JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Work, Power & Energy

A particle is projected with velocity

v_{0}

along x -axis. A damping force is acting on the particle which is proportional to the square of the distance from the origin i.e.

ma = - α x^{2}

. The distance at which the particle stops:

Answer: (D)

Solution

Given:

ma = - α x^{2}

, so

m \frac{d v}{d t} = - α x^{2}

.

Using

a = v \frac{d v}{d x}

, we get:

m v \frac{d v}{d x} = - α x^{2}

Integrating from initial conditions (at

x = 0

v = v_{0}

) to final (at

x = x_{f}

v = 0

\int_{v_{0}}^{0} m v d v = - \int_{0}^{x_{f}} α x^{2} d x

m [\frac{v ^{2}}{2}]_{v_{0}}^{0} = - α [\frac{x ^{3}}{3}]_{0}^{x_{f}}

- \frac{m v _{0}^{2}}{2} = - \frac{α x _{f}^{3}}{3}

x_{f}^{3} = \frac{3 m v _{0}^{2}}{2 α}

x_{f} = (\frac{3 v _{0}^{2}}{2 α / m})^{1/3}

Since the force equation is given as

ma = - α x^{2}

, the parameter

α

already includes mass dependence. The answer is:

x_{f} = (\frac{3 v _{0}^{2}}{2 α})^{1/3}

JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Single correct option