Edexcel M3 2018 June — Question 3 7 marks

Exam BoardEdexcel
ModuleM3 (Mechanics 3)
Year2018
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAdvanced work-energy problems
TypeInverse-square gravitational force
DifficultyChallenging +1.2 This is a standard M3 inverse-square law gravitation problem requiring energy conservation with variable force. While it involves integration and algebraic manipulation beyond basic mechanics, the approach is methodical: set up work-energy equation, integrate F·dx from initial to final position, and solve the resulting equation. The given force formula and speeds guide students through a well-established technique, making it moderately above average difficulty but not requiring novel insight.
Spec6.02c Work by variable force: using integration6.02i Conservation of energy: mechanical energy principle

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 \(\frac { m g R ^ { 2 } } { x ^ { 2 } }\). When \(P\) is at a distance \(2 R\) from the surface of the Earth, the speed of \(P\) is \(\sqrt { \frac { g R } { 3 } }\). Assuming that air resistance can be ignored, find the distance of \(P\) from the surface of the Earth when the speed of \(P\) is \(2 \sqrt { \frac { g R } { 3 } }\).

Question 3:
AnswerMarks Guidance
Working/AnswerMarks Guidance
\(mv\dfrac{\text{d}v}{\text{d}x} = -\dfrac{mgR^2}{x^2}\)M1 Equation of motion with correct number of terms and acceleration \(v\dfrac{\text{d}v}{\text{d}x}\); allow with minus missing
\(\dfrac{1}{2}v^2 = gR^2x^{-1}\ (+c)\)dM1A1 Attempt integration of both sides; \(x^{-2} \to x^{-1}\); correct equation after integration; constant of integration may be missing
\(x = 3R,\ v = \sqrt{\dfrac{gR}{3}} \Rightarrow c = \dfrac{1}{2}\cdot\dfrac{gR}{3} - \dfrac{gR^2}{3R} = -\dfrac{gR}{6}\)dM1 Substitute \(x=3R\), \(v=\sqrt{\dfrac{gR}{3}}\) and obtain expression for \(c\)
\(v = 2\sqrt{\dfrac{gR}{3}}\): \(\dfrac{2gR}{3} = \dfrac{gR^2}{x} - \dfrac{gR}{6}\), solve for \(x\)dM1 Substitute \(v = 2\sqrt{\dfrac{gR}{3}}\) and solve for \(x\); depends on first M mark
\(x = \dfrac{6R}{5}\)A1 Correct \(x\); double sign error scores A0
Distance from surface \(= \dfrac{6R}{5} - R = \dfrac{R}{5}\)A1cso [7] Correct answer from completely correct working
# Question 3:

| Working/Answer | Marks | Guidance |
|---|---|---|
| $mv\dfrac{\text{d}v}{\text{d}x} = -\dfrac{mgR^2}{x^2}$ | M1 | Equation of motion with correct number of terms and acceleration $v\dfrac{\text{d}v}{\text{d}x}$; allow with minus missing |
| $\dfrac{1}{2}v^2 = gR^2x^{-1}\ (+c)$ | dM1A1 | Attempt integration of both sides; $x^{-2} \to x^{-1}$; correct equation after integration; constant of integration may be missing |
| $x = 3R,\ v = \sqrt{\dfrac{gR}{3}} \Rightarrow c = \dfrac{1}{2}\cdot\dfrac{gR}{3} - \dfrac{gR^2}{3R} = -\dfrac{gR}{6}$ | dM1 | Substitute $x=3R$, $v=\sqrt{\dfrac{gR}{3}}$ and obtain expression for $c$ |
| $v = 2\sqrt{\dfrac{gR}{3}}$: $\dfrac{2gR}{3} = \dfrac{gR^2}{x} - \dfrac{gR}{6}$, solve for $x$ | dM1 | Substitute $v = 2\sqrt{\dfrac{gR}{3}}$ and solve for $x$; depends on first M mark |
| $x = \dfrac{6R}{5}$ | A1 | Correct $x$; double sign error scores A0 |
| Distance from surface $= \dfrac{6R}{5} - R = \dfrac{R}{5}$ | A1cso [7] | Correct answer from completely correct working |

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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 $\frac { m g R ^ { 2 } } { x ^ { 2 } }$. When $P$ is at a distance $2 R$ from the surface of the Earth, the speed of $P$ is $\sqrt { \frac { g R } { 3 } }$.

Assuming that air resistance can be ignored, find the distance of $P$ from the surface of the Earth when the speed of $P$ is $2 \sqrt { \frac { g R } { 3 } }$.

\hfill \mbox{\textit{Edexcel M3 2018 Q3 [7]}}