I was recently asked the following question in a physics contest:

For some odd reason, you decide to throw baseballs at a car of mass $M_i$, which is free to move frictionlessly on the ground. You throw the balls at the back of the car at speed $u$, and at a mass rate of $σ$ kg/s (assume the rate is continuous, for simplicity). If the car starts at rest, find its speed and mass as a function of time, assuming that the balls rest inside the car after they colide with it.

I found out after the contest that this is a variation of a problem posed by David Morin in his book of mechanics, and adressed in this question. But the change of the problem has a reason to be as it introduces complexity, since the mass changes over time. At the time I answered with a somewhat unfit aproximation, which I will provide later. This is my atempt at solving it. It gives rise to an ODE that I can't solve, and I believe there is an analitical solution based on what the problem makers said to me.

After a colision of some infinitesimal mass $m$ with the car with mass $\small M(t)$, by conservation of momentum and dividing by some infinitesimal $ \Delta t$ we have that

$$\normalsize v_f\cdot (M(t)+m) = v_i \cdot M(t) + u\cdot m \Leftrightarrow \\ \Leftrightarrow(v_f-v_i) / \Delta t = \frac {m(u-v_i)}{M(t)+m} / \Delta t \Leftrightarrow \\ \Leftrightarrow v'(t) = M'(t) \frac{u-v(t)}{M(t)}$$

Since $\small m/\Delta t = M'(t)$. To relate the derivative of the mass with the velocity we use a sort of "dopler efect" of the mass rate of the font. We then have:

$$M'(t)=\frac {u-v(t)}{u} σ $$

Which gives us two ODEs to solve one. If we replace $M(t)$ and $M'(t)$ as funtions of the position $x(t)$ of the car, we get, if my algebra is correct:

$$x''(t) = (u-x'(t))^2 \ \frac {σ}{u} (σt - \frac {σ}{u} x(t) + M_i)$$

Which looks absolutly hideous. Differentiating the former equation for the derivative of the mass in respect to the velocity, we get that $-M''(t) \frac {u}{σ} = v'(t)$ and so: $$M''(t) = - (M'(t))^2 M(t) $$

Which looks much more manageable. Even so, I have no idea how to solve this. Using wolprham alpha I get as an answer some expression envolving the inverse error function and some complex numbers, that I can't use to get the velocity.

P.S. My resolution for the contest envolved assuming $M_i $>>$ \sum m $, and so $ M(t) \approx M_i$ Then: $$v'(t) = (u-v(t))^2 \frac{σ}{(u M_i)} $$ which has the solution (knowing that $v(0)=0$) $$v(t)= \frac {σut}{σt+M_i}$$ That surprisingly converges to u. (Surprising not because it's wrong, it's correct, but I would believe this aproximation would only make sence for small t.) And then the mass has equation: $$ M(t)=M_i(1+ln(1+σt/M_i))$$ Which makes no sense for large $t$.

P.P.S. After Mark H.'s answer we can solve the diferential equation for the velocity and we get: $$v(t)=\frac{-\sqrt 2\sqrt {a^2 k + at} + 2aku+2tu}{2(ak+t)}$$ For some $k$ and $a=M_i u^2/σ$. Knowing that $v(0)=0$ we get that $k=1/2u^2$ and so $$v(t)=u(1-\frac{\sqrt b}{\sqrt{2t+b}})$$ Where $b=M_i/σ$. It's a very clean solution, and one can solve the remaning elementary intregral to calculate the expression for the mass.

  • $\begingroup$ Re, "assume the rate is continuous, for simplicity." IDK. If you treat each baseball as a discrete event, then the velocity of the car is a function of the number of baseballs thrown. I'm guessing, a 2nd degree polynomial with coefficients that depend on the mass of the car, the mass of a baseball, and the velocity with which the balls are thrown. You should be able to infer them by inductive reasoning about each next ball thrown. Whether that is more or less simple than assuming a continuous stream of baseball-stuff, and solving differential equations is a matter of opinion, I guess. $\endgroup$ – Solomon Slow Feb 21 at 19:17
  • $\begingroup$ To see why the asymptotic value is $u$, look at the problem from the "car" reference frame. $\endgroup$ – Jerry Schirmer Feb 21 at 19:36
  • $\begingroup$ @SolomonSlow, Not a function of the number of baseballs thrown, but of the number that have hit the car. Each ball takes longer to reach the car than the previous ball. $\endgroup$ – S. McGrew Feb 21 at 19:40
  • $\begingroup$ @S.McGrew. Ha ha! Foolish of me to assume that not all of the balls would hit the car. Also, I'm assuming that they all would stick to the car, as specified in the OP's problem statment. It would start to get ugly if you were given statistical distributions for the elasticities and the rebound angles of the collisions. $\endgroup$ – Solomon Slow Feb 21 at 19:44
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    $\begingroup$ Since the assumption $M(t) \approx M_i $ gets more and more incorrect as time goes by. $\endgroup$ – RicardoMM Feb 21 at 20:39

Part of the strategy for solving these problems is not trying to solve for more than the question asks. The problem only asks for velocity as a function of time, not position, so there's one integral we don't need to do. We already need to do two integrations since we have $v'(t)$ and $M'(t)$, so let's not add more.

We'll start from $$v'(t) = M'(t) \frac{u-v(t)}{M(t)}$$ By rearranging and using the product rule, we can get one integration over and done with easily: $$M(t)v'(t) + M'(t)v(t) - M'(t)u = 0$$ $$\frac{d}{dt}\left[M(t)v(t)\right] - M'(t)u = 0$$ $$\frac{d}{dt}\left[M(t)v(t) - M(t)u\right] = 0$$ Integrating: $$M(t)v(t) - M(t)u = k$$ where $k$ is some constant. Using $M(0) = M_i$ and $v(0) = 0$, we get $k = -M_iu$ and $$M(t)\left[v(t) - u\right] = -M_iu$$ That's one integration down.

Now, we don't care about $M(t)$, so let's get rid of it. $$M(t) = -\frac{M_iu}{\left[v(t) - u\right]}$$ $$M'(t) = \frac{M_iu}{\left[v(t) - u\right]^2}v'(t)$$ Combining this with the "Doppler" equation $$M'(t)=\frac {u-v(t)}{u}\sigma$$ yields $$\frac{M_iu}{\left[v(t) - u\right]^2}v'(t) = \frac{u-v(t)}{u}\sigma$$ or $$\left(\frac{M_iu^2}{\sigma}\right)v'(t) = \left[u-v(t)\right]^3$$ Notice that everything here is a constant except for $v(t)$ and $v'(t)$. This is a tractable differential equation.

  • $\begingroup$ Great answer! I guess that's they had in mind. $\endgroup$ – RicardoMM Feb 21 at 22:50

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