Where Does Lost Energy from Golf Ball Go on Carpet?

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When a golf ball bounces on carpet, it loses energy compared to bouncing on concrete due to the carpet's ability to absorb some of that energy. The energy is transferred to the carpet, causing it to deform and store potential energy temporarily. As the carpet recoils, it releases some of that energy, but not all of it is returned to the ball, resulting in a lower bounce. The interaction can be likened to the mechanics of a crumple zone in a car, where energy absorption occurs during impact. Overall, the energy lost by the ball is primarily converted into kinetic energy within the carpet and dissipated as heat.
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Homework Statement


A golf ball will bounce much higher on concrete than on carpet. What happens to this lost energy?


Homework Equations





The Attempt at a Solution


The carpet stores it as potential energy?
 
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What happens to the carpet as the ball hits it?
 
It sinks in, but then it recoils, so I would think that it would relinquish any potential energy it had then.
 
Gawd... I'm not actually too sure myself come to think of it. My understanding is that some of the kinetic energy of the ball would be transferred to kinetic energy in the carpet.

It will be related to impulse somehow; working in a similar way to a crumple zone on a car.

I could be totally astray though; will be interested to see someone else's response.
 
Well, when it recoils, it moves, so I suppose that if enough of it moved and if it was massive enough, that would make sense.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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