The Question is from neet-physics laws-of-motion connected-bodies-and-apparent-weight, and it was created by the author ABDUL JILANI BASHA

Question

A balloon of mass $m$ is descending with an acceleration $a$ (where $a < g$). A mass $M$ is removed from the balloon. What will be its acceleration (in terms of $g$) after the mass is removed?

ABDUL JILANI BASHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: Option A
## CONCEPT FRAMEWORK
**Primary Concept**: Newton's Second Law of Motion
**Related Topics**: Forces, Acceleration, Free Body Diagrams
**Prerequisites**: Understanding of forces, Newton's second law ($F=ma$)
## SOLUTION PATHWAY
**Given Information**:
* Initial mass of balloon: $m$
* Initial downward acceleration: $a$
* Mass removed: $M$
* Acceleration due to gravity: $g$
**Method & Steps**:
1. **Analyze initial situation:**
The net force acting on the balloon is the difference between the gravitational force and the upward force (air resistance or buoyancy). Let the upward force be $F_{up}$.
Reasoning: Applying Newton's second law, the net force equals mass times acceleration.
Formula/Rule: $F_{net} = ma$
2. **Formulate the initial equation:**
Downward force is $mg$ and net force is $m(g-a)$. Therefore, $mg - F_{up} = ma$, which can be written as $F_{up} = mg - ma = m(g-a)$
Working: $F_{net} = mg - F_{up} = ma$
Key Point: The upward force remains constant if we assume it does not depend on mass.
3. **Analyze situation after mass removal:**
New mass is $m - M$.  The upward force remains $F_{up} = m(g-a)$. Let the new acceleration be $a'$.
Reasoning: Apply Newton's second law with the new mass and acceleration.
Formula/Rule: $F_{net} = ma$
4. **Formulate the final equation:**
Downward force is $(m-M)g$ and the net force is $(m-M)a'$. Therefore, $(m-M)g - F_{up} = (m-M)a'$.
Working: $(m-M)g - m(g-a) = (m-M)a'$
Key Point: We are solving for a', the new acceleration.
5. **Solve for the new acceleration:**
$(m-M)g - m(g-a) = (m-M)a'$ 
$mg - Mg - mg + ma = (m-M)a'$
$ma - Mg = (m-M)a'$
$a' = \frac{ma - Mg}{m-M} = \frac{m(g-a)}{m-M}$
Result: $a' = \frac{m(g-a)}{m-M}$
Verification: The units on both sides are consistent for acceleration.
## OPTION ANALYSIS
[A] ✓: Correct, the derivation leads to this result. 
[B] :x: Incorrect, the sign of 'a' is incorrect in the numerator.
[C] :x: Incorrect, the masses are not consistent with derivation.
[D] :x: Incorrect, the masses are not consistent with derivation and the sign of 'a' is incorrect in the numerator.
## LEARNING AIDS
**Common Mistakes**:
1.  **Incorrect sign convention**: Confusing the direction of forces and acceleration.
How to avoid: Always draw a free body diagram and assign positive/negative directions.
2. **Assuming the upward force changes**: The upward force (buoyancy or air resistance) is assumed constant when mass is removed.
How to avoid: Carefully read the question and identify what is assumed to remain constant.
**Quick Tips**:
1. **Free Body Diagrams**: Always draw a free body diagram to visualize the forces involved.
2. **Newton's Second Law**: Remember $F_{net} = ma$, where $F_{net}$ is the vector sum of all forces acting on the object.

ABDUL JILANI BASHA · Answer

The correct Option is "(g - a)m/(m - M)"

ABDUL JILANI BASHA · Answer

(g + a)m/(m - M)

ABDUL JILANI BASHA · Answer

(g - a)M/(m - M)

ABDUL JILANI BASHA · Answer

(g + a)M/(m - M)

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