Question

# A man of mass m climbs a rope of length L suspended below a balloon of mass M. The balloon is stationary with respect to ground. If the man begins to climb up the rope at a speed vrel (relative to the rope), in what direction and with what speed (relative to the ground) will the balloon move?

A
Downwards, m vrelm+M
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B
Upwards, m vrelm+M
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C
Downwards, m vrelM
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D
Downwards, (M+m) vrelM
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Solution

## The correct option is A Downwards, m vrelm+M Initially the system of balloon and man is at rest and when the man starts climbing up on the rope, Fext=0 on the system. m = mass of the man M = mass of the balloon Thus applying law of momentum conservation in vertical direction on the system of man and balloon. pf=pi ⇒m vm−M vb=0 .....(i) Taking upwards as +ve direction. Here vm is the velocity of man w.r.t ground and vrel is the man's velocity w.r.t the rope. vb is velocity of balloon w.r.t ground ⇒vm=vrel+(−vb) or, vm=vrel−vb .....(ii) Since pi=0, the balloon will tend to move downwards in response of the man climbing upwards such that the final momentum remains zero. From Eq. (i) & (ii). m(vrel−vb)−Mvb=0 or, mvrel=mvb+Mvb ⇒vb=m vrelM+m Why this question?Caution: While applying momentum conservation, pay attention to keep all the velocities w.r.t ground along with the proper signs as per the reference direction.

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