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The following is a fairly straightforward problem about momentum conservation in inelastic collisions (I am only interested in part (b) of the exercise). The question asks whether the kinetic energy of the system of wagon and persons changes due to the collision. The answer is that yes, the kinetic energy changes, because x-momentum is conserved. However, I am confused about where exactly the energy goes, or which force is doing work. It must be a static friction force, but does static friction dissipate energy? Problem description

The solution goes as follows. Conservation of x-momentum yields $$ Mv_{1x} = (M + m)v_{2x}, $$ where $M$ is the mass of the cart and $m$ is the combined mass of the two persons. Thus, the final kinetic energy is \begin{align} K_2 &= \frac{1}{2} (M + m) v_{2x}^2\\ &= \frac{1}{2} \frac{M^2}{M + m}v_{1x}^2\\ &= \frac{M}{M + m} K_1, \end{align} where $K_1 = \frac{1}{2}Mv_{1x}^2$ is the initial kinetic energy.

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    $\begingroup$ physics.stackexchange.com/questions/568455/… $\endgroup$
    – BowlOfRed
    Commented Jun 5, 2023 at 7:57
  • $\begingroup$ @BobD I think this can be neglected because to slow them down a friction force (e.g. in their knees) only needs to act vertically, so it does no work in the x-direction. In the official solution they just assume they had a height of ≈0 so no potential energy. $\endgroup$ Commented Jun 5, 2023 at 12:24

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Kinetic energy is not a conserved quantity. Only the momentum of an isolated system. In this example kinetic energy is lost due to the collision between the heroes and wagon being inelastic. Both the heroes and the wagon/gold "stick together" following the collision moving together with the same final velocity.

Hope this helps.

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  • $\begingroup$ I don’t know why this is downvoted because KE is not conserved in an inelastic collision. The velocity changes, and is non-linearly related to KE = 1/2 mv^2. $\endgroup$ Commented Jun 5, 2023 at 13:36
  • $\begingroup$ @ErolBakkalbasi I didn't see a downvote $\endgroup$
    – Bob D
    Commented Jun 5, 2023 at 13:39
  • $\begingroup$ Thanks! I agree that kinetic energy is not conserved here. However, the change of kinetic energy is given by the work-energy theorem. Some force is doing work to slow down the wagon. If there is no kinetic friction involved, i.e. the heroes really just “stick” to the wagon, then what force is doing this work? $\endgroup$ Commented Jun 5, 2023 at 14:06
  • $\begingroup$ In the inelastic collision macroscopic KE is converted to microscopic (internal) KE (increased molecular motion) in the inelastically colliding bodies . If you include the increase in internal KE, the work energy theorem still applies. $\endgroup$
    – Bob D
    Commented Jun 5, 2023 at 14:38
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    $\begingroup$ @OnnoEberhard according to Newton's 2nd law, it takes a force to accelerate the Lone Ranger and Tonto in a horizontal direction. This force comes from the wagon, and that force is applied over a distance, meaning that the wagon did work on the two men. Due to Newton's 3rd law, the men also did negative work on the wagon, slowing it down. Work is equivalent to energy, so there is no "missing energy" in this problem. $\endgroup$ Commented Jun 5, 2023 at 15:05
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Where does the energy go in this problem ?

In an inelastic collision the "lost" kinetic energy is dissipated into the environment as sound and heat.

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