A steel ball of mass `0.5 kg` is fastened to a cord `20 cm` long and fixed at the far end and is released when the cord is horizontal. At the bottom of its path the ball strikes a `2.5 kg` steel block initially at rest on a frictionless surface. The collision is elastic. The speed of the block just after the collision will be.

Options

A. a.`10/3ms^(-1)`
B. b. `20/3ms^(-1)`
C. c. `5ms^(-1)`
D. d. `5/3ms^(-1)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the speed of the steel ball just before the collision

When the steel ball is released from the horizontal position, it converts its potential energy at the height of the cord into kinetic energy at the bottom of its swing.

1. **Potential Energy (PE) at the top**:
\[ PE = mgh \]
where:
- \( m = 0.5 \, \text{kg} \) (mass of the ball)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 0.2 \, \text{m} \) (height, which is the length of the cord)

\[ PE = 0.5 \times 9.8 \times 0.2 = 0.98 \, \text{J} \]

2. **Kinetic Energy (KE) at the bottom**:
\[ KE = \frac{1}{2} mv^2 \]
Setting \( PE = KE \):
\[ 0.98 = \frac{1}{2} \times 0.5 \times v^2 \]
\[ 0.98 = 0.25 v^2 \implies v^2 = \frac{0.98}{0.25} = 3.92 \implies v = \sqrt{3.92} \approx 1.98 \, \text{m/s} \]

### Step 2: Apply conservation of momentum

In an elastic collision, the momentum before the collision equals the momentum after the collision.

Let:
- \( m_1 = 0.5 \, \text{kg} \) (mass of the ball)
- \( m_2 = 2.5 \, \text{kg} \) (mass of the block)
- \( u_1 = 1.98 \, \text{m/s} \) (initial speed of the ball)
- \( u_2 = 0 \, \text{m/s} \) (initial speed of the block)
- \( v_1 \) = final speed of the ball after collision
- \( v_2 \) = final speed of the block after collision

Using conservation of momentum:
\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]
Substituting the known values:
\[ 0.5 \times 1.98 + 2.5 \times 0 = 0.5 v_1 + 2.5 v_2 \]
\[ 0.99 = 0.5 v_1 + 2.5 v_2 \quad \text{(1)} \]

### Step 3: Apply conservation of kinetic energy

In an elastic collision, the total kinetic energy before the collision equals the total kinetic energy after the collision.

\[ \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]
Substituting the known values:
\[ \frac{1}{2} \times 0.5 \times (1.98)^2 + 0 = \frac{1}{2} \times 0.5 v_1^2 + \frac{1}{2} \times 2.5 v_2^2 \]
\[ 0.5 \times 0.5 \times 3.92 = 0.25 v_1^2 + 1.25 v_2^2 \]
\[ 0.98 = 0.25 v_1^2 + 1.25 v_2^2 \quad \text{(2)} \]

### Step 4: Solve the equations

From equation (1):
\[ 0.5 v_1 + 2.5 v_2 = 0.99 \implies v_1 = \frac{0.99 - 2.5 v_2}{0.5} = 1.98 - 5 v_2 \]

Substituting \( v_1 \) in equation (2):
\[ 0.98 = 0.25 (1.98 - 5 v_2)^2 + 1.25 v_2^2 \]

This is a quadratic equation in terms of \( v_2 \). Solving this will give us the value of \( v_2 \).

### Step 5: Calculate \( v_2 \)

After substituting and simplifying, you will find:
\[ v_2 \approx \frac{2}{3} \, \text{m/s} \]

### Final Answer

The speed of the block just after the collision is approximately \( \frac{2}{3} \, \text{m/s} \).

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