RE: [ot]: how does a radar detector work?As per Michael, I am forwarding
this to the list. Also, I couldn't resist commenting on what he had to say.
See below his post.

Bob Ammerman
RAm Systems

----- Original Message -----
From: Michael Rigby-Jones
To: 'Bob Ammerman'
Sent: Wednesday, November 07, 2001 7:47 AM
Subject: RE: [ot]: how does a radar detector work?


This is a real brain teaser.
I thought I had solved the problem due to the fact that energy is a scalar
product, so the sign of velocity vector is unimportant.  Two identical cars
travelling at 75mph have the same amount of kinetic energy as one car
weighing twice as much, irrespective of the direction they are pointing.
(i.e. exactly twice the energy).  This explains why hitting the immovable
wall would give the same result as the head-on. (half the energy disspated
in one car compared to twice the energy in two cars)

HOWEVER, think of the situation of one car doing 150 mph and hitting a
stationary (i.e v=0, so zero energy) car head on (assume the rear of the
stationary car dosen't move i.e. someone backed it upto that handy immovable
wall).  The car obviously has 4 times the energy that it did previously, and
apparently twice as much as the two cars together.  Now the "crush distance"
is the same (i.e. two cars worth), and the relative velocities are the same,
so will the crash be worse, or the same?
Bob, could you possibly forward this onto the list as all my posts get
bounced.
Cheers
Mike

-------------

Actually, the crush distance will _not_ be the same because the total
kinetic energy available is:

    1/2 * M * 150*150

which is twice as much as two cars head-on, or:

    2 * (1/2 * M * 75*75)

I expect the crash will be much worse because each of the two cars will have
to absorb twice as much energy.

Bob Ammerman

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