I have 2 cubes where one mass is greater than the other. $M$ and $m$, where $M>m$, and there is a hill that is symmetrical on both sides, and has a friction factor of $k$ between the object and the surface.

If we place both objects on one side of the hill, point 1, that has no slope, so it is flat, and give both of the masses the same velocities in the direction of the top of the hill so they finally land on the other side, on the flat area, point 2 (symmetrical to the point 1).

How can I prove that the speed of the heavier cube is smaller than the lighter one in the point 2? (If I'm correct) I tried equating $E_{kinetic}$ and $E_{potential}$... I see there is work from the friction, but whenever I try anything I cross out the mass.

Heres a diagram: http://www.wolframalpha.com/input/?i=sin%5E2(x)%2C+0%3Cx%3Cpi&x=0&y=0

Where point 1 is at (0, 0), and point 2 at (pi, 0).

What I have so far: where $g$ is constant of gravity and $k$ friction factor, and $d$ distance.

$E_{kinetic.start} - W_{friction} = E_{kinetic.end}$ $M(\frac{v_0^2}{2}-gkd)=M(\frac{v_1^2}{2})$

I could change $M$ with $m$ and nothing would change, they'd both have same speeds...

So it's obvious that the masses cancel out and the energy is proportionally lower at both masses, the distance for friction is the same and starting velocities are the same, therefore the velocity should be the same for both bodies at the end. This confuses me as I believe the smaller body should be faster.

(Cant noone help me with this? Im now doubting the solution i've provided first is right. I cannot prove otherwise.)

  • 2
    $\begingroup$ Please provide an image of the scenario. $\endgroup$
    – guntbert
    Commented May 29, 2014 at 20:12
  • $\begingroup$ wolframalpha.com/input/… where point one would be (0, 0) and point 2 (pi, 0). And curve represents the hill. $\endgroup$
    – user30948
    Commented May 29, 2014 at 20:29
  • $\begingroup$ For a start: omit the hill - what happens energetically (with friction), when you run that experiment on a flat surface? $\endgroup$
    – guntbert
    Commented May 29, 2014 at 20:37
  • $\begingroup$ The energy is lost due to work of the friction? I would equate $E_{starting.kin}-W_{friction}=E_{ending.kin}$ and again I can cross out the masses so I'm left with nothing. $\endgroup$
    – user30948
    Commented May 29, 2014 at 20:46
  • $\begingroup$ Don't forget: the bodies have different $E_{kin}$ from the beginning. $\endgroup$
    – guntbert
    Commented May 29, 2014 at 20:52

2 Answers 2


Your line of thought and the way you are arguing in the second part is correct.

Lets separate the task: one with friction but on a flat surface (no hill), which you already did (1) and one without friction but with the hill on the way (2).

1. Friction, flat surface

I am disregarding energy here - I concentrate on velocity and deceleration. Both bodies start with $v_0$. Both bodies are decelerated due to the frictional force $F_{FR}$.

$F_{FR,1}$ = $M \cdot g \cdot k \Rightarrow a_1= {F_{FR,1} \over {m_1}} = g \cdot k$, (the same applies for the smaller body)

So the deceleration due to friction is again independent from the mass of the bodies. Both bodies will have the same velocity at every given time.

2. No Friction, hill

You are well aware that without friction both bodies must have constant overall mechanic energy throughout the whole process. Lets look at what this means for their velocity at the hilltop (I am using $_0$ to denote values at the beginning, $_{top}$ for the hilltop, $h$ for gained height, $m$ for (general) mass:

$E_{kin, top} = m \cdot {{v_{top}^2} \over 2}= E_{kin,0} - E_{pot}= { {m \cdot v_0^2} \over 2} - m \cdot g \cdot h = m \cdot \left( \frac {v_0^2} {2} - g \cdot h \right)$

Once more the mass cancels out and (without really calculating $v_{top}$) we can see that both bodies will have the same velocity at the top of the hill (and everywhere else between).

So in both cases we see that the velocity of the body is completely indepenent of its mass. Your intuition led you astray :-)).

Please keep in mind that we did not take air resistance into the calculation.

  • $\begingroup$ I actually imagined the hill as a stronger type of friction and therefore velocities should stay the same. I wrote the opposite because my book said so, not because it was my intuition. ;) Danke schön! $\endgroup$
    – user30948
    Commented May 30, 2014 at 10:37

This is a question, where I feel that your intuition is going against your modelling of situation. I believe that your calculations are correct and the modelling is approximately giving you that the velocities should be same.

However, is there any flaw in your arguments? Let's analyse step by step.

Is the mass of the bodies different? Yes. Is the hill really a curve of $\sin^2 x$ ? Okay, it can be. Is the coefficient of friction same for both of them, really? No, not really.(it depends upon a lot of things,usually we assume it to be approximately constant, but then again, we get approximate result as this question shows! )

I think this is the reason I believe that the intuition is going against the physical modelling.

To support my argument: Have you heard about the Galileo's experiment with falling objects? If yes, I would like to state the result in a bit different way. Two particles, passing through same force field, should end up with the same velocity, irrespective of their mass. In that experiment, the drag or air resistance was the one which causes the difference in velocity, which is similar to friction here.

This is my suggestion to the problem. Please correct me if I'm wrong!

  • $\begingroup$ I improvised on the curve since I didnt want to draw imperfectly, so I took that function... but it doesnt really matter. Second of all, friction coeffitient is the same for all in all times. Because we imply the hill is made of the same material in any point and so is with the bodies. I found in another book that the velocity is bigger on the lighter body thats what had supported my main idea. All physics (i did) confirms is that they will have the same velocity at the end. Thats why im confused. $\endgroup$
    – user30948
    Commented May 30, 2014 at 7:42
  • $\begingroup$ @user2684291 Then I don't think that the velocities would change. I guess your calculation is correct. I think I should read the book you are mentioning to get exactly what the author wants to say. Can you name the book please? $\endgroup$
    – puru
    Commented May 30, 2014 at 13:52
  • $\begingroup$ Yes, Nada Brković, Planinka Pećina - Fizika u 24 lekcije, 6th lesson, 26th assignment. The book doesnt provide an explanation, only the answers at the back. The book is in croatian language and is a prep for our "matura exam in physics." Ive found lots of wrong answers like this one so its fine. $\endgroup$
    – user30948
    Commented May 30, 2014 at 15:27

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