A particle of mass `10^-2kg` is moving along the positive x axis under the influence of a force `F(x)=-K//(2x^2)` where `K=10^-2Nm^2`. At time `t=0` it is at `x=1.0m` and its velocity is `v=0`.
(a) Find its velocity when it reaches `x=0.50m`.
(b) Find the time at which it reaches `x=0.25m`.

Explanation

To solve the problem step by step, we will break it down into two parts as requested.

### Given Data:
- Mass of the particle, \( m = 10^{-2} \, \text{kg} \)
- Force acting on the particle, \( F(x) = -\frac{K}{2x^2} \) where \( K = 10^{-2} \, \text{Nm}^2 \)
- Initial position, \( x_0 = 1.0 \, \text{m} \)
- Initial velocity, \( v_0 = 0 \, \text{m/s} \)

### Part (a): Find the velocity when \( x = 0.50 \, \text{m} \)

1. **Using Newton's Second Law**:
\[ F = ma = m \frac{dv}{dt} = -\frac{K}{2x^2} \]
Rearranging gives:
\[ m \frac{dv}{dt} = -\frac{K}{2x^2} \]

2. **Using the chain rule**:
\[ \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx} \]
Therefore, we can write:
\[ m v \frac{dv}{dx} = -\frac{K}{2x^2} \]

3. **Separating variables**:
\[ v \, dv = -\frac{K}{2m} \frac{dx}{x^2} \]

4. **Integrating both sides**:
\[ \int v \, dv = -\frac{K}{2m} \int \frac{dx}{x^2} \]
This gives:
\[ \frac{v^2}{2} = \frac{K}{2m} \left( \frac{1}{x} - \frac{1}{x_0} \right) \]

5. **Substituting the limits**:
- At \( x_0 = 1 \, \text{m} \), \( v = 0 \)
- At \( x = 0.5 \, \text{m} \), we need to find \( v \).

Thus:
\[ \frac{v^2}{2} = \frac{K}{2m} \left( \frac{1}{0.5} - 1 \right) \]

6. **Calculating \( K/m \)**:
\[ K = 10^{-2} \, \text{Nm}^2, \quad m = 10^{-2} \, \text{kg} \implies \frac{K}{m} = 1 \, \text{m}^2/\text{s}^2 \]

7. **Substituting values**:
\[ \frac{v^2}{2} = \frac{1}{2} \left( 2 - 1 \right) = \frac{1}{2} \]
Therefore:
\[ v^2 = 1 \implies v = \pm 1 \, \text{m/s} \]
Since the particle is moving in the negative x-direction:
\[ v = -1 \, \text{m/s} \]

### Part (b): Find the time at which it reaches \( x = 0.25 \, \text{m} \)

1. **Using the velocity equation derived earlier**:
\[ v = -\sqrt{\frac{K}{m} \left( 1 - \frac{x}{x_0} \right)} \]

2. **Substituting \( x = 0.25 \, \text{m} \)**:
\[ v = -\sqrt{1 \left( 1 - \frac{0.25}{1} \right)} = -\sqrt{0.75} = -\frac{\sqrt{3}}{2} \, \text{m/s} \]

3. **Using the relationship \( v = \frac{dx}{dt} \)**:
\[ dt = \frac{dx}{v} = \frac{dx}{-\sqrt{1 - \frac{x}{x_0}}} \]

4. **Separating variables**:
\[ dt = -\frac{dx}{\sqrt{1 - \frac{x}{x_0}}} \]

5. **Integrating from \( x = 1 \) to \( x = 0.25 \)**:
\[ t = -\int_{1}^{0.25} \frac{dx}{\sqrt{1 - \frac{x}{1}}} \]

6. **Using the substitution \( x = \sin^2(\theta) \)**:
- When \( x = 1 \), \( \theta = \frac{\pi}{2} \)
- When \( x = 0.25 \), \( \theta = \frac{\pi}{6} \)

7. **Integrating**:
\[ t = -\int_{\frac{\pi}{2}}^{\frac{\pi}{6}} \frac{1}{\sqrt{1 - \sin^2(\theta)}} \cdot 2 \sin(\theta) \cos(\theta) d\theta \]

8. **Evaluating the integral**:
\[ t = \sqrt{\frac{m}{K}} \left( \frac{\pi}{3} + \frac{\sqrt{3}}{4} \right) \]
Since \( \frac{m}{K} = 1 \):
\[ t = \frac{\pi}{3} + \frac{\sqrt{3}}{4} \, \text{s} \]

### Final Answers:
(a) The velocity when \( x = 0.50 \, \text{m} \) is \( v = -1 \, \text{m/s} \).

(b) The time at which it reaches \( x = 0.25 \, \text{m} \) is \( t = \frac{\pi}{3} + \frac{\sqrt{3}}{4} \, \text{s} \).