Question

A small ball of mass m starts at a point A with speed $v_0$ and moves along a fricionless track AB as shown. The track BC has co-efficient of friction $\mu$. The ball comes to stop at C after travelling distance L which as

• A. $\frac{2h}{\mu }+\frac{v_0^2}{2\mu g}$
• B. $\frac{h}{\mu }+\frac{v_0^2}{2\mu g}$
• C. $\frac{h}{2\mu }+\frac{v_0^2}{\mu g}$
• D. $\frac{h}{2\mu }+\frac{v_0^2}{2\mu g}$

Answer 1

The correct option is B $\frac{h}{\mu }+\frac{v_0^2}{2\mu g}$

If we $v_o$ be the initial velocity and h be the initial height,

Applying conservation of energy,

$mgh+\frac{1}{2}m{v}_o^2=\frac{1}{2}m{v}^2$

$v^2=v_o^2+2gh$

Now, if a be the decelaration produced because of friction,

Using, Newton's third equation of motion.

$v^2=u^2+2as$

$0=v_o^2+2gh-2aL$

$L=\frac{v_o^2+2gh}{2a}$

Now, $f=\mu N$

where, $N=mg$

$f=\mu mg$

$a=\mu g$

Putting back this value,

$L=\frac{v_o^2+2gh}{2\mu g}$

$=\frac{h}{\mu }+\frac{v_o^2}{2\mu g}$.