Question

A balloon and its content having mass \( M \) is moving up with an acceleration \( a \). The mass that must be released from the content so that the balloon starts moving up with an acceleration \( 3a \) will be:

• A. \( \frac{2Ma}{3a + g} \)
• B. \( \frac{3Ma}{2a - g} \)
• C. \( \frac{3Ma}{2a + g} \)
• D. \( \frac{2Ma}{3a - g} \)

Hint:

In problems involving forces and accelerations, remember to apply Newton’s second law for both the initial and final conditions, and use the relationship between mass and acceleration carefully.

Answer 1

The Correct Option is A

Step 1: Force Equation for Initial Condition 

Let the force \( F \) be the force acting on the balloon. The force equation for the initial condition (with mass \( m \)) is: \[ F - mg = ma \]

Step 2: Force When Mass \( x \) is Released

The force when the mass \( x \) is released becomes: \[ F = ma + mg \]

Step 3: Force After Releasing Mass \( x \)

After releasing mass \( x \), the equation becomes: \[ F - (m - x)g = (m - x) 3a \]

Step 4: Substitute the Value of \( F \)

Substituting the value of \( F \) from the previous equation: \[ Ma + mg - mg + xg = 3ma - 3xa \]

Step 5: Solve for \( x \)

Solving for \( x \): \[ x = \frac{2ma}{g + 3a} \]

Final Answer: \[ x = \frac{2ma}{g + 3a} \]

Answer 2

Step 1: Understanding the setup.
Let the total mass of the balloon (including gas and payload) be \( M \). The balloon is moving upward with acceleration \( a \). The forces acting on it are:
- Upward buoyant force = \( F_B \)
- Downward weight = \( Mg \)

The net force on the balloon is given by Newton’s second law:
\[ F_B - Mg = Ma \] ⟹ \( F_B = M(g + a) \).

Step 2: After releasing some mass.
Let the mass released from the balloon be \( m \). The new mass of the balloon becomes \( (M - m) \). The buoyant force remains the same (since the displaced air volume is the same).

Now, the balloon moves upward with acceleration \( 3a \). So the new equation of motion is:
\[ F_B - (M - m)g = (M - m)(3a) \]

Step 3: Substitute \( F_B = M(g + a) \) from Step 1.
\[ M(g + a) - (M - m)g = (M - m)(3a) \] Simplify:
\[ Mg + Ma - Mg + mg = 3a(M - m) \] \[ Ma + mg = 3aM - 3am \] Rearranging terms:
\[ Ma - 3aM = -3am - mg \] \[ -2aM = -m(3a + g) \] \[ m = \frac{2Ma}{3a + g} \]

Step 4: Final Answer.
The mass that must be released from the balloon so that it moves up with acceleration \( 3a \) is:
\[ \boxed{m = \frac{2Ma}{3a + g}} \]