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A man of mass m climbs to a rope ladder suspended below a balloon of mass M. The balloon is stationary with respect to the ground.
a) If the man begins to climb the ladder at a speed v (with respect to ladder), in what direction and with what speed (with respect to ground) will the balloon move?
b) What is the state of the motion after the man stops climbing?
According to the law of conservation of center of mass,
=> mcmvcm = m1v1 + m2v2
=> 0= Mvbal + mvman
Now we know that the velocity of the man with respect to the ground = v - vbal
=> 0= Mvbal + m(v - vbal) => 0= Mvbal + mv -mvbal => vbal(M-m)+mv=0
=> -mv/(M-m) ----(ans)
Direction : Downwards
After the man stops climbing, the ballon will again be at rest
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