| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2021 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Inverse-square gravitational force |
| Difficulty | Standard +0.8 This M3 question requires understanding of variable gravitational force and energy methods beyond standard constant-g mechanics. Part (a) is a standard 'show that' using inverse square law and surface conditions. Part (b) requires integrating variable force or applying energy conservation with non-constant gravity—a step up from typical A-level mechanics but standard for M3 students who have learned these techniques. |
| Spec | 1.08a Fundamental theorem of calculus: integration as reverse of differentiation3.02h Motion under gravity: vector form6.02e Calculate KE and PE: using formulae6.06a Variable force: dv/dt or v*dv/dx methods |
| VIAV SIHI NI III IM I ON OC | VIAV SIMI NI III M M O N OO | VIUV SIMI NI JIIYM ION OC |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(F = \dfrac{k}{(x+R)^2}\) | M1 | Setting up inverse square relationship between \(F\) and \((x+R)\). Can be negative. Allow \(d = x+R\) or \(k = GMm\). |
| \(x=0,\ F=mg \Rightarrow mg = \dfrac{k}{R^2}\) | M1 | Clear use of \(x=0\) and \(F=mg\) to find value of constant (\(k\) or \(GM\)). |
| \(k = mgR^2 \Rightarrow F = \dfrac{mgR^2}{(x+R)^2}\) | A1* | Given result reached with both M marks clearly earned. Must be positive. |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(mv\dfrac{dv}{dx} = -\dfrac{mgR^2}{(x+R)^2}\) or \(m\dfrac{d}{dx}\!\left(\tfrac{1}{2}v^2\right) = -\dfrac{mgR^2}{(x+R)^2}\) | M1 | Use of \(mv\frac{dv}{dx}\) or \(m\frac{d}{dx}(\frac{1}{2}v^2)\). Condone sign error. |
| \(\tfrac{1}{2}v^2 = -\int \dfrac{gR^2}{(x+R)^2}\,dx\) | dM1 | Separate variables to produce form ready for integration. Condone sign error. |
| \(\tfrac{1}{2}v^2 = \dfrac{gR^2}{x+R}\ (+c)\) | A1 | Correct integration. Sign must be correct now. Constant of integration not needed. |
| \(x=R,\ v=U\) | M1 | Use of initial conditions in result of integration to find constant. |
| \(\dfrac{U^2}{2} = \dfrac{gR^2}{2R} + c \Rightarrow c = \dfrac{U^2 - gR}{2}\) | A1 | Correct value of \(c\). |
| \(x=0 \Rightarrow \tfrac{1}{2}v^2 = gR + \dfrac{U^2-gR}{2}\) | ||
| \(v^2 = U^2 + gR \Rightarrow v = \sqrt{U^2 + gR}\) | M1, A1 | M1: Finding \(v\) using \(x=0\), \(v^2\) from dimensionally correct expression. A1: Correct expression for \(v\). |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\dfrac{mv^2}{2} - \dfrac{mU^2}{2} = -m\int_R^0 \dfrac{gR^2}{(x+R)^2}\,dx\) | M1 | Equates change in KE to integral of \(F\). Condone sign error. |
| \(\dfrac{v^2}{2} - \dfrac{U^2}{2} = \left[\dfrac{gR^2}{x+R}\right]_R^0\) | dM1 A1 | dM1: Integrates (power of \((x+R)\) must increase). A1: Correct integration, sign must be correct for LHS. |
| \(\dfrac{v^2}{2} - \dfrac{U^2}{2} = \dfrac{gR^2}{R} - \dfrac{gR^2}{2R}\) | M1 A1 | M1: Substitution of both limits \(R\) and \(0\). A1: Correct limits, correct way round. |
| \(v^2 = U^2 + gR \Rightarrow v = \sqrt{U^2+gR}\) | M1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(mv\dfrac{dv}{dx} = -\dfrac{mgR^2}{(x+R)^2}\) | M1 | Use of \(mv\frac{dv}{dx}\). Condone sign error. |
| \(\int_U^V v\,dv = -\int_R^0 \dfrac{gR^2}{(x+R)^2}\,dx\) | dM1 | Separate variables, form ready for integration. Condone sign error. Limits not needed. |
| \(\left[\dfrac{v^2}{2}\right]_U^V = \left[\dfrac{gR^2}{x+R}\right]_R^0\) | A1 | Correct integration. Sign must be correct now. |
| \(\dfrac{v^2}{2} - \dfrac{U^2}{2} = \dfrac{gR^2}{R} - \dfrac{gR^2}{2R}\) | M1 A1 | M1: Substitution of both limits; must be \(0, R, v\) and \(U\). A1: Correct limits, correct way round. |
| \(v^2 = U^2 + gR \Rightarrow v = \sqrt{U^2+gR}\) | M1, A1 |
## Question 2:
### Part (a)
| Working | Mark | Guidance |
|---------|------|----------|
| $F = \dfrac{k}{(x+R)^2}$ | M1 | Setting up inverse square relationship between $F$ and $(x+R)$. Can be negative. Allow $d = x+R$ or $k = GMm$. |
