Question

$n$ elastic balls are placed at rest on a smooth horizontal plane which is circular at the ends with radius $r$ as shown in the figure. The masses of the balls are $m,\frac{m}{2},\frac{m}{2^2},...,\frac{m}{2^{n-1}}$ respectively. What is the minimum velocity which should be imparted to the first ball of mass $m$ such that this $n^{th}$ ball will complete the vertical circle ?

• A. ${\left(\frac{3}{4}\right)}^{n-1}\sqrt{5gr}$
• B. ${\left(\frac{4}{3}\right)}^{n-1}\sqrt{5gr}$
• C. ${\left(\frac{3}{2}\right)}^{n-1}\sqrt{5gr}$
• D. ${\left(\frac{2}{3}\right)}^{n-1}\sqrt{5gr}$

Answer 1

The correct option is A ${\left(\frac{3}{4}\right)}^{n-1}\sqrt{5gr}$

In head - on elastic collision
$v_1^{\prime }=\left(\frac{\frac{m}{2}-m}{\frac{m}{2}+m}\right)\left(0\right)+\left[\frac{2\left(m\right)}{\frac{m}{2}+m}\right]\left(v\right)$
$=\frac{4}{3}v$

Finally
Velocity of $n^{th}$ ball is $v_n^{\prime }={\left(\frac{4}{3}\right)}^{n-1}v$
For looping a loop
$v_n^{\prime }={\left(\frac{4}{3}\right)}^{n-1}v=\sqrt{5gr}$
$\therefore v={\left(\frac{3}{4}\right)}^{n-1}\sqrt{5gr}$