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In a rocket-propulsion problem the mass is variable. Another such problem is a raindrop falling through a cloud of small water droplets. Some of these small droplets adhere to the raindrop, thereby increasing its mass as it falls. The force on the raindrop is $$F_{\mathrm{ext}}=\frac{d p}{d t}=m \frac{d v}{d t}+v \frac{d m}{d t}$$ Suppose the mass of the raindrop depends on the distance \(x\) that it has fallen. Then \(m=k x,\) where \(k\) is a constant, and \(d m / d t=k v\) . This gives, since \(F_{\text { ext }}=m g .\) $$m g=m \frac{d v}{d t}+v(k v)$$ Or, dividing by \(k\) $$x g=x \frac{d v}{d t}+v^{2}$$ This is a differential equation that has a solution of the form \(v=a t,\) where \(a\) is the acceleration and is constant. Take the initial velocity of the raindrop to be zero. (a) Using the proposed solution for \(v,\) find the acceleration \(a\) . (b) Find the distance the raindrop has fallen in \(t=3.00 \mathrm{s}\) . (c) Given that \(k=2.00 \mathrm{g} / \mathrm{m}\) , find the mass of the raindrop at \(t=3.00 \mathrm{s}\) . For many more intriguing aspects of this problem, see \(\mathbf{K} .\) S. Krane, Amer Jour. Phys, Vol. 49\((1981)\) pp. \(113-117\)

Step by step solution

01

Substitute the solution form for velocity

According to the problem, the solution form for velocity is given by \(v = at\) where \(a\) is constant acceleration. Substitute \(v = at\) into the differential equation \(xg = x \frac{dv}{dt} + v^2\).

02

Express velocity and acceleration terms

Since \(v = at\), \(\frac{dv}{dt} = a\). Substitute these into the differential equation: \[xg = xa + (at)^2\] which simplifies to \[xg = xa + a^2 t^2\].

03

Relate distance and time

Remember that \(x = \frac{1}{2}at^2\) for an object starting from rest under constant acceleration, and replace \(x\) in the equation from Step 2: \[\left(\frac{1}{2}at^2\right)g = \left(\frac{1}{2}at^2\right)a + a^2 t^2\].

04

Simplify and solve for acceleration

Rewriting the equation \[\frac{1}{2}agt^2 = \frac{1}{2}a^2t^2 + a^2t^2\], this simplifies to \[\frac{1}{2}ag = \frac{3}{2}a^2\]. Cancel \(a\) (assuming \(aeq0\)) and solve for \(a\): \[g = 3a \Rightarrow a = \frac{g}{3}\].

05

Calculate the fallen distance in 3 seconds

Using \(a = \frac{g}{3}\) and \(x = \frac{1}{2}at^2\), calculate \(x\) when \(t = 3.00\,\text{s}\). Substitute \(a\) and \(t\) into the equation: \[x = \frac{1}{2} \left(\frac{g}{3}\right)(3^2)\]. With \(g \approx 9.81\,\text{m/s}^2\), \[x \approx \frac{1}{2}\left(\frac{9.81}{3}\right)(9) \approx \frac{1}{2}(32.67)\approx 16.34\,\text{m}\].

06

Calculate the mass of the raindrop at t = 3.00 s

Given \(m = kx\) and \(k = 2.00\,\text{g/m}\), substitute \(x = 16.34\,\text{m}\) from the previous step into the mass equation: \(m = 2.00 \times 16.34 = 32.68\,\text{g}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equations

Differential equations are mathematical tools used to describe how things change over time or space. In the context of variable mass systems, like a raindrop collecting water droplets as it falls, differential equations can describe the change in mass and velocity. The key equation in this problem is

• \(F_{\mathrm{ext}} = m \frac{dv}{dt} + v \frac{dm}{dt}\)
• This equation considers the external force acting on the raindrop and how its momentum changes due to its velocity \(v\) and the rate of mass change \(\frac{dm}{dt}\).
• In our raindrop problem, the mass \(m\) depends on the distance \(x\) fallen, represented by \(m = kx\), where \(k\) is a constant.

This setup allows us to define the movement of the raindrop through the differential equation

• \(xg = x \frac{dv}{dt} + v^2\)

By solving this equation, we can derive relations for velocity and acceleration as the system evolves over time.

Rocket Propulsion

Rocket propulsion problems often deal with systems where mass changes. Just like with raindrops, understanding how a rocket's mass changes as fuel burns is crucial. Rocket propulsion deals primarily with two equations: the conservation of momentum and the Tsiolkovsky rocket equation. In these systems, the change in mass impacts how force is applied.

• In the raindrop problem, mass change due to adhered droplets is similar to a rocket losing mass as fuel burns.
• The mass change rate \(\frac{dm}{dt}\), plays a key role, affecting the velocity \(v\) and subsequently the acceleration \(a\).

In both rockets and raindrops, calculating how mass influences acceleration helps predict their motion. Understanding these principles allows engineers to design efficient propulsion systems for space travel and helps meteorologists forecast weather related phenomena.

Acceleration Calculations

Acceleration is the rate at which an object's velocity changes. For the raindrop problem, calculating the acceleration involves understanding that it depends on gravitational force and the change in mass over time. From our differential equation, we can deduce that

• \(a = \frac{g}{3}\)

This shows that, under the given conditions, the raindrop accelerates at one-third of the gravitational pull of Earth.

• Using this constant acceleration, we can calculate the distance \(x\) fallen using the formula \(x = \frac{1}{2}at^2\).
• After 3 seconds, the raindrop falls approximately 16.34 meters.

These calculations not only show how far the raindrop falls but also determine the mass it accumulates, based on the given constant \(k = 2.00 \, \text{g/m}\).