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$v_{2}_{β²}=m_{1}+m_{2}2m_{1}v_{1}β+m_{1}+m_{2}m_{1}βm_{2}βv_{2}$;

Velocity of the second ball $v_{2}=0$;

Let velocity of the first ball $v_{1}=v$;

For collision between the first and second ball,

$v_{2}_{β²}=m+2mβ2mvβ=34βv$;

The second ball will have a elastic collision with the third ball.

Since the ratio of consecutive masses of the balls is same, the velocity ratio of the ball will be same.

$v_{3}_{β²}=34βv_{3}=(34β)_{2}v$;

Continuing this for the n-th ball, we get

$v_{n}_{β²}=34βv_{n}=(34β)_{nβ1}v$;

From motion of a particle in a vertical circle, the minimum velocity required for the particle to reach the highest point on the vertical circle is $(5gR)_{21β}$;

$v_{n}_{β²}=(5gR)_{21β}$;

$β(5gR)_{21β}=(34β)_{nβ1}v$;

$βv=(43β)_{nβ1}(5gR)_{21β}$.

$βv=(43β)_{nβ1}5grβ$

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