Question

A particle is released at point $P(-2\text{m},2\text{m})$ on a smooth parabolic track given by $y=\frac{x^2}{2}$. It slides down the track and leaves the contact with the track at point $Q(1\text{m},0.5\text{m})$, and then slides till point $R$ along $QR$. Distance $QR$ in m equals $(g=10{\text{m/s}}^2)$

Answer 1

$y=\frac{x^2}{2}$
$(\text{T.M.E}{)}_P=(\text{T.M.E}{)}_Q$
$\Rightarrow mg\times \frac{3}{2}=\frac{1}{2}m{v}^2$
$\Rightarrow v^2=3g$

Also $a=-\mu g$
$v\left(\frac{dv}{dx}\right)=-\mu g\Rightarrow \underset{\sqrt{3}g}{\overset{0}{\int }}vdv=-\mu g\underset{0}{\overset{x}{\int }}dx$
${\left[\frac{v^2}{2}\right]}_{\sqrt{3g}}^0=-\mu g{\left[x\right]}_0^x\Rightarrow \mu x=\frac{3}{2}$
$\therefore x=\frac{3}{2\mu }=\frac{3}{2\times \frac{1}{2}}=3\text{m}$