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A hot air balloon of mass M is stationary (with respect to the ground) in mid-air. A passenger of mass m climbs out and slides down a rope with constant

velocity v with respect to the balloon. With what velocity (magnitude and direction) relative to the ground does the balloon move? What happens if the passenger then stops sliding? Now, answer the following questions:  Check, mathematically, whether energy is conserved while the passenger is sliding. Does your answer make sense? Shouldn’t there be some rubbing between the passenger’s hand and the rope?  It seems that energy is “created” when both person and balloon start moving, and “lost” when both stop. How does that happen? Show, mathematically, that whatever mechanism you come up with “creates” the required energy.


See question number 8 in this document.

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We apply the conservation of momentum principle.
M V + m v = 0                  where V is the velocity of the ballon
                             R H S is 0 as both of them were stationary initially.
V = - m v / M.                  V is in the direction opposite to v of the passenger.
                               As passenger is sliding down, balloon climbs up.

When the passenger stops sliding , he/she exerts a force on the balloon to stop himself/herself. That force will stop the balloon too. With the force balloon exerts on him, he will stop sliding.

Energy is conserved as, passenger loses some height and the balloon climbs up gaining potential energy,  The kinetic energy comes from the potential energy lost by person.

as MV = - mv

To prove    Mg H + 1/2 M V² = mg h + 1/2m v²

H = height climbed by baloon in time t = V * t
h =  height lost by person in time t = v * t  =  MV/m t

So to prove  Mg V t + 1/2 M V² =   m g  (MV/m)  t + 1/2 m v²
                                               =   Mg V t + 1/2 M²V² / m      as - mv = MV

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