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,m2,m22.............m2n−1 respectively. What is the minimum velocity which should be imparted to the first ball of mass m such that nth ball completes the vertical circle

Velocity of second ball=(m/2−mm/2+m)(0)+(2×mm+m/2)v =43vVelocity of third ball will become, 43 times 43v or (43)2v∴ vn=(43)n−1v=5gr ∴ v=(34)n−15gr

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,m2,m22.............m2n−1 respectively. What is the minimum velocity which should be imparted to the first ball of mass m such that nth ball completes the vertical circle

Velocity of second ball=(m/2−mm/2+m)(0)+(2×mm+m/2)v =43vVelocity of third ball will become, 43 times 43v or (43)2v∴ vn=(43)n−1v=5gr ∴ v=(34)n−15gr

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$n$ 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 $n^{t h}$ ball completes the vertical circle

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${(\frac{2}{3})}^{n - 1} \sqrt{5 g r}$

${(\frac{3}{4})}^{n - 1} \sqrt{5 g r}$

${(\frac{4}{3})}^{n - 1} \sqrt{5 g r}$

${(\frac{3}{2})}^{n - 1} \sqrt{5 g r}$

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Detailed Solution

Velocity of second ball

$= (\frac{m / 2 - m}{m / 2 + m}) (0) + (\frac{2 \times m}{m + m / 2}) v = \frac{4}{3} v$

Velocity of third ball will become,

$\frac{4}{3}$ times $\frac{4}{3} v$ or ${(\frac{4}{3})}^{2} v$

$∴ v_{n} = {(\frac{4}{3})}^{n - 1} v = \sqrt{5 g r} ∴ v = {(\frac{3}{4})}^{n - 1} \sqrt{5 g r}$

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