| $x=0,\ F=mg \Rightarrow mg = \dfrac{k}{R^2}$ | M1 | Clear use of $x=0$ and $F=mg$ to find value of constant ($k$ or $GM$). |
| $k = mgR^2 \Rightarrow F = \dfrac{mgR^2}{(x+R)^2}$ | A1* | Given result reached with both M marks clearly earned. Must be positive. |
### Part (b) — Main Scheme
| Working | Mark | Guidance |
|---------|------|----------|
| $mv\dfrac{dv}{dx} = -\dfrac{mgR^2}{(x+R)^2}$ or $m\dfrac{d}{dx}\!\left(\tfrac{1}{2}v^2\right) = -\dfrac{mgR^2}{(x+R)^2}$ | M1 | Use of $mv\frac{dv}{dx}$ or $m\frac{d}{dx}(\frac{1}{2}v^2)$. Condone sign error. |
| $\tfrac{1}{2}v^2 = -\int \dfrac{gR^2}{(x+R)^2}\,dx$ | dM1 | Separate variables to produce form ready for integration. Condone sign error. |
| $\tfrac{1}{2}v^2 = \dfrac{gR^2}{x+R}\ (+c)$ | A1 | Correct integration. Sign must be correct now. Constant of integration not needed. |
| $x=R,\ v=U$ | M1 | Use of initial conditions in result of integration to find constant. |
| $\dfrac{U^2}{2} = \dfrac{gR^2}{2R} + c \Rightarrow c = \dfrac{U^2 - gR}{2}$ | A1 | Correct value of $c$. |
| $x=0 \Rightarrow \tfrac{1}{2}v^2 = gR + \dfrac{U^2-gR}{2}$ | | |
| $v^2 = U^2 + gR \Rightarrow v = \sqrt{U^2 + gR}$ | M1, A1 | M1: Finding $v$ using $x=0$, $v^2$ from dimensionally correct expression. A1: Correct expression for $v$. |
### Part (b) — ALT1 (Change in KE = Work Done)
| Working | Mark | Guidance |
|---------|------|----------|
| $\dfrac{mv^2}{2} - \dfrac{mU^2}{2} = -m\int_R^0 \dfrac{gR^2}{(x+R)^2}\,dx$ | M1 | Equates change in KE to integral of $F$. Condone sign error. |
| $\dfrac{v^2}{2} - \dfrac{U^2}{2} = \left[\dfrac{gR^2}{x+R}\right]_R^0$ | dM1 A1 | dM1: Integrates (power of $(x+R)$ must increase). A1: Correct integration, sign must be correct for LHS. |
| $\dfrac{v^2}{2} - \dfrac{U^2}{2} = \dfrac{gR^2}{R} - \dfrac{gR^2}{2R}$ | M1 A1 | M1: Substitution of both limits $R$ and $0$. A1: Correct limits, correct way round. |
| $v^2 = U^2 + gR \Rightarrow v = \sqrt{U^2+gR}$ | M1, A1 | |
### Part (b) — ALT2 (Definite Integration)
| Working | Mark | Guidance |
|---------|------|----------|
| $mv\dfrac{dv}{dx} = -\dfrac{mgR^2}{(x+R)^2}$ | M1 | Use of $mv\frac{dv}{dx}$. Condone sign error. |
| $\int_U^V v\,dv = -\int_R^0 \dfrac{gR^2}{(x+R)^2}\,dx$ | dM1 | Separate variables, form ready for integration. Condone sign error. Limits not needed. |
| $\left[\dfrac{v^2}{2}\right]_U^V = \left[\dfrac{gR^2}{x+R}\right]_R^0$ | A1 | Correct integration. Sign must be correct now. |
| $\dfrac{v^2}{2} - \dfrac{U^2}{2} = \dfrac{gR^2}{R} - \dfrac{gR^2}{2R}$ | M1 A1 | M1: Substitution of both limits; must be $0, R, v$ and $U$. A1: Correct limits, correct way round. |
| $v^2 = U^2 + gR \Rightarrow v = \sqrt{U^2+gR}$ | M1, A1 | |2. A particle $P$ of mass $m$ is at a distance $x$ above the surface of the Earth. The Earth exerts a gravitational force on $P$. This force is directed towards the centre of the Earth. The magnitude of this force is inversely proportional to the square of the distance of $P$ from the centre of the Earth.
At the surface of the Earth the acceleration due to gravity is $g$.
The Earth is modelled as a fixed sphere of radius $R$.
\begin{enumerate}[label=(\alph*)]
\item Show that the magnitude of the gravitational force on $P$ is $\frac { m g R ^ { 2 } } { ( x + R ) ^ { 2 } }$
A particle is released from rest from a point above the surface of the Earth. When the particle is at a distance $R$ above the surface of the Earth, the particle has speed $U$.
Air resistance is modelled as being negligible.
\item Find, in terms of $U , g$ and $R$, the speed of the particle when it strikes the surface of the Earth.\\
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VIAV SIHI NI III IM I ON OC & VIAV SIMI NI III M M O N OO & VIUV SIMI NI JIIYM ION OC \\
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\hfill \mbox{\textit{Edexcel M3 2021 Q2 [10]}}