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 `upsilon_(rel)` (relative to rope) in what direction and with what speed ( relative to ground) will the balloon move ?

Options

A. `vec(V)=m/M vec(v)_(rel.)`
B. `vec(V)=-m/Mvec(v)_(rel.)`
C. `vec(V)=-m/((m+M))vec(v)_(rel.)`
D. `vec(V)=+m/((m+M))vec(v)_(rel.)`

Answer

C

Explanation

Given that initially the system is at rest
ie, `vec(v)_(CM)=0," so "vec(v)_(CM)="constant"=0., " ie," (mvec(v)+Mvec(V))/(m+M)=0`
or `mvec(v)+Mvec(V)=0` [as (m + M) = finite], ie, `Mvec(V)=-mvec(v)` -----(1)
Further more, here it is given thatl `vec(v)_(rel)=vec(v)-vec(V)` -----(2)
Putting the value of `vec(v)` from eqn. (2) is eqn. (1), we get,
`Mvec(V)=-m(vec(v)_(rel)+vec(V))," "vec(V)=-(mvec(v)_(rel))/((m+M))`
Thus it is clear that the direction of motion of balloon is opposite to that of climbing `(vec(v)_(rel))`, ie, down