A small ball of mass $\mathrm{m}$ starts at a point $\mathrm{A}$ with speed $\mathrm{v}_{\mathrm{o}}$ and…

A small ball of mass $\mathrm{m}$ starts at a point $\mathrm{A}$ with speed $\mathrm{v}_{\mathrm{o}}$ and moves along a frictionless track $\mathrm{AB}$ as shown. The track $\mathrm{BC}$ has coefficient of friction $\mu$. The ball comes to stop at $\mathrm{C}$ after travelling a distance $L$ which is:
  1. $\frac{2 \mathrm{~h}}{\mu}+\frac{\mathrm{v}_{\mathrm{o}}^2}{2 \mu \mathrm{g}}$
  2. $\frac{\mathrm{h}}{\mu}+\frac{\mathrm{v}_0^2}{2 \mu \mathrm{g}}$
  3. $\frac{\mathrm{h}}{2 \mu}+\frac{\mathrm{v}_{\mathrm{o}}^2}{\mu \mathrm{g}}$
  4. $\frac{\mathrm{h}}{2 \mu}+\frac{\mathrm{v}_{\mathrm{o}}^2}{2 \mu \mathrm{g}}$

Solution

Initial speed at point A, $u=v_0$ Speed at point $\mathrm{B}, v=$ ? $ \begin{aligned} &v^2-u^2=2 g h \\ &v^2=v_0^2+2 g h \end{aligned} $ Let ball travels distance ' $S$ ' before coming to rest $ \begin{aligned} &S=\frac{v^2}{2 \mu g}=\frac{v_0^2+2 g h}{2 \mu g} \\ &=\frac{v_0^2}{2 \mu g}+\frac{2 g h}{2 \mu g}=\frac{h}{\mu}+\frac{v_0^2}{2 \mu g} \end{aligned} $

Asked in: JEE Main 2014 (11 Apr Online)

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