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### Colliding Balls

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Date: 04/28/98 at 23:10:11
From: Amy de Garavilla
Subject: Conservation of kinetic energy

Our physics teacher assigned us the following problem:

A 5 kg ball is attached to a pendulum with a string of length 5 m.
The pendulum is let go from a 45 degree angle causing the 5 kg ball
to collide with an adjacent 3 kg ball. The 3 kg ball, we'll call it
ball x, falls a certain distance. Upon landing it continues its path
in a straight line and goes up and around a loop (as in a roller
coaster.) We are asked to determine the maximum diameter of the loop.

The only other information given is the distance from the edge of the
area to the midpoint of the loop, which is 20 m. The coefficient of
friction for this same area is 20 N.

The only hint given by the teacher was "Think conservation of kinetic
energy." If and how do we apply that to this problem?

I am having trouble overall with this problem. I would appreciate it
if you could you tell me how to go about solving for the maximum
diameter of the loop so I can get started in the right direction?

-Amy de Garavilla
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Date: 04/30/98 at 19:50:37
From: Doctor Ujjwal
Subject: Re: Conservation of kinetic energy

Dear Amy,

This indeed looks like a very interesting problem. Unfortunately
there are lots of 'missing links', but I can lay out some steps that
can help you solve it (provided the missing data are filled in).

There are many events occuring in this problem. Each is governed by
different laws of physics.

Event 1
-------
A ball raised to a certain height 'picks up speed' by converting its
potential energy (PE = mgh = 5 * 9.81 * 5 * (1 - cos 45)) into kinetic
energy (KE = mv^2/2). According to CONSERVATION OF ENERGY we can
equate the two to get the velocity v of the 5 kg ball at the time
of impact:

v = square root (2gh)
= square root (2 * 9.81 * 5 * (1 - cos 45))
= 5.36 m/s

Event 2
-------
The two balls collide. This interaction can be analysed using
CONSERVATION OF MOMENTUM and CONSERVATION OF KINETIC ENERGY (Assuming
an 'elastic' collision).

If:

m1, m2 = masses of the balls
u1, u2 = their initial velocities respectively
v1, v2 = their final velocities respectively

then:

m1u1 + m2u2 = m1v1 + m2v2

Substitute in:

m1 = 5, m2 = 3, u1 = 5.36, u2 = 0

The other equation that helps us here comes from the property of
elastic collisions. In simple words it means the balls will fly apart
as fast they came together! Thus the difference of their velocities
before and after the collision is the same.

Event 3
-------
The second ball falls through some distance. It is important to know
how much distance. Because the greater this distance the higher the
speed it gains. (Just like event 1 above)

Event 4
-------
It travels some distance along a straight line. Again the distance is
important, because you are now 'spending' the kinetic energy to 'fight
against' the friction. By the way, the coefficient of friction is not
related to area, is always less than 1, and has no units. So it cannot
be 40 N (Newtons?). Maybe that was a typo.

Event 5
-------
The second ball 'loops a loop'. Assuming the ball makes it to the top

1. the centripetal force mv^2/r must be at least equal g
2. the potential energy of the ball is 2mgR

At this stage it is important to know whether the loop also has some
friction. In that case the problem can be pretty involved (but still
solvable).

These are only some broad brush strokes. Depending upon what more is
known, the depth of analysis can be determined. Do write us back with
the missing data. It sure promises to be a fun problem to solve!

Good luck!

-Doctor Ujjwal Rane,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/
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Associated Topics:
High School Physics/Chemistry

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