A block of mass `m=0.1 kg` is connected 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 equilibrium 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 step by step, we will analyze the situation involving the spring and the blocks.

### Step 1: Understand the initial conditions
A block of mass \( m = 0.1 \, \text{kg} \) is connected to a spring and is compressed to a distance \( x \) from its equilibrium position. When released from rest, it moves towards the equilibrium position.

**Hint:** Identify the initial potential energy stored in the spring when compressed.

### Step 2: Determine the final conditions after the collision
After the block moves half the distance \( \left(\frac{x}{2}\right) \) from the equilibrium position, it collides with another block and comes to rest momentarily. The second block moves with a velocity of \( 3 \, \text{m/s} \).

**Hint:** Use the conservation of momentum to relate the velocities before and after the collision.

### Step 3: Apply conservation of momentum
Before the collision, the momentum of the system is given by:
\[ p_{\text{initial}} = m \cdot v \]
where \( v \) is the velocity of the first block before the collision.

After the collision, the momentum is:
\[ p_{\text{final}} = 0 + m_2 \cdot v_2 \]
where \( m_2 \) is the mass of the second block and \( v_2 = 3 \, \text{m/s} \).

Setting the initial momentum equal to the final momentum:
\[ m \cdot v = m_2 \cdot 3 \]

**Hint:** Solve for \( v \) (the initial velocity of the first block).

### Step 4: Relate velocity to spring potential energy
The velocity of the block can also be expressed in terms of the spring's potential energy. The speed of the block when it is at a distance \( \frac{x}{2} \) from the equilibrium position can be found using the conservation of energy:
\[ \frac{1}{2} k x^2 = \frac{1}{2} k \left(\frac{x}{2}\right)^2 + \frac{1}{2} mv^2 \]

**Hint:** Rearrange the equation to find \( v \) in terms of \( k \) and \( x \).

### Step 5: Substitute and solve for the spring constant \( k \)
From the previous step, we can express \( k \) in terms of \( v \):
\[ \frac{1}{2} k x^2 = \frac{1}{2} k \frac{x^2}{4} + \frac{1}{2} mv^2 \]
This simplifies to:
\[ \frac{1}{2} k x^2 - \frac{1}{8} k x^2 = \frac{1}{2} mv^2 \]
\[ \frac{3}{8} k x^2 = \frac{1}{2} mv^2 \]

**Hint:** Isolate \( k \) and express it in terms of \( m \), \( v \), and \( x \).

### Step 6: Calculate the initial energy of the spring
The initial potential energy stored in the spring when compressed to distance \( x \) is given by:
\[ E = \frac{1}{2} k x^2 \]

Substituting the expression for \( k \) from the previous step into this equation will allow us to find the total initial energy of the spring.

**Hint:** Substitute the values of \( m \) and \( v \) to find the numerical value of \( E \).

### Final Calculation
After substituting the values and simplifying, we find:
\[ E = 0.6 \, \text{J} \]

Thus, the total initial energy of the spring is \( 0.6 \, \text{J} \).