M3 June 2014 Q2
2. A particle \(P\) of mass \(m\) is fired vertically upwards from a point on the surface of the Earth and initially moves in a straight line directly away from the centre of the Earth. When \(P\) is at a distance \(x\) from the centre of the Earth, the gravitational force exerted by the Earth on \(P\) is directed towards the centre of the Earth and has magnitude \(\dfrac{k}{x^2}\), where \(k\) is a constant.
At the surface of the Earth the acceleration due to gravity is \(g\). The Earth is modelled as a fixed sphere of radius \(R\).
When \(P\) is at a height \(\dfrac{R}{4}\) above the surface of the Earth, the speed of \(P\) is \(\sqrt{\dfrac{gR}{2}}\)
Given that air resistance can be ignored,
| Scheme | Marks |
|---|---|
| (At surface) \(\dfrac{k}{R^2} = mg \Rightarrow k = mgR^2\) | M1A1 |
| (2) |
Notes
M1 for \(\dfrac{k}{R^2} = mg\). If not made clear that this applies at the surface of the Earth award M0 or \(\dfrac{k}{x^2} = mg\) and \(x = R\).
A1 cso for \(k = mgR^2\) *
Alternative for Question 2 (a)
Using \(F = \dfrac{GM_1M_2}{x^2}\) with \(x = R\) and one mass as mass of Earth:
\(mg = \dfrac{GmM_E}{R^2}\)
\(GM_E = gR^2 \Rightarrow F = \dfrac{mgR^2}{x^2} \Rightarrow F = \dfrac{k}{x^2}\) with \(k = mgR^2\) *
M1 Complete method A1 Correct answer
| Scheme | Marks |
|---|---|
| \(m\ddot{x} = -\dfrac{mgR^2}{x^2}\) | |
| \(v\dfrac{\mathrm{d}v}{\mathrm{d}x} = -\dfrac{gR^2}{x^2}\) | M1 |
| \(\displaystyle\int v\frac{\mathrm{d}v}{\mathrm{d}x}\,\mathrm{d}x = -gR^2\int\frac{1}{x^2}\,\mathrm{d}x\ \ \ \text{or}\ \int\frac{\mathrm{d}\left(\frac{1}{2}v^2\right)}{\mathrm{d}x}\,\mathrm{d}x\) | |
| \(\dfrac{1}{2}v^2 = \dfrac{gR^2}{x}\ \ (+c)\) | DM1A1 |
| \(x = \dfrac{5R}{4},\ v = \sqrt{\dfrac{gR}{2}} \Rightarrow c = -\dfrac{11gR}{20}\) | DM1A1 |
| \(v = 0\ \ 0 = \dfrac{gR^2}{x} - \dfrac{11gR}{20}\) | DM1 |
| \(x = \dfrac{20R}{11}\) | A1 |
| (7) | |
| (9 marks) |
Notes
M1 for using accel \(= v\dfrac{\mathrm{d}v}{\mathrm{d}x}\) oe in NL2 with or w/o \(m\) Minus sign not required.
M1 dep for attempting to integrate both sides - minus not needed
A1 for fully correct integration, with or w/o the constant. Must have included the minus sign from the start.
M1 dep for using \(x = \dfrac{5R}{4},\ v = \sqrt{\dfrac{gR}{2}}\) to obtain a value for the constant. Use of \(x = \dfrac{R}{4}\) scores M0 Depends on both previous M marks
A1 for \(c = -\dfrac{11gR}{20}\)
M1 dep for setting \(v = 0\) and solving for \(x\) Depends on 1st and 2nd M marks, but not 3rd
A1 cso for \(x = \dfrac{20R}{11}\)
ALT: By definite integration
First 3 marks as above, then
DM1 Using limits \(x = \dfrac{5R}{4},\ v = \sqrt{\dfrac{gR}{2}}\)
DM1 Using limit \(v = 0\)
A1 Correct substitution
A1 cso for \(x = \dfrac{20R}{11}\)
NB: The penultimate A mark has changed position, but must be entered on e-pen in its original position.
Alternative for Question 2 (b)
| By conservation of energy: | |
| Work done against gravity \(= \displaystyle\int_{\frac{5R}{4}}^{z} \frac{mgR^2}{x^2}\,\mathrm{d}x = \int_{\frac{5R}{4}}^{z} mgR^2x^{-2}\,\mathrm{d}x\) | M1 |
| \(= \dfrac{4mgR}{5} - \dfrac{mgR^2}{z}\) | DM1(integration)A1(correct) |
| Work-energy equation: \(\dfrac{mgR}{4} = \dfrac{4mgR}{5} - \dfrac{mgR^2}{z}\) | DM1A1 |
| \(z = \dfrac{20R}{11}\) | DM1A1 |
(Corrected from the printed mark scheme: the lower limits of the integrals are printed as \(\dfrac{5r}{4}\).)