Re: Physics Answer to a Question...

["A. R. Marlow" <marlow@beta.loyno.edu>]

Your answer is correct, granted the rather nonstandard assumption that you 
have some means of keeping the old man's mass constant (resupplying him
with water) without contributing any additional change to his speed.  As
someone mentioned, possibly a nurse running alongside and continually
feeding him enough water (always adjusted to be at his instantaneous
speed) to just keep his mass constant?

Your answer follows from a general equation I derived and published in
The Physics Teacher, Vol. 33, Feb. 1995, pp. 124 - 125, "The Pedagogy of
the Rocket."  The equation, in suitably simplified one dimensional 
notation, is eq. 3 of the paper:

        dvr/dt = u/mr dmr/dt -(me/mr) dve/dt,

where vr and ve are the speeds of old man (rocket) and tobacco pieces 
(exhaust particles) respectively, u is the exhaust speed (speed at 
which tobacco pieces are launched relative to old man), and mr and me
are the masses of the old man and tobacco pieces, respectively.  The 
usual rocket equation follows from throwing away the last term when
me/mr approaches zero relative to dve/dt (e. g., in a smoothly burning
well constructed rocket, where me is of molecular size and << mr).

As noted in the paper, the equation also covers "the motion of a 
rowboat propelled by hurling bowling balls off the stern ... a 
skateboarder jettisoning coat, shoes, protective pads ..., or any
other system in which the instantaneous "exhaust" mass me is not
negligible relative to the mass mr of the propelled system."

In the situation you have assumed for the old man, the last term of the
equation is all there is, since dmr/dt = 0 by hypothesis, and, using

      dve/dt = - 5(m/s)/te

where te is the time between spits, and me/mr = .001kg/50kg = 2 x 10-5,
and integrating the equation over a time T = n te, we find

      vr = 1 m/s = 2 x 10-5 x 5 m/s x n 

or n = 10,000 as you claim.

As others have pointed out, a more standard problem would be to allow the
old man to lose mass as he spits.  Even then, though, your problem does
not reduce to the well constructed rocket problem, since both terms in
the equation above contribute almost equally, so that the old man would
then need only about 5000 spits. 


On Sat, 8 Feb 1997, Dwight K. Souder wrote:

> ...
> Imagine good ol' Grandpa Ester sitting in a frictionless wheeled chair,
> with the inevitable wad of chewing tobacco lodged in his mouth.  Assuming
> he weighs 50Kg (including the wheel-chair), and is skilled enough in the
> art of spitting to launch 1g of "projectile" at 5 m/s each time he spits,
> how many "projectiles" would be needed to be launched to get him moving
> at 1 m/s?  Also, assume he periodically drinks some water to replace the
> mass he is launching.
>
> To solve it, I used a momentum formula:  M1 x V1 = M2 x V2
> M1 = mass of first object in Kilograms
> V1 = Velocity of first object in meters per second
> M2 = mass of second object in Kilograms
> V2 = velocity of second object in meters per second
>
> M1= 50Kg
> V1 = ?
> M2 = 1g = 0.001 Kg
> V2= 5 m/s
>
> Solving for V1 I get = 0.0001 m/s
> Therefore, each time he spits, his velocity is increasing by 1x10^-4 m/s,
> which means  he would have to spit 10 000 times.
>
> Is this correct?  Any help would be greatly appreciated.  If it is wrong,
> could you please tell me where I went wrong?
>
> Thanks,
> Dwight

A. R. Marlow                          E-MAIL:  marlow@beta.loyno.edu
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