Re: impulse/momentum

[Bob Sciamanda <trebor@VELOCITY.NET>]

1) If m, travelling at v, collides elastically with M, at rest, the
subsequent speed of M is
U = (2v) / ( 1 + M/m) and the subsequent momentum of M is MU = 2mMv/(m+M).

2) In the limit M/m => infinity these become U=0 and MU = 2mv, respectively.

3) In addition, the subsequent KE of M is .5MU^2 =  (2M(mv)^2)/((m+M)^2)
This goes to zero in the above limit.

Bob Sciamanda
Physics, Edinboro Univ of PA (Em)
http://www.velocity.net/~trebor/
trebor@velocity.net
----- Original Message -----
From: "Brian Whatcott" <betwys1@SBCGLOBAL.NET>
To: <PHYS-L@lists.nau.edu>
Sent: Friday, December 12, 2003 1:17 PM
Subject: Re: impulse/momentum
> . . .
> There is something unnatural about the propensity for modeling
> infinite  masses. If a turbine wheel is locked, the conditions are
satisfied.
> If a turbine wheel is unlocked, then the startup phase is also
approximated
> by the model.
> I therefore conclude that the sentences
>   "the momentum change of the turbine plus the momentum change of the
water
> is zero. This means the turbine does not have a zero change of momentum"
.....
> are true and false, respectively.
> Brian W