If two objects, one light and one heavy but otherwise identical, are given the same impulse from rest and begin to travel horizontally through a resistive fluid, which object makes it farther?
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$\begingroup$ Do the objects have the same dimensions, so that they experience the same drag force? $\endgroup$– Michael SeifertMar 1, 2022 at 22:39
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$\begingroup$ @MichaelSeifert Yes, edited to reflect that. $\endgroup$– EtheraxMar 2, 2022 at 1:34
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1 Answer
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This feels a bit "homework-like", and there are also some assumptions missing, so I'll just point you in the direction I would go to solve this problem and let you fill in the blanks yourself.
Assuming that the drag force is a function of velocity, $F_\text{drag} = -f(v)$, we have $$ m \frac{dv}{dt} = - f(v) \quad \Rightarrow \quad m \frac{dv}{dx} v = - f(v) $$ which is separable and yields $$ \Delta x = m \int_0^{v_0} \frac{v \, dv}{f(v)} $$ where $v_0$ is the initial velocity of the object.
From here, the answer will depend on the exact dependence of the drag force on the velocity. The case where the drag force is strictly a linear function of $v$ (i.e., pure viscous drag) leads to a particularly elegant result. But for cases where inertial drag as well as viscous drag need to be taken into account, the relationship between the initial impulse and the distance traveled will not be as straightforward.
answered Mar 2, 2022 at 14:36
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$\begingroup$ If you really want to know the context, there's an old anime called Dragon Ball where a character throws a stone pillar then jumps forward to ride it. I wanted to know if that was actually a valid way of traveling farther as opposed to just applying the same impulse to yourself by jumping. I tried to solve it with Newtonian drag but ran into issues, and Stoke drag didn't seem valid for high speeds. Thanks for the response. $\endgroup$– EtheraxMar 2, 2022 at 15:01
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$\begingroup$ @Etherax: You can use the above equation to show that the distance traveled is infinite for "pure" inertial drag, $f(v) \propto v^2$. A more physically reasonable approximation would be to use a combination of viscous and inertial drags, $f(v) = \alpha v + \beta v^2$ for some coefficients $\alpha$ and $\beta$. In this model, the viscous drag takes over once the object slows down enough and cures the logarithmic divergence that would happen with pure inertial drag. $\endgroup$ Mar 2, 2022 at 15:18
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$\begingroup$ Also, as far as the cartoon physics goes, it sounds like the character is providing two impulses to the system consisting of themself and the stone pillar (one when they throw the pillar, one when they jump forwards). So that might well yield a longer distance on its own. $\endgroup$ Mar 2, 2022 at 15:20