A block of mass `2kg` is attached with a spring of spring constant `4000Nm^-1` and the system is kept on smooth horizontal table. The other end of the spring is attached with a wall. Initially spring is stretched by `5cm` from its natural position and the block is at rest. Now suddenly an impulse of `4kg-ms^-1` is given to the block towards the wall.
Find the velocity of the block when spring acquires its natural length

Options

A. (a) `5ms^-1`
B. (b) `3ms^-1`
C. (c) `6ms^-1`
D. (d) None of these

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given data
- Mass of the block, \( m = 2 \, \text{kg} \)
- Spring constant, \( k = 4000 \, \text{N/m} \)
- Initial stretch of the spring, \( x_i = 5 \, \text{cm} = 0.05 \, \text{m} \)
- Impulse given to the block, \( J = 4 \, \text{kg m/s} \)
- Initial velocity of the block, \( u = 0 \, \text{m/s} \) (since it is at rest)

### Step 2: Calculate the change in velocity due to the impulse
The impulse-momentum theorem states:
\[ J = m \Delta v \]
Where \( \Delta v = v - u \). Since the initial velocity \( u = 0 \):
\[ 4 = 2(v - 0) \]
\[ v = \frac{4}{2} = 2 \, \text{m/s} \]
So, the velocity of the block after the impulse is \( v = 2 \, \text{m/s} \).

### Step 3: Apply the work-energy theorem
When the spring returns to its natural length, the work done by the spring will equal the change in kinetic energy of the block. The work done by the spring can be expressed as:
\[ W = \frac{1}{2} k x_i^2 - \frac{1}{2} k x_f^2 \]
Where \( x_f = 0 \) (natural length of the spring). Thus:
\[ W = \frac{1}{2} k x_i^2 \]
Substituting the values:
\[ W = \frac{1}{2} \times 4000 \times (0.05)^2 = \frac{1}{2} \times 4000 \times 0.0025 = 5 \, \text{J} \]

### Step 4: Calculate the change in kinetic energy
The initial kinetic energy \( KE_i \) when the spring is stretched is:
\[ KE_i = \frac{1}{2} m u^2 = \frac{1}{2} \times 2 \times 0^2 = 0 \, \text{J} \]
The final kinetic energy \( KE_f \) when the spring reaches its natural length is:
\[ KE_f = \frac{1}{2} m v_f^2 \]
Using the work-energy theorem:
\[ W = KE_f - KE_i \]
\[ 5 = \frac{1}{2} \times 2 \times v_f^2 - 0 \]
\[ 5 = v_f^2 \]
\[ v_f = \sqrt{5} \approx 3.16 \, \text{m/s} \]

### Step 5: Conclusion
The velocity of the block when the spring acquires its natural length is approximately \( 3 \, \text{m/s} \).