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, let's break it down into parts (a) and (b) as given in the question.

### 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, \( v_0 = 0 \)

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

1. **Using the work-energy principle**:
The work done by the force when the particle moves from \( x_0 \) to \( x \) is equal to the change in kinetic energy.

\[ W = \Delta KE = \frac{1}{2} m v^2 - \frac{1}{2} m v_0^2 \]

Since \( v_0 = 0 \), we have:

\[ W = \frac{1}{2} m v^2 \]

2. **Calculate the work done**:
The work done by the force can be calculated as:

\[ W = \int_{x_0}^{x} F(x) \, dx = \int_{1}^{0.5} -\frac{K}{2x^2} \, dx \]

Evaluating the integral:

\[ W = -\frac{K}{2} \int_{1}^{0.5} \frac{1}{x^2} \, dx = -\frac{K}{2} \left[-\frac{1}{x}\right]_{1}^{0.5} \]

\[ W = -\frac{K}{2} \left(-2 + 1\right) = -\frac{K}{2} \cdot (-1) = \frac{K}{2} \]

Substituting \( K = 10^{-2} \):

\[ W = \frac{10^{-2}}{2} = 5 \times 10^{-3} \, \text{J} \]

3. **Setting the work equal to the change in kinetic energy**:

\[ \frac{1}{2} m v^2 = 5 \times 10^{-3} \]

Substituting \( m = 10^{-2} \):

\[ \frac{1}{2} (10^{-2}) v^2 = 5 \times 10^{-3} \]

\[ 10^{-2} v^2 = 10^{-2} \]

\[ v^2 = 1 \implies v = -1 \, \text{m/s} \quad (\text{negative because the particle is moving in the opposite direction}) \]

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

1. **Using the velocity expression**:
We derived the velocity in terms of position earlier:

\[ v = -\sqrt{\frac{K}{m} \left(\frac{1}{x} - 1\right)} \]

For \( x = 0.25 \):

\[ v = -\sqrt{\frac{10^{-2}}{10^{-2}} \left(\frac{1}{0.25} - 1\right)} = -\sqrt{1 \cdot (4 - 1)} = -\sqrt{3} \]

2. **Relating velocity to time**:

\[ v = \frac{dx}{dt} \implies dt = \frac{dx}{v} \]

\[ dt = \frac{dx}{-\sqrt{3}} \quad \text{(as velocity is negative)} \]

3. **Integrate from \( x_0 = 1 \) to \( x = 0.25 \)**:

\[ t = \int_{1}^{0.25} \frac{dx}{-\sqrt{3}} = -\frac{1}{\sqrt{3}} \left[x\right]_{1}^{0.25} \]

\[ t = -\frac{1}{\sqrt{3}} (0.25 - 1) = \frac{0.75}{\sqrt{3}} \approx 0.433 \, \text{s} \]

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

(b) The time at which it reaches \( x = 0.25 \, \text{m} \) is approximately \( 0.433 \, \text{s} \).