
A hot air balloon of mass m is descending vertically with downward acceleration a. The amount of mass that must be thrown out to give the balloon an upward acceleration a is?
(A) \[\frac{g+a}{2ma}\]
(B) \[\frac{2ma}{g+a}\]
(C) \[\frac{2mg}{g+a}\]
(D) \[\frac{ma}{g+a}\]
Answer
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Hint: the balloon is coming downwards with a constant acceleration a. the mass of the balloon is m. We want it to move upwards with acceleration.
The balloon is constantly feeling a force called gravitational force acting downwards. For it to move up it must feel a force acting upwards and whose magnitude must be greater than the gravitational force.
Complete step by step answer:While the balloon is coming downwards it feels not only gravitational force acting downwards but also a thrust. Thrust is a reaction force. When the body accelerates in one direction, the accelerated body will cause a force of equal magnitude but opposite direction.
Let us represent thrust force by R. so, the net force acting on the body is mg-R=ma
Let m’ is removed from the body so that it started to move now upwards with an acceleration a.
New gravitational force =\[(m-m')g\]
If the body moves up net force is given as \[R-(m-m')g=(m-m')a\]
\[\begin{align}
& R-mg+m'g=ma-m'a \\
& -ma+m'g=ma-m'a \\
& m'(g+a)=2ma \\
& m'=\frac{2ma}{g+a} \\
\end{align}\]
So, the correct option is (B)
Additional Information: Thrust is a mechanical force.
Note:When a body is being acted upon by multiple forces then we have to add the forces vectorially by taking into account the direction of each force. When forces act in the same direction they get added up.
The balloon is constantly feeling a force called gravitational force acting downwards. For it to move up it must feel a force acting upwards and whose magnitude must be greater than the gravitational force.
Complete step by step answer:While the balloon is coming downwards it feels not only gravitational force acting downwards but also a thrust. Thrust is a reaction force. When the body accelerates in one direction, the accelerated body will cause a force of equal magnitude but opposite direction.
Let us represent thrust force by R. so, the net force acting on the body is mg-R=ma
Let m’ is removed from the body so that it started to move now upwards with an acceleration a.
New gravitational force =\[(m-m')g\]
If the body moves up net force is given as \[R-(m-m')g=(m-m')a\]
\[\begin{align}
& R-mg+m'g=ma-m'a \\
& -ma+m'g=ma-m'a \\
& m'(g+a)=2ma \\
& m'=\frac{2ma}{g+a} \\
\end{align}\]
So, the correct option is (B)
Additional Information: Thrust is a mechanical force.
Note:When a body is being acted upon by multiple forces then we have to add the forces vectorially by taking into account the direction of each force. When forces act in the same direction they get added up.
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