text{cm} \]

### Final Answer
The maximum height reached by the system is **1.25 cm**.

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