A block of mass `m=0.1 kg` is connceted to a spring of unknown spring constant k. It is compressed to a distance x from its equilibrium position and released from rest. After approaching half the distance `((x)/(2))` from the euilibrium position, it hits another block and comes to rest momentarily, while the other block moves with velocity `3ms^(-1)`. The total initial energy of the spring is :

Options

A. 0.6 J
B. 0.8 J
C. 1.5 J
D. 0.3 J

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the System
A block of mass \( m = 0.1 \, \text{kg} \) is connected to a spring compressed by a distance \( x \) from its equilibrium position. When released, it moves back towards the equilibrium position, colliding with another block after reaching a distance of \( \frac{x}{2} \).

### Step 2: Apply Conservation of Momentum
When the first block collides with the second block, it comes to rest momentarily, and the second block moves with a velocity of \( 3 \, \text{m/s} \). By conservation of momentum:
\[ m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2' \]
Here, \( m_1 = 0.1 \, \text{kg} \), \( v_1 = 3 \, \text{m/s} \) (the speed of the first block before collision), \( v_1' = 0 \) (the speed of the first block after collision), and \( m_2 \) is the mass of the second block (which we will assume is equal to \( m_1 \) for simplicity).

Substituting the values:
\[ 0.1 \cdot 3 + m_2 \cdot 0 = 0 + m_2 \cdot 3 \]
This implies:
\[ 0.3 = 3m_2 \]
So,
\[ m_2 = \frac{0.3}{3} = 0.1 \, \text{kg} \]

### Step 3: Determine the Speed at Half Compression
At the point where the block has moved to \( \frac{x}{2} \), we can use the formula for the speed in simple harmonic motion (SHM):
\[ v = \omega \sqrt{A^2 - x^2} \]
Where:
- \( A = x \) (the amplitude),
- \( x = \frac{x}{2} \) (the distance from the mean position).

Substituting into the equation:
\[ 3 = \omega \sqrt{x^2 - \left(\frac{x}{2}\right)^2} \]
Calculating \( \left(\frac{x}{2}\right)^2 = \frac{x^2}{4} \):
\[ 3 = \omega \sqrt{x^2 - \frac{x^2}{4}} = \omega \sqrt{\frac{3x^2}{4}} = \frac{\omega x \sqrt{3}}{2} \]
Thus,
\[ 3 = \frac{\omega x \sqrt{3}}{2} \implies \omega x \sqrt{3} = 6 \implies \omega x = \frac{6}{\sqrt{3}} = 2\sqrt{3} \]

### Step 4: Relate Spring Constant to Angular Frequency
The spring constant \( k \) is related to the mass \( m \) and angular frequency \( \omega \) by:
\[ k = m \omega^2 \]
Substituting \( \omega = \frac{2\sqrt{3}}{x} \):
\[ k = 0.1 \left(\frac{2\sqrt{3}}{x}\right)^2 = 0.1 \cdot \frac{12}{x^2} = \frac{1.2}{x^2} \]

### Step 5: Calculate Initial Energy of the Spring
The initial energy stored in the spring when compressed to distance \( x \) is given by:
\[ E = \frac{1}{2} k x^2 \]
Substituting for \( k \):
\[ E = \frac{1}{2} \left(\frac{1.2}{x^2}\right) x^2 = \frac{1.2}{2} = 0.6 \, \text{J} \]

### Final Answer
The total initial energy of the spring is:
\[ \boxed{0.6 \, \text{J}} \]