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A balloon of mass M is stationary in air. It has a ladder on which a man of mass m is standing. If the man starts climbing up the ladder with a velocity v relative to the ladder, the velocity of the balloon is

\[\begin {array} {1 1} (a)\;\frac{m}{M}v \;upwards & \quad (b)\;\frac{mv}{(m+M)}\;downwards \\ (c)\;\frac{mv}{(m+M)}\;upwards & \quad  (d)\;\frac{Mv}{(M+m)}\;downwards \end {array}\]

Can you answer this question?
 
 

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Let $V_b$ be the velocity of ladder downwards .
Velocity of the man with respect to ground $=V-V_b$
$M(V_b)=m(V-V_b)$
$V_b=\large\frac{mv}{m+M}$
hence b is the answer
answered Nov 21, 2013 by meena.p
edited Jun 17, 2014 by lmohan717
 

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