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Problem 36 Easy Difficulty

If we ignore air resistance, we can conclude that heavier objects fall no faster than lighter objects. But if we take air resistance into account, our conclusion changes. Use the expression for the velocity of a falling object in Exercise 35(a) to find $ dv/dm $ and show that heavier objects $ do $ fall faster than lighter ones.

Answer

$$
e^{c t / m}=\sum_{n=0}^{\infty} \frac{(c t / m)^{n}}{n !}
$$

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Video Transcript

caution given this velocities mg by C one minus It is two minus CTV i m and we have to show that velocity is dependent on mass. Okay, so that can be sown by differentiating velocity with respect to time. And if velocities function off em, then differentiation will remain. If not, the differentiation will be zero. Okay, so let's start with different stations so D V by d t Sorry, m so d by d m off m g o r c times one minus Syria's two minus city by m So you're G bisys constant. So this will be d by d m and we can multiply m or fine if you don't. Since either of these two cases will give us a method to use you interview so we can directly use that. So m times one minus serious to minus C t or am So we'll apply you into we rule So G o s c Now you're into these days you keep them out, that is you. Out differentiate we Plus you take out a V and differentiate them. Great! So G o s c. Now I am will remain as it is since it's in left hand Off left hand side in the left Off the by D m Now different station off one is zero The transition off areas two X is series two X into differentiation off X. In this case, we have to differentiate Setif. I m It's a change. All right. Okay, bless one minus. It is two minus city. Who would him and the transitional oneness Emmys one. Okay, great. So we are onwards to getting an idea if the we depends on him or not, since many times air remaining and we cannot make it zero. But let's complete it first and then we'll discuss it. So Okay, so m times zero minus. So we don't have to write that. So it is. It is minus. It is two minus city over him. Times different station off city by m So since cities constant, it will come out and different Station off one by MX minus one by M squared. And plus, here we have one minus. It is two minus e d a what? And and GCS as it is. Okay, so yeah, a little bit them back plane. So g c o S c. This gives us. We can cancel one off the M. So it's CD by M this minus. And this minus become plus society by m it is two minus C t I m okay. Plus one minus. It is two minus city or mm. Okay, So they said Devi by the now, If you see, we cannot eliminate this m here or this entire time is not going to go to zero. This is not equals to zero. Therefore, we is function off m and hence we depends on moss. Look to check if mass increase in agreement off mass, resulting the agreement off. More velocity or not, that this lighter mass lighter object falls faster or heavier Object for faster. We can see the term mass in the exponential terms which is decaying since it's minus and we know graph off areas to minus excess decaying one as the mass increases as the mass increases in cream int, what happens is it is two minus value increases. And since it is two minus value increases, it will drop faster. And since this value is coming at faster rate, when masses high it represent, that velocity will be higher. Okay. Thank you.