A ballon of mass M with light rope having monkey on the rope is in equilibrium.If the monkey starts moving up with acceleration 'a' w.r.t. rope. Then the acceleration of centre of mass of the system is

\frac{a (M - m)}{(M + m)}

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zero

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a

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\frac{a M}{m}

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Solution

The correct option is A $\frac{a (M - m)}{(M + m)}$
A balloon of mass M with light rope making monkey on the rope is in equilibrium. monkey is moving up with
acceleration $a$
The acceleration of center of mass of the system is,

$a_{c m} = \frac{m_{1} a_{1} + m_{2} a_{2}}{m_{1} + m_{2}}$

$a_{c m} = \frac{M a - m a}{M + m}$

$a_{c m} = \frac{a (M - m)}{M + m}$
The correct option is A.

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A ballon of mass M with light rope having monkey on the rope is in equilibrium.If the monkey starts moving up with acceleration 'a' w.r.t. rope. Then the acceleration of centre of mass of the system is

The correct option is A a(M−m)(M+m)A balloon of mass M with light rope making monkey on the rope is in equilibrium. monkey is moving up with acceleration a The acceleration of center of mass of the system is,acm=m1a1+m2a2m1+m2acm=Ma−maM+macm=a(M−m)M+mThe correct option is A.