To solve the problem step by step, let's break it down into manageable parts.
### Step 1: Understanding the System
We have two blocks, A and B, both with a mass of 1 kg. Block A is on an inclined plane with a coefficient of friction of \( \frac{1}{\sqrt{3}} \), while block B has no friction with the plane. The system starts from rest and travels a distance of 30 m before the cord connecting the two blocks is burnt.
### Step 2: Determine the Forces Acting on the Blocks
1. **For Block A**:
- The forces acting on block A are:
- Gravitational force down the incline: \( mg \sin \theta \)
- Frictional force opposing the motion: \( f = \mu mg \cos \theta \)
- Tension in the cord (before it is burnt): \( T \)
2. **For Block B**:
- The only force acting on block B is the gravitational force down the incline: \( mg \sin \theta \).
### Step 3: Set Up the Equations of Motion
1. For Block B:
\[
mg \sin \theta - T = ma \quad \text{(1)}
\]
2. For Block A:
\[
mg \sin \theta + T - f = ma \quad \text{(2)}
\]
where \( f = \mu mg \cos \theta \).
### Step 4: Combine the Equations
Adding equations (1) and (2):
\[
mg \sin \theta - T + mg \sin \theta + T - f = 2ma
\]
This simplifies to:
\[
2mg \sin \theta - f = 2ma
\]
Substituting \( f = \mu mg \cos \theta \):
\[
2mg \sin \theta - \mu mg \cos \theta = 2ma
\]
Dividing through by \( 2m \):
\[
g \sin \theta - \frac{\mu g \cos \theta}{2} = a
\]
### Step 5: Calculate the Acceleration
Given \( g = 10 \, \text{m/s}^2 \), \( \mu = \frac{1}{\sqrt{3}} \), and \( \theta = 30^\circ \):
- \( \sin 30^\circ = \frac{1}{2} \)
- \( \cos 30^\circ = \frac{\sqrt{3}}{2} \)
Substituting these values:
\[
a = 10 \cdot \frac{1}{2} - \frac{\frac{1}{\sqrt{3}} \cdot 10 \cdot \frac{\sqrt{3}}{2}}{2}
\]
\[
= 5 - \frac{5}{2} = 5 - 2.5 = 2.5 \, \text{m/s}^2
\]
### Step 6: Find the Velocity After 30 m
Using the equation of motion:
\[
v^2 = u^2 + 2as
\]
where \( u = 0 \) (starts from rest), \( a = 2.5 \, \text{m/s}^2 \), and \( s = 30 \, \text{m} \):
\[
v^2 = 0 + 2 \cdot 2.5 \cdot 30 = 150
\]
Thus,
\[
v = \sqrt{150} \, \text{m/s}
\]
### Step 7: Analyze the Situation After the Cord is Burnt
After the cord is burnt, block A continues to move under the influence of gravity and friction. The net force on block A is:
\[
F_{\text{net}} = mg \sin \theta - f
\]
Substituting the values:
\[
F_{\text{net}} = mg \sin \theta - \mu mg \cos \theta
\]
This results in zero net force, meaning block A will continue with the same velocity of \( \sqrt{150} \, \text{m/s} \) as it travels an additional 20 m.
### Conclusion
The velocity of block A after it travels a further distance of 20 m is:
\[
\sqrt{150} \, \text{m/s}
\]
### Final Answer
**Option C: \( \sqrt{150} \, \text{m/s} \)**
