Circus Performer Falling into a Safety Net

[kd001]

Can this be explained in terms of inelastic collisions? This is what I think:

The collision is perfectly inelastic because the person and the net move together with the same velocity. This means that the initial kinetic energy of the system (and therefore the kinetic energy of the person) is reduced. Basically the safety net acts by transferring kinetic energy from the person to the net.

Do you think this is a valid argument?

Thanks

 

[A.T.]

Can this be explained in terms of inelastic collisions? This is what I think:

Not really. Without gravity a Performer hitting the net would be shoot back again. That would be pretty elastic. On Earth gravity forces him to stay in the Net, just like a bouncing ball stays on the ground after some jumps. This doesn't make the ball/ground collision inelastic.

Usually when considering elastic vs. inelastic collisions you don't have any forces from outside acting on the bodies.

 

[rcgldr]

Trapeze nets are semi-elastic. Like an inefficient trampolene. The peformers can sometimes bounce back up quite aways, but this may be adjustable via tension, or it's a different, more elastic material used for those acts where the peformers bounce back up again. Note that the competition trampolines are also net like (Austrilian or Ozzie bed), unlike the canvas strap ones we're used to seeing, or the outdoor black mesh ones.

 

[vanesch]

Not really. Without gravity a Performer hitting the net would be shoot back again. That would be pretty elastic. On Earth gravity forces him to stay in the Net, just like a bouncing ball stays on the ground after some jumps. This doesn't make the ball/ground collision inelastic.

Actually, it does! If the collision were perfectly elastic, the performer (or the bouncing ball) would jump up exactly as high as the height it originally fell from (conservation of energy). Only because there is dissipation (heat), the rebounce is not as high. But that's exactly what it means, an inelastic collision: that there is dissipation.

 

[A.T.]

Actually, it does! If the collision were perfectly elastic,

Yeah, I meant: "This doesn't make the ball/ground collision perfectly inelastic." which is what the OP was asking about.

But if you consdider the whole bouncing less & less process as one collision, you could call it perfectly inelastic.

 

[vanesch]

Yeah, I meant: "This doesn't make the ball/ground collision perfectly inelastic."

I think you're digging yourself in a hole.

I don't know what would be a perfectly inelastic collision, honestly...

An elastic collision is a collision where the sum of kinetic energies before and after collision is conserved. An inelastic collision is one which is not elastic, that is, where this sum is not conserved (most of the time, the sum is less after than before though not always).

So a perfectly inelastic collision is ... (?)

 

[A.T.]

I think you're digging yourself in a hole.

I don't know what would be a perfectly inelastic collision, honestly...

http://en.wikipedia.org/wiki/Inelastic_collision#Perfectly_inelastic_collision

So a perfectly inelastic collision is ... (?)

Two clumps of clay?

 

[vanesch]

http://en.wikipedia.org/wiki/Inelastic_collision#Perfectly_inelastic_collision



Two clumps of clay?

Ok, my apologies, I really never heard that term.