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 $x$ :

$k v^{2} = g (1 - e^{- 2 k x})$

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 $δ t$ ).

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

$\frac{1}{\sqrt{1 - e^{- 2 k x}}} dx = \int \sqrt{\frac{g}{k}} dt$

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

$(1 / k) arsech e^{- k x} = t \sqrt{g / k} + C$

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

$x = k^{- 1} \ln \cosh (t \sqrt{g k})$

The book gives $x = k^{- 1} \ln \cosh (t \sqrt{g / k})$

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 $t \sqrt{g k} = T \sqrt{L T^{- 2} . L^{- 1}} =$ dimensionless

But $t \sqrt{g / k} = T \sqrt{L T^{- 2} / L^{- 1}} = L$

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