Question

A Body of mass $m$ slides down an incline and reaches the bottom with a speed $v.$ if the same mass in the form of a ring rolls down this incline, what will be the velocity of the ring at the bottom?

• A. $v\sqrt{2}$
• B. $\frac{v}{\sqrt{2}}$
• C. $2\sqrt{2}v$
• D. $\frac{v}{\sqrt{2}+1}$

Answer 1

The correct option is B $\frac{v}{\sqrt{2}}$

Given: a Body of mass m slides down an incline and reaches the bottom with a speed v. if the same mass in the form of a ring rolls down this incline,

To find the velocity of the ring at the bottom

Solution:

We know,

Total energy at the bottom of the inclined plane = translational kinetic energy + rotational kinetic energy

And total energy at the top of the plane = potential energy

Now, as the mass slides down the inclined plane then potential energy gets converted into kinetic energy

In the first case when it is not a ring, if h is the height of the plane, then

$mgh=\frac{1}{2}m{v}^2⟹h=\frac{v^2}{2g}.............\left(i\right)$

Now according to the second case, if v'is the velocity at the bottom of the plane then

$mgh=\frac{1}{2}m{v}^{\prime 2}+\frac{1}{2}I{\omega }^2$

$⟹mgh=\frac{1}{2}m{v}^{\prime 2}+\frac{1}{2}m{r}^2\times {\left(\frac{v^{\prime 2}}{r}\right)}^2$ (as $v^{\prime }=r\times \omega$)

$⟹mgh=m{v}^{\prime 2}⟹h=\frac{v^{\prime 2}}{g}.........\left(ii\right)$

From eqn(i) and eqn(ii), we get

$\frac{v^{\prime 2}}{g}=\frac{v^2}{2g}⟹v^{\prime 2}=\frac{v^2}{2}⟹v^{\prime }=\frac{v}{\sqrt{2}}$

is the velocity of the ring at the bottom