Question

A particle of mass $\frac{10}{7} Kg$ is moving in the positive direction of $x$. Its initial position $x=0$ & initial velocity is $1 \,m / s$. The velocity at $x =10$ is (in $m / s$ )

Answer 1

$P =m V^{2} \frac{d V}{d x} $
$ P =2+\frac{2}{10} \times x $
$P =2+\frac{x}{5} $
$P =m V \frac{d V}{d x} \times V $
$=\frac{10}{7} \times V^{2} \frac{d V}{d x}$
$\frac{10}{7} \int\limits_{1}^{ v } V ^{2} dV =\int\limits_{ x =0}^{ x =10}\left(2+\frac{ x }{5}\right) dx$
$\frac{10}{7}\left[\frac{ V ^{3}-1}{3}\right]=[2 x ]_{0}^{10}+\frac{1}{5}\left[\frac{ x ^{2}}{2}\right]_{0}^{10}$
$\frac{10}{21}\left[V^{3}-1\right]=20+\frac{1}{10} \times 100$
$V^{3}-1=30 \times \frac{21}{10}$
$V^{3}=64$
$V=4\, m /\sec .$