A recent discussion started by Lokesh S. dealt with conservation of momentum. This reminded me of a series of simple questions that confuses many beginning students.

1) An object is moving in some potential energy. Give me a potential energy function such that the total mechanical energy of the object is conserved but the momentum is not. (This should be straightforward, just think of a spherical cow sliding down a frictionless hill under the influence of gravity.)

2) Now, give me a potential energy function such that the momentum of the object is conserved but the total mechanical energy of the object is not.

3) What does this tell you about conservation laws and symmetry?

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## Comments

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TopNewestConsider the contrived potential, \(U(t,x)=t\).

The momentum of a particle in this potential will be conserved \(\left(\displaystyle\frac{dp}{dt} = F = \frac{dU}{dx} = 0\right)\) while the energy of the particle will increase with time \(\left(\displaystyle\frac{dE}{dt} = \frac{dU}{dt} = 1\right)\)

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Hi, if the mechanical energy of a system is not conserved, then ,

\(K + U \neq constant\)

Derivate with respect to x to get,

\(\frac{dK}{dx} + \frac{dU}{dx} \neq 0\)

Hence, \(F \neq \frac{-dU}{dx} , \Rightarrow \boxed{\frac{dU}{dx} \neq 0}\)(as\(F=0\)), but you are saying that \(\frac{dU}{dx} = 0\)

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Your first line does not actually imply the second line always...

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Nowhere ,i used it as a function of \(x\) only.

Say \(U(t,x) = t\) , then i would be saying \(\frac{dU}{dx} = \frac{dt}{dx} \neq 0\)(and actually it is not!)

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Josh is correct. Now, can anyone use this to answer part 3 of the question?

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In the second case we can sight the example of raindrops falling .. Initially they have a lot of energy due to the mgh (huge Height) whereas while coming down the whole of the potential energy is not converted into kinetic .. rather they fall with some terminal velocity (thus momentum can be conserved only after they have covered some distance) ..

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3) We should conserve energy when the work done by the external forces is 0. We should conserve momentum when no external (impulsive) force acts ..

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True, but there are non-conservative forces at work here. Just think of a single particle in a potential.

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I have a question: Can potential energy be assigned to a particle which is under the influence of both conservative and non-conservative force?

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Hey Santanu, nice example. I was wondering what would be the potential energy function for such a situation. Would it be U =

mgyor something else?Log in to reply

Yes U = mgy, but you can't say that it is a potential energy function such that the momentum is conserved but the mechanical energy is not, actually, here is an external work doing force which balances the change in potential energy such that the net change in kinetic energy is \(0\).

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Momentum conservation has to do with space symmetry and energy with time. For example, if \(V(t,x) = f(t)\), then \(-dV/dx = 0 = F = dp/dt\) which means momentum is not changing. Similarly, if \(V(t,x) = g(x)\) and \(T = p^2/2m\), then \(dE/dt = dV/dt + d/dt(p^2/2m) = (p/m)(dp/dt) = (p/m)F\) which is 0 as long as there aren't outside forces, right? I feel like I might have just begged the question.... If you transform the time or space variables but the potential doesn't change (symmetry), that will correspond to conservation. Maybe I need a fancier formulation of Newton's Laws to make the math work out?

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No, you simply need to realize that you just essentially rediscovered Noether's Theorem. Nicely done! :)

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1)

U = k*x(object falling from a certain height )2)

U = ???( I don't think it exists )3)

Momentum need not to be conserved when mechanical energy is conserved but when momentum is conserved, mechanical energy has to be conserved.Is that right?

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The reason for second answer: When momentum of the object is conserved, kinetic energy has to be conserved. Now, since the total mechanical energy is not conserved, the potential energy should also not be conserved. Consequently, this would imply a force is being applied on the object as \(U = -\int F dx\). This contradicts our first condition that momentum is conserved. This situation can't exist.

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If on a body which converts energy to mass (the converted mass adds to its own mass) a force is applied, then consider the equation p=mv , if m increases by a certain factor and v decreases by the same factor (most probably for this the force might need to be non constant) which cancels out then momentum is conserved. However kinetic energy is decreasing as 2KE=mv^2 , m increases by a certain factor and v decreases by the same factor twice thus KE decreases by that factor .. Please Comment

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I think I am wrong. As said by Santanu B, the second situation exists. The conclusion I think now is that

Momentum need not to be conserved when mechanical energy is conserved and symmetrically, mechanical energy need not to conserved when momentum is conserved.Right?Log in to reply

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For the given case 2,

\( \displaystyle F=\frac{dP}{dt}=0 \Rightarrow U=\text{constant} \)

This means that total energy must stay constant and hence, case 2 is not possible.

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\( F = \frac {dP} {dt} \),

F is the net force which is not necessarily the same F used to calculate U. Consider the example of rain falling down, the net force on the rain particle is 0 though there is potential energy associated due to gravitational force but not due to viscous drag, that's the flaw in your argument.

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Nope. It's the other way around for the second part of the third one. Imagine we have an object that can explode into two parts (suppose its a bomb) and each part moving in the opposite direction to one another. Here, clearly, the

mechanicalenergyisnotconserved; simply because initially the block is stationary and finally both blocks move with different velocities and each has a new kinetic energy. However, in all cases, including this one, the momentum isalwaysconserved. The momenta of the two parts are such that both of themcanceleach other out (as they move in opposite directions). The exact same laws would apply if, for example, two bodies stick together at impact and move with a common velocity there-on. In this case, mechanical energy isdestroyed, or maybe even completely destroyed! But, here momentum conservation still applies, the body would either move or not movedependingupon initial momentum states.Log in to reply

Yeah, I get it. Thanks.

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Hi, in the exploding of a bomb, there must be some repulsive forces inside an exploding object leading to a positive potential energy, getting converted to kinetic energy , heat, and sound.

This is the reason for the fact that a stone doesn't explode on its own.

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By this logic, all forms of energies whether it be chemical, magnetic, radiant or heat etc should be a part of potential energy. I am not defying your logic (rather I agree with it), but almost all these energies are due to their configuration of position and should be a part of potential energy. And since mechanical energy is sum of potential energy and kinetic energy, mechanical energy should consist of all forms of energy. But I haven't seen such a thing. Chemical, magnetic and all other forms of energies are not treated under mechanical energy.

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totalenergy of the system isalwaysconserved. But formechanicalenergy to be conserved ineverysituation, all the mechanical energy must be potential. Then, there will be no exception to the law of conservation of mechanical energy. However, as we have seen, kinetic mechanical energy can sometimes bedestroyed(meaning it is being converted to non-mechanical energy forms). Thus, mechanical energy needn't be conserved always.Log in to reply

1) \(U = f(x) \neq constant\), given that there is no non conservative dissipative force.

2)There is no such \(U\) in general, there must be a dissipative non -conservative force.

3) Momentum can be conserved in a direction in which there is no net external force acting on the system.Mechanical energy can be conserved if there is no non conservative force on the system

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The answer to 2 is incorrect.

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