STEP I 1997 question 11 solution - The Student Room
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STEP I 1997 question 11 solution

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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