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A wedge of mass $M$ rests on a smooth horizontal surface. The face of the wedge is a smooth plane inclined at an angle α to the horizontal. A particle of mass $kM$ slides down the face of the wedge, starting from rest. At a later time $t$, the speed of the wedge is $V$, the speed of the particle $v$ and the angle of the velocity of the particle to the horizontal β.

The solution states that conservation of energy gives

  • $mgy = {m \over 2} v^2 + {m \over 2} V^2$.

I don't understand how intuitively this is true. I would have thought, for the particle, $mgy = {m \over 2} v^2$ and then for the wedge the horizontal force on it would be the force doing the work and be the thing producing ${m \over 2} V^2$. So the equation I get would be similar to • but the left hand side of the equation would have the work done by the horizontal force on the wedge.

I have only ever done conservation of energy with a single particle, never when two are interacting, so sorry if the question is confused.

Any help would be appreciated!

David G. Stork
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milanios
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  • Conservation of energy is a property of a system. So the sum of total energy of the particle and wedge remains a constant, which is what the equation says. – Macavity Aug 24 '15 at 10:01
  • Individually, the wedge increases in energy (kinetic) while the particle loses the same energy - its loss in potential energy is more than what it gains in kinetic. To solve the system you will also need conservation of momentum. – Macavity Aug 24 '15 at 10:03

2 Answers2

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At the instant before things start moving, the wedge is on the horizontal surface. At time $t$ the wedge is still on the horizontal surface. So the wedge is at the same height in the "before" and "after" pictures, therefore it has not gained or lost any potential energy. But its kinetic energy "before" was zero, and now it is $\frac 12 MV^2$. So where did that kinetic energy come from?

The answer is that the kinetic energy came from the force of the particle against the inclined surface of the wedge. Since that force is not perpendicular to the horizontal surface, it pushes the wedge aside.

But in pushing the wedge aside, the particle does work, thereby expending energy. Where did it get that energy? It got the energy by falling, that is, it used some of its potential energy to push the wedge. The energy it used to push the wedge is no longer available to be invested in the particle's own kinetic energy.

That's what we mean by conservation of energy. If you observe kinetic energy somewhere in the system at time $t$, it had to come from somewhere where it existed at time zero.

David K
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  • Thanks for the reply, but what I don't understand is why the work done by the reaction forces on each object all cancel out to zero, leaving just the work done by the weight of the particle.. – milanios Aug 24 '15 at 19:29
  • What do you mean by "work done by the reaction forces"? Can you give an example of a reaction force and the work it does? – David K Aug 24 '15 at 19:59
  • The particle exerts a horizontal force of Rsin(α) on the wedge, so I would have thought the work done there would equal Rsin(α)d, where d is the distance moved by the block. equal and oppositve puts the same R force on the particle, which I would have thought does no work since it is perpendicular to the motion of the particle. So why is the Rsin(α)d component of work done on the block not anywhere in the equation of energy conservation? – milanios Aug 24 '15 at 20:10
  • First, the particle moves sideways; the horizontal force is not perpendicular to its motion except at the very start. Second, the horizontal forces are caused by the force of gravity pushing the particle down on the wedge. The work done by those forces is not ignored; it's just part of the work done by gravity, which is what $mgy$ represents. If you pull a rope to lift a weight, you can say you do work on the rope, or that the rope does work on the weight, but it's not right to add those two quantities of work. They're really the same thing measured in two different ways. – David K Aug 25 '15 at 01:32
  • Sorry for the late reply. I understand what you're saying now I think. The only bit which I don't get is this bit. 'the particle moves sideways; the horizontal force is not perpendicular to its motion except at the very start.' While the particle moves down the wedge is the reaction force from the wedge not always perpendicular to the plane and therefore perpendicular to the motion of the particle? – milanios Aug 27 '15 at 17:34
  • At the start of all movement, the slope of the particle's motion is exactly equal to the slope of the wedge. But at every time after that, the wedge is moving to the right. As a result, the particle is not pushed as far to the left as it would have been if the wedge were stationary; and the particle ends up following a path that is sloped steeper than the wedge. It's not the easiest thing to visualize! – David K Aug 27 '15 at 20:03
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Since at the initial condition, the particle & the wedge are in the state of rest (both have zero velocity)

Now, taking the horizontal line as the reference line,

The total energy of the system (particle +wedge) in initial condition is $$E_1=[\text{P.E. (potential energy) of particle}+\text{K.E. (kinetic energy) of particle}]+[\text{P.E. (potential energy) of wedge}+\text{K.E. (kinetic energy) of wedge}]$$ $$=[mgy+0]+[0+0]=mgy$$ Where, $y$ is the vertical height of particle from the horizontal (reference) line in initial state of rest.

Now, at any time $t$, the particle with mass $m$ is moving with the velocity $v$ & the particle with mass $M$ is moving with the velocity $V$

Hence, the total energy of the system (particle +wedge) at the time $t$ is $$E_2=[\text{P.E. (potential energy) of particle}+\text{K.E. (kinetic energy) of particle}]+[\text{P.E. (potential energy) of wedge}+\text{K.E. (kinetic energy) of wedge}]$$ $$=\left[0+\frac{1}{2}mv^2\right]+\left[0+\frac{1}{2}MV^2\right]=\frac{1}{2}mv^2+\frac{1}{2}MV^2$$

Now, neglecting the losses, the total energy of the system (particle +wedge) by must be constant according to the law of conservation of the energy hence

$$E_1=E_2$$ $$mgy=\frac{1}{2}mv^2+\frac{1}{2}MV^2$$

Harish Chandra Rajpoot
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