They are small elastic balls placed at rest on a smooth horizontal surface which are circulated at the extremes with radius 2m. The masses of balls are M,M2,M22....M2n−1 . Find the least velocity which should be provided to the first ball of mass M such that 5th ball completes a vertical circle (nearly) [in ms−1] .

V2=2m1u1m1+m2 V2=2M(M+M2)u=43u V3=2M2(M2+M4)×43u=(43)2uBetween M4 and M8V4=(43)3uSimilarly V5=(43)4uGiven, V5=5gR=5×2×10=10⇒10=(43)4u⇒u=3.16 m/s

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They are small elastic balls placed at rest on a smooth horizontal surface which are circulated at the extremes with radius 2m. The masses of balls are $M, \frac{M}{2}, \frac{M}{2^{2}} .... \frac{M}{2^{n - 1}}$ . Find the least velocity which should be provided to the first ball of mass M such that 5^th ball completes a vertical circle (nearly) $[i n m s^{- 1}]$ .

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answer is 3.16.

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$V_{2} = \frac{2 m_{1} u_{1}}{m_{1} + m_{2}} V_{2} = \frac{2 M}{(M + \frac{M}{2})} u = \frac{4}{3} u$

$V_{3} = \frac{2 \frac{M}{2}}{(\frac{M}{2} + \frac{M}{4})} \times \frac{4}{3} u = {(\frac{4}{3})}^{2} u$

Between $\frac{M}{4} a n d \frac{M}{8}$

$V_{4} = {(\frac{4}{3})}^{3} u$

Similarly $V_{5} = {(\frac{4}{3})}^{4} u$

Given, $V_{5} = \sqrt{5 g R} = \sqrt{5 \times 2 \times 10} = 10$

$\Rightarrow 10 = {(\frac{4}{3})}^{4} u \Rightarrow u = 3.16 m / s$

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