Mark Durkee

Posted on Monday, 17 March, 2003 - 08:27 pm:

I have got an answer for a question in the Edexcel M5 book that is different to that in the answers and would like to see whether I have made a mistake or the book is wrong (or both).

The question (Review Ex1 Q31) is:

A particle falls from rest under gravity through a stationary cloud. The mass of the particle increases by accretion from the cloud at a rate which at any time is m k v where m is the mass and v is the speed of the particle, and k is constant. Show that after the particle has fallen a distance x:

k v²=g(1-e^{-2k x})

and find the distance that the particle has fallen after time t.

The first part of this comes out fine (by saying that Impulse = Change of Momentum over the interval dt).

I have solved the second part by writing v as dx/dt and separating variables to give

1

Ö

1-e^{-2k x}

dx= ó
õ æ
ú
Ö

g
k

dt

I then solved this by the substitution sech y = e^{-k x} which gives (after some rearranging):

(1/k) arsech e^{-k x}=t ___
Ög/k

+C

The initial conditions give C=0 so this rearranges to:

x=k^-1ln cosh(t __
Ög k

)

The book gives
x=k^-1 ln cosh (t ___
Ög/k

)

David Loeffler

Posted on Monday, 17 March, 2003 - 08:44 pm:

I get the same. I suspect that the book is wrong, as you suggest.

David

Andre Rzym

Posted on Tuesday, 18 March, 2003 - 08:18 am:

The book can be seen to be incorrect simply from a dimensional analysis perspective. Look to express the various quantities (m, k, x etc) in terms of the units M L T (mass length time). So we have:

g=L T^-2

x=L

t=T

v=L T^-1

k=L^-1

[the last can be obtained from the dimensions of the equation dm/dt=m k v]

Now
t __
Ög k

=T
Ö

L T^-2.L^-1

=

dimensionless

But
t ___
Ög/k

=T
Ö

L T^-2/L^-1

=L

Since the argument to cosh must be dimensionless, the latter cannot be correct.

Andre

Mark Durkee

Posted on Tuesday, 18 March, 2003 - 01:41 pm:

Thanks