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In a thought experiment, we have a trampoline on the ground (well it's fancier than a spring) and a ball which is made to undergo free-fall from a height $h$. In a non ideal scenario, will the ball bounce back to the same height $h$ ?Its definitely not higher, so does the strength of the trampoline(or spring force constant $k$) change the max height? That is, can it account for the drag loss due to air? Possible eqns: $$mgh+\frac{1}{2}kx^2=F_{drag}(h+x)+mgh'$$

Will the energy loss due to this air drag be compensated by the trampoline? Is $h=h'$ ? This is done in a real setting, not an ideal one.

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  • $\begingroup$ Note: Please correct me if I have gotten my equations or concept wrong. I still am in Highschool $\endgroup$
    – evamPUNdit
    May 26, 2019 at 5:46

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Energy losses:

  • The single largest one is the air resistance of the mat. Each bounce (of a person sized mass) pushes a volume of air, down, then up. Competition grade mats are over 50% space. Instead of moving a parcel of air feet, you are moving it fractions of a cm.

  • The second source of loss: On the rebound when the mat is nearing it's flat position, it stops accelerating. At this point the mass leaves the mat, but the mat is still moving. A competition mat is about 10 Kg. A garden trampoline mat about twice this. The springs are heavy, but they don't move much. Some people claim they can feel the difference using a shorter, fatter, but lighter spring. Such a spring has less mass, and being shorter, less movement. Preliminary back of the envelope calcs suggest this is a VERY small effect.

  • On a cheap trampoline, the fabric of the mat is stretchy. Most stretchy fabrics have a high hysteresis, so a lot of the energy put in isn't recovered.

I have a good garden tramp. With a reasonable jump it takes about 7 bounces with straight locked legs to damp to the point where I'm not leaving the trampoline.

I also use an olympic tramp at my gym. The height I can get at maximum effort at home, I can get just using my calf muscles at the gym. It feels like I'm getting more than 90% of my height back, but this may be an illusion. I do know that it takes a lot longer to come to rest on a crash and burn.

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First think about the ideal system:

State 1) the ball is at a height h, the trampoline is at rest. So the ball has potential energy and the trampoline has none.

State 2) The ball just hit the trampoline, the ball has kinetic energy and potential energy (yes the ball will still travel a bit until maximal deformation). the trampoline is still at rest so it has no potential energy.

State 3) the ball has deformed the trampoline to its maximal deformation, the ball has no kinetic energy nor potential energy. All the energy is "stored" in the trampoline deformation so the trampoline has some potential energy.

State 4) The trampoline has bounce back the ball and the ball is just about leaving the trampoline. So the ball has some kinetic energy, but the trampoline is now also moving so it also have a bit of kinetic energy!

State 5) the ball goes to its higher point so it has now only potential energy, but the trampoline is now bouncing so it also have some energy...

Therefore even in the ideal case (without friction and without heat transfer due to the deformation of the trampoline) the ball will never be able to reach back the same height...

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  • $\begingroup$ Why isn't all of the energy transferred back to the ball? Since you mentioned ideal, a spring / trampoline would transfer all the energy back to the ball right? $\endgroup$
    – evamPUNdit
    May 26, 2019 at 7:33
  • $\begingroup$ Nope, as it is not an elastic choc. If you have a ball that hit the ground then yes, in the ideal case the ball go back at the same height that it started, but in that case you "think" the ground to be a complete static object which does not absorb any energy, and the trampoline does (as it get deformed). What would "force" the trampoline spring to not bounce? $\endgroup$
    – David Daverio
    May 26, 2019 at 7:58
  • $\begingroup$ Well, I had previously thought that a spring with a high spring constant would come to rest after the initial bounce and all energy goes to ball $\endgroup$
    – evamPUNdit
    May 26, 2019 at 7:59
  • $\begingroup$ If the spring constant is infinite, then yes the ball will go back to the same height as the spring will not deform and you will end with a elastic choc. $\endgroup$
    – David Daverio
    May 26, 2019 at 8:01
  • $\begingroup$ Ok then thanks for the answer $\endgroup$
    – evamPUNdit
    May 26, 2019 at 8:02

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