The potential energy of a point particle with mass $ m $ moving along the $ x $-axis is described by:

$ V(x) = A x + B x^2 $

We need to determine the maximum speed attained by the particle if it starts from rest at $ x = \frac{A}{B} $.

The particle achieves its maximum speed when its kinetic energy is at its maximum, which occurs when its potential energy is at a minimum. To find where the potential energy is minimized, we calculate the derivative of $ V(x) $ and set it to zero:

$ \frac{d(V(x))}{dx} = A + 2Bx = 0 $

Solving for $ x $ gives:

$ x = -\frac{A}{2B} $

The second derivative, $\frac{d^2(V(x))}{dx^2} = 2$, confirms a minimum at this point.

Applying the work-energy theorem, we calculate the kinetic energy $ KE_f $ when the potential energy is minimized:

Using the initial and final potential energy values:

Initial potential energy at $ x = \frac{A}{B} $:

$ PE_i = A\left(\frac{A}{B}\right) + B\left(\frac{A}{B}\right)^2 = \frac{A^2}{B} + \frac{A^2}{B} = \frac{2A^2}{B} $

Final potential energy at $ x = -\frac{A}{2B} $:

$ PE_f = A\left(-\frac{A}{2B}\right) + B\left(\frac{-A}{2B}\right)^2 = -\frac{A^2}{2B} + \frac{A^2}{4B} = -\frac{A^2}{4B} $

The change in kinetic energy, using the work-energy theorem:

$ KE_f = PE_i - PE_f = \frac{2A^2}{B} - \left(-\frac{A^2}{4B}\right) = \frac{9A^2}{4B} $

Expressing kinetic energy in terms of velocity:

$ \frac{1}{2} mv^2 = \frac{9A^2}{4B} $

Solving for $ v $, the maximum speed:

$ v = \sqrt{\frac{9A^2}{2Bm}} = \frac{3A}{\sqrt{2Bm}} $

Mechanics

Optics

Electromagnetism

Modern Physics

Work, Energy and Power · Physics · IAT (IISER)

MCQ (Single Correct Answer)