Colliding Balls

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

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 
of the loop of radius R, two statements can be made:

  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/