M3 June 2018 Q3
3. A particle \(P\) of mass \(m\) moves in a straight line away from the centre of the Earth. The Earth is modelled as a sphere of radius \(R\). When \(P\) is at a distance \(x\), \(x \geqslant R\), from the centre of the Earth, the force exerted by the Earth on \(P\) is directed towards the centre of the Earth and has magnitude \(\dfrac{mgR^2}{x^2}\). When \(P\) is at a distance \(2R\) from the surface of the Earth, the speed of \(P\) is \(\sqrt{\dfrac{gR}{3}}\).
Assuming that air resistance can be ignored, find the distance of \(P\) from the surface of the Earth when the speed of \(P\) is \(2\sqrt{\dfrac{gR}{3}}\). (7)
| Scheme | Marks |
|---|---|
| \(mv\dfrac{\mathrm{d}v}{\mathrm{d}x} = -\dfrac{mgR^2}{x^2}\) | M1 |
| \(\dfrac{1}{2}v^2 = -\displaystyle\int gR^2x^{-2}\,\mathrm{d}x\) | |
| \(\dfrac{1}{2}v^2 = gR^2x^{-1}\ \ (+c)\) | dM1A1 |
| \(x = 3R \ \ v = \sqrt{\dfrac{gR}{3}} \ \Rightarrow c = \dfrac{1}{2}\dfrac{gR}{3} - \dfrac{gR^2}{3R} = -\dfrac{gR}{6}\) | dM1 |
| \(v = 2\sqrt{\dfrac{gR}{3}} \quad 2\dfrac{gR}{3} = \dfrac{gR^2}{x} - \dfrac{gR}{6} \qquad x = \ldots\) | dM1 |
| \(x = \dfrac{6R}{5}\) | A1 |
| Dist from surface \(= \dfrac{6R}{5} - R = \dfrac{R}{5}\) oe | A1cso |
| (7 marks) |
Notes
M1 Attempting an equation of motion with correct number of terms and acceleration \(v\dfrac{\mathrm{d}v}{\mathrm{d}x}\).
Allow with minus missing. Can be given by implication if acceleration is integrated to \(\dfrac{1}{2}v^2\)
dM1 Attempting the integration of both sides of their equation. \(x^{-2} \to x^{-1}\) Depends on the first M mark.
A1 Correct equation after correct integration. Constant of integration may be missing.
Double sign error scores A0 here.
dM1 Substitute \(x = 3R\) \(v = \sqrt{\dfrac{gR}{3}}\) and obtain an expression for \(c\). Depends on the first M mark providing an attempt at integration is seen. (eg \(x^{-2} \to x^{-3}\) could score M1M0A0dM1)
dM1 Substitute \(v = 2\sqrt{\dfrac{gR}{3}}\) in their expression for \(v^2\) and solve for \(x\) Depends on the first M mark.
A1 Correct \(x\) Double sign error scores A0 here.
A1cso Correct answer from completely correct working.
ALT 1 Definite Integration:
| \(mv\dfrac{\mathrm{d}v}{\mathrm{d}x} = -\dfrac{mgR^2}{x^2}\) | M1 |
| \(\displaystyle\int_{\sqrt{\frac{gR}{3}}}^{2\sqrt{\frac{gR}{3}}} v\,\mathrm{d}v = -\int_{3R}^{X} gR^2x^{-2}\,\mathrm{d}x\) | |
| \(\left[\dfrac{1}{2}v^2\right]_{\sqrt{\frac{gR}{3}}}^{2\sqrt{\frac{gR}{3}}} = \left[gR^2x^{-1}\right]_{3R}^{X}\) | dM1A1 |
| \(\dfrac{1}{2} \times 4\dfrac{gR}{3} - \dfrac{1}{2}\dfrac{gR}{3} = \dfrac{gR^2}{X} - \dfrac{gR}{3}\) | dM1 |
| \(\dfrac{2}{3} - \dfrac{1}{6} = \dfrac{R}{X} - \dfrac{1}{3} \qquad \dfrac{R}{X} = \dfrac{5}{6}\) | |
| \(X = \dfrac{6}{5}R\) | dM1A1 |
| Dist from surface \(= \dfrac{6R}{5} - R = \dfrac{R}{5}\) oe | A1 cso [7] |
M1 Attempting an equation of motion with correct number of terms and acceleration \(v\dfrac{\mathrm{d}v}{\mathrm{d}x}\).
Allow with minus missing.
dM1 Attempting the integration of both sides of their equation. \(x^{-2} \to x^{-1}\) Depends on the first M mark. Limits not needed (ignore any shown)
A1 Correct integration. Ignore any limits shown.
dM1 Substitute correct limits. May be as shown or both sets reversed. Depends on the first M mark.
dM1 Solve to \(X = \ldots\) Depends on the first M mark.
A1 Correct \(X\)
A1cso Correct answer from completely correct working.
ALT 2 Energy:
Variable force, so no integration implies no marks
| \(\dfrac{1}{2}mv^2 - \dfrac{1}{2}mu^2 = \displaystyle\int F\,\mathrm{d}x = \int -\frac{mgR^2}{x^2}\,\mathrm{d}x\) | M1 (minus and limits may be missing) |
| \(\dfrac{1}{2}mv^2 - \dfrac{1}{2}mu^2 = \left[\dfrac{mgR^2}{x}\right]_{3R}^{X}\) | dM1A1 Integrate RHS (as above) Inconsistent signs scores dM1A0) |
| \(\dfrac{1}{2}m \times \dfrac{4gR}{3} - \dfrac{1}{2}m \times \dfrac{gR}{3} = \dfrac{mgR^2}{X} - \dfrac{mgR}{3}\) | dM1 Sub correct limits |
| \(X = \dfrac{6}{5}R, \qquad\) Dist from surface \(= \dfrac{6R}{5} - R = \dfrac{R}{5}\) oe | dM1A1, A1cso As alt 1 |