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I get Momentum and Impulse as well as Work and energy but struggle when it comes to connecting the two ideas. I understand that an objects KE can change without its momentum changing, Like in a inelastic collision were the KE of the system decreases but the amount of matter moving to left with some speed - the amount of matter moving to the right with some speed will be equal before and after the collision. Hence Momentum is Conserved. My confusion starts with the question is an impulse of a force on an object always accompanied by some work done on the object by that force?

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Impusle is a property defined as: $$\vec{I} = \int \sum \vec{F}.dt$$ and by virtue of this definition, it need not always be accompanied by work.

Consider an example of you standing on a skateboard and pushing on a wall setting yourself in motion away from the wall.

In this case, the force from the wall on your hands moves through no displacement; the force is always located at the interface between the wall and your hands. Hence, no work is done on the system of you and the skateboard. (The resulting kinetic energy however comes from the potential energy stored in your body).

The impulse-momentum theorem will suggest that: $$\Delta \vec p_{tot} = \vec I = m \Delta \vec v = \int \vec F_{\text{wall}}.dt$$ where $\vec F_{\text{wall}}$ is the force exerted by the wall on your hands. Hence, the impulse does not result in a work.

Hope this helps.

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  • $\begingroup$ @MariusLadegårdMeyer, Not really. Realistically speaking, any wall would deform to a certain degree. $\endgroup$
    – Cross
    Commented Oct 1, 2021 at 16:01
  • $\begingroup$ If there were a net force on the wall it would accelerate. $\endgroup$
    – Bob D
    Commented Oct 1, 2021 at 16:02
  • $\begingroup$ Ok so what your saying is I push on the wall but because it doesn't move, But if the surface of the wall and my hand heat up there must be some transform of energy happing and therefore some amount of work right? $\endgroup$
    – CatsOnAir
    Commented Oct 1, 2021 at 16:10
  • $\begingroup$ @MariusLadegårdMeyer, I completely edited the answer, and made it more detailed $\endgroup$
    – Cross
    Commented Oct 1, 2021 at 16:13
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    $\begingroup$ @CatsOnAir, My answer talked about how the wall did no work on the system of you on a skateboard, but it resulted in an impulse. This is an example of a deformable system, where work need not be done, but there can be a change in kinetic energy. As Peter points out, there may also be energy transfer by heat, resulting in an increase in the internal energy of the wall ( This is a thermodynamic property, you may look it up). And yes, your generalization is also correct, but that's not the example I used. $\endgroup$
    – Cross
    Commented Oct 1, 2021 at 16:43
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The standard example of a force doing no work is when an object is moving in a circle at constant speed - the force points towards the centre of the circle. Over 180 degrees there is a non-zero impulse; the direction of movement has changed so the momentum vector has changed. However the speed has not changed, so no work has been done.

It is actually impossible to change the kinetic energy of a single object without changing its momentum, because if you change the kinetic energy you must change the velocity. In an interaction between more than one object the total momentum stays fixed though, even if the total kinetic energy changes.

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  • $\begingroup$ See this makes scenes to me because (And might be wrong about this statement and please tell me if I am) That work is the only method in which energy can be transferred. In your example the energy of the ball isn't changing so therefor no work is being done and it makes sense to me but for example when pushing on a wall the wall doesn't move but energy is transferred for the surface I'm pushing on heats up and I get confused again. $\endgroup$
    – CatsOnAir
    Commented Oct 1, 2021 at 16:15
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    $\begingroup$ If you just lean on a wall there is no work done, and no momentum change. The wall might get warm from the heat of your hand, but that is not mechanical energy, and it is not the result of the force, only of the contact. $\endgroup$
    – Peter
    Commented Oct 1, 2021 at 16:21

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