A particle of mass m moves on positive x-axis under the influence of force acting towards the origin given by `-kx^2 hat i.` If the particle starts from rest at `x=a,` the speed it will attain when it crosses the origin is

Options

A. `(sqrtk/(ma))`
B. `(sqrt2k/(ma))`
C. `(sqrt(ma)/2k)`
D. None of these

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the Force Acting on the Particle
The force acting on the particle is given by:
\[ F = -kx^2 \hat{i} \]
This force is directed towards the origin and depends on the position \( x \).

### Step 2: Relate Force to Acceleration
Using Newton's second law, we can relate force to acceleration:
\[ F = ma \]
Thus, we have:
\[ ma = -kx^2 \]
From this, we can express acceleration \( a \):
\[ a = \frac{-kx^2}{m} \]

### Step 3: Use the Relation Between Acceleration and Velocity
Acceleration can also be expressed in terms of velocity \( v \) and position \( x \):
\[ a = v \frac{dv}{dx} \]
Setting the two expressions for acceleration equal gives:
\[ v \frac{dv}{dx} = \frac{-kx^2}{m} \]

### Step 4: Integrate to Find Velocity
We will integrate both sides. The left side integrates from \( 0 \) to \( v \) and the right side integrates from \( a \) to \( 0 \):
\[ \int_{0}^{v} v \, dv = \int_{a}^{0} \frac{-kx^2}{m} \, dx \]
The left side becomes:
\[ \frac{v^2}{2} \]
The right side becomes:
\[ \frac{-k}{m} \left[ \frac{x^3}{3} \right]_{a}^{0} = \frac{-k}{m} \left( 0 - \frac{a^3}{3} \right) = \frac{ka^3}{3m} \]
Thus, we equate both sides:
\[ \frac{v^2}{2} = \frac{ka^3}{3m} \]

### Step 5: Solve for Velocity
Now, we can solve for \( v^2 \):
\[ v^2 = \frac{2ka^3}{3m} \]
Taking the square root gives:
\[ v = \sqrt{\frac{2ka^3}{3m}} \]

### Final Answer
The speed the particle will attain when it crosses the origin is:
\[ v = \sqrt{\frac{2ka^3}{3m}} \]