I have been working on a problem for some time but haven't found a solution yet. I am mainly looking for a hint that will help me find a breakthrough and not a full solution. Here is the statement of the problem:

A small block with mass $m$ is sitting on a large block of mass $M$ that is sloped so that the small block can slide down the larger block. There is no friction between the two blocks, no friction between the large block and the table, and no drag force. The center of mass of the small block is located a height $H$ above where it would be if it were sitting on the table, and both blocks are started at rest (so that the total momentum of this system is zero, note well!)

a) Are there any net external forces acting in this problem? What quantities do you expect to be conserved?

b) Using suitable conservation laws, find the velocities of the two blocks after the small block has slid down the larger one and they have separated."

The source of the problem is here.

For part a), I think that energy is conserved, but I am not sure about momentum conservation, as gravity seems to be acting unopposed and thus there are net, external forces on the system. I know that the initial energy of the system is $E_0=mgH$, and the final energy after the blocks have separated should be $E_f=\frac{1}{2}v_M^2+\frac{1}{2}mv_m^2$. We thus have $$mgH=\frac{1}{2}v_M^2+\frac{1}{2}mv_m^2$$ We need one more equation relating $v_m$ and $v_M$ to find a solution to the problem. Would this be conservation of momentum? The problem statement also talk about the center of mass of the small block, but I am not sure how to use it to solve for the final velocities of the masses.

Thank you for your help.


2 Answers 2


You are correct in saying that momentum is not conserved here. Certainly the only horizontal forces involved here are internal, so the horizontal component of the momentum is conserved, and hence it remains $0$. However, the vertical component of the total velocity is not conserved. You can easily see how it is not. As the top block slides down its vertical velocity changes. But the bottom block keeps its $0$ vertical velocity due to the table. Therefore the total vertical momentum has changed, and it is not conserved.$^*$

You are also correct in saying that energy is conserved here. There are no dissipative forces here (The normal forces between the two blocks actually do equal but opposite work here, so the total energy stays conserved).

Therefore you have two conserved quantities and therefore two equations to solve for your two unknown values.

$^*$One might argue that since the vertical momentum after the top block reaches the table is equal to the initial vertical momentum (both $0$) that momentum is conserved. However, this is not the definition of conserved. Conserved means it does not change for all times. If this was the definition of conserved then you could say, for example, that "position is conserved" for riders of a roller coaster since they start and end at the same position.


Yes conservation of momentum provides you with another equation. MV = mv. Or velocity of small object = v = MV/m. Momentum is a vector quantitity and was initially zero. Hence the final momentum is zero as well. Since masses are constant, the velocities must be in opposite directions to yield zero.

  • $\begingroup$ The final momentum is not $0$ $\endgroup$ Commented Jul 25, 2019 at 1:51
  • $\begingroup$ the final momentum is one body sliding horizontally to right and other block sliding horizontally to the left. Momentum is conserved in this problem. There are no frictional forces. the velocities depend on the ratio of the masses... $\endgroup$
    – Natsfan
    Commented Jul 25, 2019 at 2:55
  • $\begingroup$ @AaronStevens yes final momentum is zero. See my comment above. $\endgroup$
    – Natsfan
    Commented Jul 25, 2019 at 3:09
  • $\begingroup$ You are only considering the horizonal component of the momentum. The vertical component changes here as the top block slides down the bottom one $\endgroup$ Commented Jul 25, 2019 at 3:50
  • $\begingroup$ I suppose we are using different ideas of final. I was thinking in terms of when the top block reaches the bottom of the other block. You are thinking in terms of when the top block is sliding along the table? $\endgroup$ Commented Jul 25, 2019 at 4:12

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