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Q: Balls ( No Answer,   5 Comments )
Question  
Subject: Balls
Category: Science > Physics
Asked by: mocha28-ga
List Price: $3.00
Posted: 20 Dec 2005 08:44 PST
Expires: 08 Jan 2006 22:02 PST
Question ID: 607964
A ball dropped to the floor rises to a height less than its original
height, what are some energy conversions that might take place?
Answer  
There is no answer at this time.

Comments  
Subject: Re: Balls
From: spanky2k-ga on 20 Dec 2005 11:42 PST
 
The ball starts off with a certain amount of potential energy, PE =
mgh where m is the mass of the ball, h is the height and g is the
gravitational pull of the Earth (9.81 m/s). As the ball falls its
potential energy decreases as its kinetic energy increases, its
kinetic energy being 0.5mv^2 where v is the velocity of the ball. A
small amount of energy is lost due to friction in the air and any
unknown forces.
So as the ball falls, its energy can be described as:

KE + PE = Energy lost due to friction and unknown forces.

When the ball hits the surface it will rebound and again follow the
prior equation. However some energy will be lost in the form of heat
and again friction as the ball touches the surface. The amount of
energy lost in the actual 'bounce' is dependant on the construction of
he ball. A particularly elastic ball such as a bouncy ball will lose
less energy while a solid stone ball will lose more energy.

The biggest loss of energy in a ball's travel to the ground and back
is due to he friction in the air though.

I hope this helps! A more detailed explaination could probably be
found in a Mechanics textbook.


Spanky
Subject: Re: Balls
From: qed100-ga on 20 Dec 2005 11:47 PST
 
It's a homework question.
Subject: Re: Balls
From: qed100-ga on 20 Dec 2005 13:38 PST
 
...But actually, the loss due to air-drag will, of course, depend upon
how dense the air is, and how far through it the ball must travel. If
I drop a rubber ball just a couple of inches above the floor, the loss
to air resistance will be pretty small per bounce. In that case the
major contributor to energy loss will be through the redistributing of
the kinetic energy into internally randomised molecular vibrations.
Indeed, if I drop the ball while standing on the Moon's surface,
there'll be no loss to drag, and all loss will be due to <100%
elasticity.
Subject: Re: Balls
From: spanky2k-ga on 20 Dec 2005 15:22 PST
 
Exactly. A ball will bounce obeying damped harmonic motion slowed by
the elasticity of the ball and the density of the medium through which
it is falling. This is a very common problem covered in a plethora of
harmonic oscillator and motion modules in Physics courses.
Subject: Re: Balls
From: azdoug-ga on 07 Jan 2006 22:25 PST
 
e

Amplitude_next_bounce = Amplitude_current_bounce * e^(-damping_coefficient * time)

or something like that...  vibe class was a few years ago...  The
damping coefficient is what describes the energy loss per bounce.  It
will be large at first, and smaller after every bounce.  It'll
resemble the graph of y = e^-x.

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