Re: momentum transfer

[Richard Tarara <rtarara@SAINTMARYS.EDU>]

To simplify the discussion (somewhat) let's change the comparison to a
collision between two moving cars and one moving car and one parked car.
Now certainly a collision with the parked car at speed v is less serious
than a head-on with both cars travelling at v.  In Michael's analysis below,
the assumption of equal times to come to rest is, I believe, the problem.
In the head-on, t will be smaller (assuming the center of mass will not move
once the cars make contact) than in the collision with the parked car (even
if that car doesn't move).

Rick

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Richard W. Tarara
Department of Chemistry & Physics
Notre Dame, IN 46556
219-284-4664
rtarara@saintmarys.edu

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----- Original Message -----
From: Michael Moloney <moloney@NEXTWORK.ROSE-HULMAN.EDU>
To: <PHYS-L@LISTS.NAU.EDU>
Sent: Thursday, July 01, 1999 12:36 PM
Subject: momentum transfer


> If two identical cars of mass M traveling at speed V  have  head-on
collision and come
> to rest in time T, the average force on each car is MV/T.
>
> If a car of mass M traveling at speed 2V strikes a large immovable tree
and comes to
> rest in time T', the average force on that car  2MV/T'.
>
> If the stopping times T and T' are the same the average force on the car
traveling at 2V
> appears to me to be twice as great as the average force on each of the
cars in the first
> case.
>
> To say the average force are the same looks like an error in physics to
me.
> Mike
>
> ===============================================================
> Mike Moloney        http://www.rose-hulman.edu/~moloney
> moloney@nextwork.rose-hulman.edu
> Dept of Physics & Applied Optics           (812) 877 8302
> Rose-Hulman Institute of Technology   Terre Haute, IN 47803