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 the 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 nth ball completes vertical circle is

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

Answer 1

Answer: (A)

$\left( \frac{3}{4} \right)^{n - 1}\sqrt{5gr}$

Answer 2

Let u be the speed of first ball, then for a strike between first and second ball

Mv = Mv₁ + $\frac{Mv_{2}}{2}$ (i)

and$\frac{1}{2}Mv^{2} = \frac{1}{2}Mv_{1}^{2} + \frac{1}{2}\frac{M}{2}v_{2}^{2}$ (ii) solving (i) and (ii)

v₂ = $\frac{4}{3}u$

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$\frac{1}{2}\frac{M}{2}\left( u_{2} \right)^{2} + \frac{1}{2}\frac{M}{(2)^{2}}\left( u_{3} \right)^{2}$

solving v₃ = $\left( {\overrightarrow{a}}_{r} \right)u$

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

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

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

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