A simple pendulum of length 0.2m has bob of mass 5gm, it is pulled aside through an angle 60° from the vertical. A spherical body of mass 2.5 gm is placed at the lowest position of the bob. When the bob is released it strikes the spherical body and comes to rest. What is the velocity of the spherical body ? (g = 9.8 `ms^(-2)` (in m/s)

Options

A. 1.4
B. 2.8
C. 3.5
D. 4.9

Answer

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

### Step 1: Calculate the height (h) the pendulum bob rises
The pendulum bob is pulled aside at an angle of 60° from the vertical. We need to find the height (h) it rises when it is pulled aside.

Using trigonometry:
- The length of the pendulum (L) = 0.2 m
- The height (h) can be calculated using the formula:
\[ h = L - L \cos(\theta) \]
where \(\theta = 60°\).

Calculating:
\[ h = 0.2 - 0.2 \cos(60°) = 0.2 - 0.2 \times \frac{1}{2} = 0.2 - 0.1 = 0.1 \text{ m} \]

### Step 2: Calculate the velocity (v1) of the pendulum bob just before the collision
Using the conservation of energy, the potential energy at the highest point is converted into kinetic energy at the lowest point.

The potential energy (PE) at the height (h) is given by:
\[ PE = mgh \]
where:
- \(m = 5 \text{ gm} = 0.005 \text{ kg}\)
- \(g = 9.8 \text{ m/s}^2\)
- \(h = 0.1 \text{ m}\)

Calculating potential energy:
\[ PE = 0.005 \times 9.8 \times 0.1 = 0.0049 \text{ J} \]

At the lowest point, all this energy is converted to kinetic energy (KE):
\[ KE = \frac{1}{2} mv^2 \]
Setting \(PE = KE\):
\[ 0.0049 = \frac{1}{2} \times 0.005 \times v^2 \]

Solving for \(v^2\):
\[ 0.0049 = 0.0025 v^2 \implies v^2 = \frac{0.0049}{0.0025} = 1.96 \implies v = \sqrt{1.96} = 1.4 \text{ m/s} \]

### Step 3: Use conservation of momentum to find the velocity of the spherical body (v2)
When the pendulum bob strikes the spherical body, we can use the conservation of momentum:
\[ m_1 v_1 = m_2 v_2 \]
where:
- \(m_1 = 5 \text{ gm} = 0.005 \text{ kg}\)
- \(v_1 = 1.4 \text{ m/s}\)
- \(m_2 = 2.5 \text{ gm} = 0.0025 \text{ kg}\)
- \(v_2\) is what we need to find.

Substituting the values:
\[ 0.005 \times 1.4 = 0.0025 \times v_2 \]
Calculating:
\[ 0.007 = 0.0025 v_2 \implies v_2 = \frac{0.007}{0.0025} = 2.8 \text{ m/s} \]

### Final Answer
The velocity of the spherical body after the collision is **2.8 m/s**.

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