Question

There are $n$ small elastic balls placed at rest on a smooth horizontal surface which is circular of radius, $R$ at other end. The masses of balls are $M$, $\frac{M}{2}$, $\frac{M}{2^2}$, $\frac{M}{2^3}$ ......., $\frac{M}{2^{n-1}}$, respectively. The least velocity which should be provided to the first ball of mass, $M$, such that $n^{th}$ ball completed vertical circle is

• A. ${\left(\frac{4}{3}\right)}^n\sqrt{gr}$
• B. ${\left(\frac{3}{4}\right)}^{n-1}\sqrt{5gr}$
• C. $\sqrt{gr}$
• D. ${\left(\frac{4}{3}\right)}^{n-1}\sqrt{5gr}$

Answer 1

Answer: (B)

Let $u$ be the speed of first ball, then for a strike between first and second ball
$Mu=M{v}_1+\frac{M{v}_2}{2}$_____ (i)

and $\frac{1}{2}M{u}^2=\frac{1}{2}M{v}_1^2+\frac{1}{2}\frac{M}{2}{v}_2^2$_____ (ii)

Solving (i) and (ii)
$v_2=\frac{4}{3}u$

Similarly for second and third ball
$\frac{M}{2}\times \frac{4}{3}u=\frac{M}{2}{v}_2+\frac{M}{2^2}{v}_3$

and $\frac{1}{2}\frac{M}{2}{\left(\frac{4}{3}u\right)}^2$$=\frac{1}{2}\frac{M}{2}(v_2{)}^2+\frac{1}{2}\frac{M}{(2{)}^2}(v_3{)}^2$

Solving $v_3=\left(\overline{a_r}\right)u$

in this way, $v_n={\left(\frac{4}{3}\right)}^{n-1}u$

For nth ball to loop $v_n=\sqrt{5gR}$

i.e.,${\left(\frac{4}{3}\right)}^{n-1}u=\sqrt{5gR}$

or $u={\left(\frac{3}{4}\right)}^{n-1}\sqrt{5gR}$