Question

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. $ma=-\alpha x^2$. The distance at which the particle stops:

• A. ${(2v_0/3\alpha )}^{\frac{1}{3}}$
• B. ${(3v_0^2/2\alpha )}^{\frac{1}{2}}$
• C. ${(3v_0^2/2\alpha )}^{\frac{1}{3}}$
• D. ${(2v_0^2/3\alpha )}^{\frac{1}{2}}$

Answer 1

The correct option is C

${(3v_0^2/2\alpha )}^{\frac{1}{3}}$

Step 1: Given data

Damping force, $ma=-\alpha x^2$

Initial velocity, $v_i=v_0$

Final velocity, $v_f=0$

Initial distance, $X_i=0$

Final distance, $X_f=x$

Step 2: Find distance

From Damping Force,

$a=\frac{-\alpha x^2}{m}$

Acceleration is also given by

$$\begin{gathered} a=vdv/dx \\ {\int }_{v_i}^{v_f}vdv={\int }_{x_i}^{x_f}adx \\ {\int }_{v_0}^{0}vdv={\int }_0^x\frac{-\alpha x^2}{m}dx \\ -v_0^2/2=(-\alpha /m)[x^3/3] \\ x={[3m{v}_0^2/2\alpha ]}^{\frac{1}{3}} \end{gathered}$$

Hence, option C is correct.