Flower pot falling on my head...

[0919mmk]

OK all, this is a physics question that I can't find an answer to anywhere, and I know someone here will know the answer.

Let's say I am walking down the street and a flower pot falls from a balcony and hits me on the head. Will it hurt more if the flower pot bounces off my head than if it breaks on my head? In other words, is more energy imparted on my head by an elastic or an inelastic collision. I am studying MCAT physics and I seem to recall my high school physics teacher telling me that the bouncing pot hurts more, but I can't exactly figure out why...

Thanks!!

edit: I should say that specifically, I think he said that there is twice as much energy imparted on my head by the bouncing flower pot than the breaking one...does anybody have an equation and/or explanation for that? Or am I making all this up?

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[ModernAlchemist]

OK all, this is a physics question that I can't find an answer to anywhere, and I know someone here will know the answer.

Let's say I am walking down the street and a flower pot falls from a balcony and hits me on the head. Will it hurt more if the flower pot bounces off my head than if it breaks on my head? In other words, is more energy imparted on my head by an elastic or an inelastic collision. I am studying MCAT physics and I seem to recall my high school physics teacher telling me that the bouncing pot hurts more, but I can't exactly figure out why...

Thanks!!

edit: I should say that specifically, I think he said that there is twice as much energy imparted on my head by the bouncing flower pot than the breaking one...does anybody have an equation and/or explanation for that? Or am I making all this up?

Well, we know that the inelastic collision does not conserve kinetic energy, but the elastic collision does. I would say it hurts more for the elastic collision because some of the energy that could have increased the force of impact on your head was dissipated through the deformation (breaking) of the pot.

 

[SteveJMarist]

Well, we know that the inelastic collision does not conserve kinetic energy, but the elastic collision does. I would say it hurts more for the elastic collision because some of the energy that could have increased the force of impact on your head was dissipated through the deformation (breaking) of the pot.

👍

I agree.

 

[backsideatk]

OK all, this is a physics question that I can't find an answer to anywhere, and I know someone here will know the answer.

Let's say I am walking down the street and a flower pot falls from a balcony and hits me on the head. Will it hurt more if the flower pot bounces off my head than if it breaks on my head? In other words, is more energy imparted on my head by an elastic or an inelastic collision. I am studying MCAT physics and I seem to recall my high school physics teacher telling me that the bouncing pot hurts more, but I can't exactly figure out why...

Thanks!!

edit: I should say that specifically, I think he said that there is twice as much energy imparted on my head by the bouncing flower pot than the breaking one...does anybody have an equation and/or explanation for that? Or am I making all this up?

A better way to think about it is using Newton's 3rd law as opposed to momentum conservation/elastic vs inelastic collisions. How much force is being applied in the vertical direction is what determines pain in your head, so any movement in the horizontal planes are irrelevant.

When the pot breaks on your head, your head is applying enough force on the pot to take it from its velocity right before hitting you (let's call that V) to zero.

When the pot BOUNCES on your head, your head is applying enough force on the pot to take it from it's velocity right before hitting you (V) to an equal and opposite velocity upward for the bounce (-V)

Thus the change in velocity (which is related to acceleration, thus related to force, thus related to pain) is only |V| in the break case but 2|V| in the bounce case.

I hope this helps more.

 

[ModernAlchemist]

A better way to think about it is ...

Haha but ya that's another way to think about it...

 

[Swagster]

Haha but ya that's another way to think about it...

Thanks for making me spray my last sip all over my fairly new keyboard. I laughed so hard, a little made it out of my nose.

 

[goobears]

how are you guys assigning these two scenarios as elastic or inelastic? I don't understand how either conserves kinetic energy..

 

[Danlee07]

Yeah, bouncing off your head will definitely hurt more. For inelastic and elastic, before thinking bout the math, i refer back to "world's weirdest people" when a guy did shows by getting mortar bricks being sledgehammered on his head. the one that didn't break and bounced because of wrong placement of the sledgehammer was the ticket to the ER.

 

[GooseWing]

Yep. It's all about the impulse delivered by the flower pot to your head.

I = ∆p = mv(final) - mv(initial)

Because mass is the same for both cases, you can just look at it as: ∆v.

In the case where the pot bounces, the ∆v is 2x greater than where it does not.

 

[TXKnight]

OK all, this is a physics question that I can't find an answer to anywhere, and I know someone here will know the answer.

Let's say I am walking down the street and a flower pot falls from a balcony and hits me on the head. Will it hurt more if the flower pot bounces off my head than if it breaks on my head? In other words, is more energy imparted on my head by an elastic or an inelastic collision. I am studying MCAT physics and I seem to recall my high school physics teacher telling me that the bouncing pot hurts more, but I can't exactly figure out why...

Thanks!!

edit: I should say that specifically, I think he said that there is twice as much energy imparted on my head by the bouncing flower pot than the breaking one...does anybody have an equation and/or explanation for that? Or am I making all this up?

the better way to think about this is in terms of Impulse, therefore bouncing off hurts more.