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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?

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?

Grade:11

1 Answers

Swapnil Saxena
102 Points
9 years ago

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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