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Let's say I want to teach kinetic and potential energy and I'd like to use an example that students can remember. The original idea is to show that speed (being squared) in the kinetic energy formula proves to be very important in the energy being delivered to a body. If a car travels in a 30 km/h limit zone with 34 km/h, how much more impact do those 4 extra km/h make when hitting a pedestrian (no braking)? You can compute the Joules difference but it doesn't say much on how it impacts the pedestrian. I thought about making this kinetic energy equal to the potential energy of a body falling to the ground and using the computed equivalent heights to show how much more dangerous it is. Is this a too simple model to be good enough?

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  • $\begingroup$ It's not as sexy as car on ped collisions, but the comparison is if I free-fall from 10 meters up, I go 10 m/s, but I need to go 41 meters up to get to 20 m/s, and I need to go 92 meters up to get to 30m/s. That graph should look a lot like a square root function. Change units as needed... $\endgroup$ – user121330 Jun 29 '17 at 18:11
  • $\begingroup$ I'd like to go from a very practical and useful example of real life towards the theory so that most students can follow the logical association between the KE and PE. I hope it would be easy to remember because of the practical/emotional aspect. Elastic/Inelastic balls that collide and graphs I would leave for the second part where the students not interested in physics will probably space out. $\endgroup$ – MeDodo Jun 30 '17 at 8:54
  • $\begingroup$ Knife penetration depth might be sexy enough... $\endgroup$ – user121330 Jun 30 '17 at 15:35
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Asking "how much more impact" is not a well defined question so it will be difficult to associate a quantity to it. You could indeed calculate the height corresponding to the PE equal to the extra KE. It does not mean that the effects on the body are the same in the two cases. In this case, considering a 2000 kg car and 100 kg man, the extra KE of the car will correspond to the man falling from about 20 m. But the extra energy is not transferred to the man. The tricky part is to find how much energy is transferred to the man by impact and this is not trivial. The collision is neither perfectly elastic, nor perfectly inelastic (they don't stick together). So the amount of KE transferred will depend on the details of the collision. Maybe you could consider an elastic collision as a model (the energy transferred to the body is maxim in this case) and calculate the difference in energy transferred. If you convert it to PE of the body it will be less than 4 m, still significant and more realistic probably than the 20 m. People often survive accidents produced by cars running at 34 km/h but not so falls from 20 m.

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  • $\begingroup$ With the elastic collision it would be a more difficult model to understand for students who are not that interested in physics. Maybe an example with a boxer trying to improve its punch is something better since both bodies have similar mass. A boxer can try and put more weight into a punch but he still needs to stay on his feet so this mass has limits. Then, the only thing he can try to improve is to punch with more speed to deliver more kinetic energy to the opponent. Then I could compare this to a, let's say, 15 kg ball falling from a certain height like in the car example... $\endgroup$ – MeDodo Jun 30 '17 at 8:33
  • $\begingroup$ yes if you are roughly speaking, other external forces and phenomenons like elasticity, friction, dissipation of energy can be ignored. As I said you can absolutely compare the impact made physically by taking an example of a mass falling from a height, but in realistic world, the falling body would make more impact on the pedestrian than the moving car. $\endgroup$ – Ajinkya Naik Jun 30 '17 at 10:37
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Yes, the example makes sense. Assuming the poor pedestrian was at rest before the impact hence its momentum will be zero too. Therefore the car will transfer its entire Kinetic Energy to the pedestrian and he will start moving at the velocity at which the car was moving earlier(ignoring all the frictional forces). You can absolutely calculate the impact made by car due to a 4 km/h difference by computing the Joule difference. As the example of a body falling to the ground from a certain height, this scenario will not exactly match with the one of the car accident, because in this example of falling body, the Potential Energy will be converted into Kinetic Energy and hence on impact the pedestrian will feel the same force, but because of the presence of ground beneath him he will then have no Kinetic Energy and will remain at rest. The impact caused by both will be same but the final state of motion of the pedestrian will change.

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    $\begingroup$ The car won't transfer its entire KE to the pedestrian unless the car and pedestrian have the same mass and the collision is perfectly elastic. None of these conditions apply. $\endgroup$ – nasu Jun 29 '17 at 17:14
  • $\begingroup$ I think this does not have any relation with mass, but it entirely depends on the momentum of the two bodies, which is mass times its velocity, as I clearly said - "ignoring the frictional forces" I referred to external forces, elasticity should be considered if we are talking about a more realistic scenario $\endgroup$ – Ajinkya Naik Jun 30 '17 at 10:33
  • $\begingroup$ roughly speaking, I assumed that both of the bodies are rigid masses and that the pedestrian was at rest, his momentum was zero, and therefore there were no resistive forces exerted by the pedestrian on the car, and so the car will transfer its entire kinetic energy to the pedestrian, exactly like the Newton's Cradle. $\endgroup$ – Ajinkya Naik Jun 30 '17 at 10:34
  • $\begingroup$ Yes, in these condition the car will NOT transfer its entire kinetic energy. Just look up the formulas for elastic collisions. In what conditions will the incoming body remain at rest after collision? In Newton's cradle the balls have the same mass. You can play with a simulation to try yourself to find the conditions for total KE transfer. phet.colorado.edu/sims/collision-lab/collision-lab_en.html $\endgroup$ – nasu Jul 1 '17 at 21:06
  • $\begingroup$ But dont you think it also depends on the momentum of two bodies, since the pedestrian was at rest his v=0 therefore momentum = 0 $\endgroup$ – Ajinkya Naik Jul 2 '17 at 9:05

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