Step 1: Understand the Forces Involved and Principle of Constant Velocity

Before writing equations, let’s identify all the forces acting on the balloon and their directions:

• Gravitational Force (Weight): This acts vertically downwards. If the total mass of the balloon (including its contents) is $m_{total}$, then the weight is $m_{total}g$.
• Buoyant Force ($B$): This acts vertically upwards and is given as constant ($B$). This force is due to the displacement of air by the balloon.
• Air Resistance Force ($F_r$): This force opposes the direction of velocity and is proportional to the velocity. We can write it as $F_r = k \times \text{velocity}$, where $k$ is the constant of proportionality.

The crucial part of the problem statement is “constant velocity”. This implies that the net force on the balloon is zero in both scenarios, according to Newton’s First Law of Motion ($\Sigma F = 0$). This is because if velocity is constant, acceleration is zero.

Step 2: Analyze the Initial Scenario – Falling Down with Constant Velocity $v$

In this initial scenario, the balloon has a total mass $M$ and is falling downwards with a constant velocity $v$. Since the net force is zero, the sum of upward forces must equal the sum of downward forces.

Let’s define our positive direction. For falling motion, it’s often convenient to take the downward direction as positive.

• Weight ($Mg$): Acts downwards (positive).
• Buoyant Force ($B$): Acts upwards (negative).
• Air Resistance ($F_r$): Since the balloon is moving downwards, the air resistance acts upwards (opposite to velocity). Its magnitude is $kv$ (negative).

Applying Newton’s Second Law ($\Sigma F = ma$, where $a=0$):

$$\Sigma F_{\text{vertical}} = 0$$

$$Mg – B – kv = 0$$

Rearranging this equation, we get a fundamental relationship for the initial state:

$$Mg = B + kv \quad \text{(Equation 1)}$$

This equation will be crucial for determining the constant $k$ in terms of known initial conditions.

Step 3: Analyze the Final Scenario – Rising Up with Constant Velocity $u$

Now, some mass has been removed from the balloon. Let the new, reduced total mass of the balloon be $m’$. The balloon is now rising upwards with a constant velocity $u$. Again, the net force on the balloon is zero.

For rising motion, it’s convenient to take the upward direction as positive.

• Weight ($m’g$): Acts downwards (negative).
• Buoyant Force ($B$): Acts upwards (positive). This force remains constant as given.
• Air Resistance ($F_r$): Since the balloon is moving upwards, the air resistance acts downwards (opposite to velocity). Its magnitude is $ku$ (negative).

Applying Newton’s Second Law ($\Sigma F = ma$, where $a=0$):

$$\Sigma F_{\text{vertical}} = 0$$

$$B – m’g – ku = 0$$

Rearranging this equation, we get a relationship for the final state:

$$B = m’g + ku \quad \text{(Equation 2)}$$

From Equation 2, we can express the new mass $m’$ that the balloon must have to rise with constant velocity $u$:

$$m’g = B – ku \implies m’ = \frac{B – ku}{g}$$

Step 4: Determine the Constant of Proportionality ($k$)

The constant $k$ for air resistance is a property of the balloon’s shape and the medium (air), which are assumed to remain constant. We can determine $k$ from the initial scenario using Equation 1:

From Equation 1 ($Mg = B + kv$):

$$kv = Mg – B$$

Solving for $k$:

$$k = \frac{Mg – B}{v}$$

This expression for $k$ connects the initial known conditions to the property of air resistance.

Step 5: Calculate the New Mass ($m’$) by Substituting $k$

Now that we have an expression for $k$, we can substitute it into the equation for the new mass $m’$ (from Step 3). This will allow us to express $m’$ solely in terms of the given parameters ($M, B, v, u, g$).

We had: $$m’ = \frac{B – ku}{g}$$

Substitute $k = \frac{Mg – B}{v}$ into this equation:

$$m’ = \frac{B – \left(\frac{Mg – B}{v}\right)u}{g}$$

To simplify this complex fraction, multiply the numerator and denominator by $v$:

$$m’ = \frac{v \left(B – \frac{(Mg – B)u}{v}\right)}{gv}$$

$$m’ = \frac{Bv – (Mg – B)u}{gv}$$

Distribute the negative sign and $u$ in the numerator:

$$m’ = \frac{Bv – Mgu + Bu}{gv} \quad \text{(New mass of the balloon)}$$

Step 6: Calculate the Mass That Should Be Removed ($\Delta M$)

The problem asks for the amount of mass that must be removed from the balloon. This is the difference between the initial mass $M$ and the new mass $m’$ required for rising with velocity $u$.

$$\Delta M = M – m’$$

Substitute the expression for $m’$ we just derived:

$$\Delta M = M – \left(\frac{Bv – Mgu + Bu}{gv}\right)$$

To combine these terms, find a common denominator ($gv$) for $M$:

$$\Delta M = \frac{M(gv)}{gv} – \frac{Bv – Mgu + Bu}{gv}$$

Combine the numerators, being very careful with the signs when subtracting the second term:

$$\Delta M = \frac{Mgv – (Bv – Mgu + Bu)}{gv}$$

$$\Delta M = \frac{Mgv – Bv + Mgu – Bu}{gv}$$

Now, let’s group terms that contain $M$ and terms that contain $B$ in the numerator:

$$\Delta M = \frac{(Mgv + Mgu) – (Bv + Bu)}{gv}$$

Factor out common terms from each grouped pair:

$$\Delta M = \frac{Mg(v + u) – B(v + u)}{gv}$$

Notice that $(v + u)$ is a common factor in both terms in the numerator. Factor it out completely:

$$\Delta M = \frac{(Mg – B)(v + u)}{gv}$$

This is the final expression for the amount of mass that needs to be removed, stated in terms of the given parameters ($M, B, v, u, g$).

An equivalent form can be obtained by substituting $Mg – B = kv$ (from Equation 1) into the numerator:

$$\Delta M = \frac{(kv)(v + u)}{gv} = \frac{k(v+u)}{g}$$

Both forms are mathematically equivalent and correct. The first form uses only the initially given parameters, which is usually preferred as a final answer.