🎉 Announcing Numerade's $26M Series A, led by IDG Capital!Read how Numerade will revolutionize STEM Learning Numerade Educator ### Problem 116 Easy Difficulty # 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.$113-117$### Answer ## (a)$a=\frac{g}{3}$(b)$\left.x\right|_{t=3}=14.7 \mathrm{m}$(c)$\left.m\right|_{t=3}=29.4 \mathrm{g}\$

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Moment, Impulse, and Collisions

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### Video Transcript

reference. This is the problem based on very well, my sister, as we know that when the mass of the system is changing with plank, then mo mentum is a function off mask as well as well is deep. So force can be defined as I am de via quantitative. Or it will be we do, um, upon it. So on this concept, this problem is based in the first part. It is given X into G skull, toe X D v. Upon duty, you can see in the problem. It is given, say equation one. Okay, The solution off this equation is we'd skull to 80. Really? Is cartu Do you be a quantity? Can you welcome? Yeah, the is goingto be except quantity Cross Multiply it so they exist. But toe be into DT from equation to we can write the excess Kal toh 80 into duty, integrating it for the limit. X knows toe X. That is 02 teeth. So you will get Yeah X minus X not is going toe half 80 square. This is foot even substituted for we from who do you be upon duty from three and x from full? We will get half. 80 square into jeep half 80 square in tow. Eight plus a square. T square. So this is three by two Beauty square. Okay, so from this equation, you will get a stir toe G by three. Right now be Barry. Yeah, for please call 23 seconds and a any depression, for we will get X at a please call 23 seconds off 9.8 by three in 23 square. So that is 14 point seven victims. Yeah. Oh, see part. We know that the relation between them mass off raindrop and the distance right? Okay, that it has fallen is given by and it's got to be x here, k having developed to grandma permitted. So for X p Y t in tow fight. We get the mosque off. Drug must off the drop at ease, but 23 2nd between two gram per second in tow. 14.7 meter. That is 29 4th graf. There's all thanks for watching it

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#### Topics

Moment, Impulse, and Collisions

##### Andy C.

University of Michigan - Ann Arbor

##### Farnaz M.

Simon Fraser University

##### Aspen F.

University of Sheffield

##### Jared E.

University of Winnipeg

Lectures

Join Bootcamp