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A balloon of mass M is descending vertically with acceleration a. How much weight must be thrown from the case attached to the balloon so that it begins to ascend vertically with acceleration a?Asume that the upward lift of the air in the balloon does not change. Navjyot Kalra
8 years ago

Take upward direction to be the positive y-direction. If m is the mass thrown out, then the equation of motion for before and after the mass m is thrown out are

Ma = F – Mg

(M–m)a = F – (M–m)g

Solving the two equations m =2ma/(a+g)

3 years ago
Consider the balloon to move along y axis. The balloon`s mass is mThen, when it is descending,ma=F-mgAfter removing mass m`, the balloon ascends, then(m-m`)a=F-mgSolve the equations, then m`=2ma/a+g
3 years ago
Consider the balloon to move along y axis. The balloon`s mass is mThen, when it is descending,. ma=mg-F. After removing mass m`, the balloon ascends, then. (m-m`)a=F-(m-m`)g. Solve the equations, then m`=2ma/a+g
3 years ago
Let M= Mass of balloon initially
m= Mass removed
a= Acceleration (a
g= acceleration due to gravity
F=FORCE acting in upward direction due to air in balloon (air resistance)
Solution:-
Mg- F= Ma (since balloon's motion is in downward direction)-------------(1)
F -(M-m)g =(M-m)a  (since, balloon's motion is in upward direction)-----------------(2)
Put the value of F from equation (2) in (1) and you'll get the value of mass 'm' that has been removed from the balloon.
i.e. m(mass removed)= 2Ma/g+a
Like if you understood the answer
Hope it helps! Yash Chourasiya
one year ago
Dear Student
Let Fbe the upthrust of air in either case
In (i) case,Mg − F = Ma …(i)
In (ii) caseF − (M−m)g = (M−m)a …(ii)

From Equation (i) and (ii)
(Mg−Mg) − Mg + mg = Ma − ma
orm(g+a) = 2Ma
m = 2Ma/(g+a)