STEP I 1997 question 11 solution

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STEP I 1997 question 11 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP 1997 Solutions > STEP I 1997 question 11 solution

Change in kinetic energy = work done on the particle:

$\displaystyle \Delta\left(\frac{v^2}{2}\right) = \int_{s_1}^{s_2} \left(- kv^2 - g\right)ds$

Differentiating:

$\displaystyle \frac{d}{ds}v^2 = -2kv^2 - 2g$

$\displaystyle \frac{d}{ds}v^2 + 2kv^2 = -2g$

One particular solution to the equation is v^2 = -g/k . The corresponding homogeneous equation

$\displaystyle \frac{d}{ds}v^2 + 2kv^2 = 0$

has solutions

$\displaystyle v^2 = Ce^{-2ks}$

Thus the general solution is

$\displaystyle v^2 = Ce^{-2ks} - \frac{g}{k}$

At s=0 we have v^2 = u^2 :

$\dislaystyle u^2 = C - \frac{g}{k}$

$\displaystyle C = u^2 + \frac{g}{k}$

Thus

$\displaystyle v^2 = \left(u^2 + \frac{g}{k}\right)e^{-2ks} - \frac{g}{k}$

$\displaystyle v^2 = u^2e^{-2ks} + \frac{g}{k}\left(e^{-2ks} - 1\right)$

as required.

The maximum height is when v = 0 :

$\displaystyle 0 = \left(u^2 + \frac{g}{k}\right)e^{-2ks} - \frac{g}{k}$

$\displaystyle \frac{g}{k} = \left(u^2 + \frac{g}{k}\right)e^{-2ks}$

$\displaystyle e^{2ks} = \frac{u^2 + g/k}{g/k}$

$\displaystyle e^{2ks} = u^2\frac{k}{g} + 1$

$\displaystyle s = \frac{\ln \left(u^2k/g + 1\right)}{2k}$ .

On the downward path, gravity is increasing the particle's kinetic energy instead, and we get the differential equation

$\displaystyle \frac{d}{ds}v^2 = 2g - 2kv^2$

which clearly has a solution that's different from the given equation. To see this, the general solution is

$\displaystyle Ce^{-2ks} + \frac{g}{k}$

instead. We have that v=0 at

$\displaystyle s = \frac{\ln \left(u^2k/g + 1\right)}{2k}$ .

and so

$\displaystyle 0 = C\frac{1}{u^2k/g + 1} + \frac{g}{k}$

$\displaystyle C = -\frac{g/k}\left(u^2k/g + 1\right)$

$\displaystyle C = u^2 + \frac{g}{k}$

and

$\displaystyle v^2 = \left(u^2 + \frac{g}{k}\right)e^{-2ks} + \frac{g}{k}$

$\displaystyle v^2 = u^2e^{-2ks} + \frac{g}{k}\left(e^{-2ks} + 1\right)$

thus the difference from when the particle was heading upwards is only that there is a plus sign instead of a minus signe before the '1'. But nevertheless it's different and so the equation (*) does not hold on the downward path.

Solution by ukgea

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