| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2023 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Inverse-square gravitational force |
| Difficulty | Standard +0.8 This M3 question involves inverse-square gravitational force and requires deriving an energy equation using calculus (part a), then applying it to find distances and escape velocity. While the techniques are standard for M3 (using v dv/dx = a and integration), the inverse-square law context and the conceptual understanding needed for escape velocity (part c) elevate this above routine mechanics questions. It's moderately challenging but within expected M3 scope. |
| Spec | 3.02h Motion under gravity: vector form6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v\frac{dv}{dx} = -\frac{gR^2}{x^2}\) | M1 A1 | M1: equation with or without -ve sign and any derivative form for acceleration. A1: correct equation with -ve sign |
| \(\int v\, dv = -\int \frac{gR^2}{x^2}\, dx\) or \(\frac{1}{2}V^2 = -\int \frac{gR^2}{x^2}\, dx\) | M1 | Separate variables and clear attempt to integrate acceleration in terms of \(v\) and \(x\) |
| \(\frac{1}{2}v^2 = \frac{gR^2}{x} + C\) | A1 | Correct equation; allow without \(C\) |
| Use of \(x = R\), \(v = U\) to find \(C\) where \(C = \frac{1}{2}u^2 - gR\) | M1 | Use of initial conditions or limits |
| \(v^2 = \frac{2gR^2}{x} + U^2 - 2gR\) | A1* | Given answer correctly obtained |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2}mv^2 - \frac{1}{2}mU^2 = \int F\, dx = -\int \frac{mgR^2}{x^2}\, dx\) | M1 A1 | Energy equation with variable force; sign may be missing for M mark |
| \(\frac{1}{2}mv^2 - \frac{1}{2}mU^2 = \left[\frac{mgR^2}{x}\right]_R^x\) | M1 A1 | Clear attempt to integrate; limits may be missing or incorrect |
| \(\frac{1}{2}mv^2 - \frac{1}{2}mU^2 = \frac{mgR^2}{x} - \frac{mgR^2}{R}\) | M1 | Correct limits substituted the right way round |
| \(v^2 = \frac{2gR^2}{x} + U^2 - 2gR\) | A1* | Given answer correctly obtained |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{4}gR = \frac{2gR^2}{x} + gR - 2gR\) | M1 | Substitution of \(v^2\) and \(U^2\) into (a) to produce a correct equation |
| \(x = \frac{8R}{5}\) | A1 | Correct value of \(x\) |
| \(AB = \frac{3R}{5}\) oe | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct statement regarding \(\frac{2gR^2}{x}\), e.g. \(\frac{2gR^2}{x} > 0\) for \(x \geq R\), or as \(x \to \infty\), \(\frac{2gR^2}{x} \to 0\) | M1 | Accept \(x = \infty\); \(x \to \infty\), \(\frac{2gR^2}{x} = 0\) |
| Correct reasoning: \(U^2 - 2gR = 0\), or \(U^2 \to 2gR\), or \(U^2 \geq 2gR\) | dM1 | Dependent on previous M. Correct reasoning leading to correct equation or inequality |
| \(U_{\min} = \sqrt{2gR}\) | A1 | cso |
## Question 5(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v\frac{dv}{dx} = -\frac{gR^2}{x^2}$ | M1 A1 | M1: equation with or without -ve sign and any derivative form for acceleration. A1: correct equation with -ve sign |
| $\int v\, dv = -\int \frac{gR^2}{x^2}\, dx$ or $\frac{1}{2}V^2 = -\int \frac{gR^2}{x^2}\, dx$ | M1 | Separate variables and clear attempt to integrate acceleration in terms of $v$ and $x$ |
| $\frac{1}{2}v^2 = \frac{gR^2}{x} + C$ | A1 | Correct equation; allow without $C$ |
| Use of $x = R$, $v = U$ to find $C$ where $C = \frac{1}{2}u^2 - gR$ | M1 | Use of initial conditions or limits |
| $v^2 = \frac{2gR^2}{x} + U^2 - 2gR$ | A1* | Given answer correctly obtained |
**ALT 5a (Energy approach):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}mv^2 - \frac{1}{2}mU^2 = \int F\, dx = -\int \frac{mgR^2}{x^2}\, dx$ | M1 A1 | Energy equation with variable force; sign may be missing for M mark |
| $\frac{1}{2}mv^2 - \frac{1}{2}mU^2 = \left[\frac{mgR^2}{x}\right]_R^x$ | M1 A1 | Clear attempt to integrate; limits may be missing or incorrect |
| $\frac{1}{2}mv^2 - \frac{1}{2}mU^2 = \frac{mgR^2}{x} - \frac{mgR^2}{R}$ | M1 | Correct limits substituted the right way round |
| $v^2 = \frac{2gR^2}{x} + U^2 - 2gR$ | A1* | Given answer correctly obtained |
---
## Question 5(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{4}gR = \frac{2gR^2}{x} + gR - 2gR$ | M1 | Substitution of $v^2$ and $U^2$ into (a) to produce a correct equation |
| $x = \frac{8R}{5}$ | A1 | Correct value of $x$ |
| $AB = \frac{3R}{5}$ oe | A1 | cao |
---
## Question 5(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct statement regarding $\frac{2gR^2}{x}$, e.g. $\frac{2gR^2}{x} > 0$ for $x \geq R$, or as $x \to \infty$, $\frac{2gR^2}{x} \to 0$ | M1 | Accept $x = \infty$; $x \to \infty$, $\frac{2gR^2}{x} = 0$ |
| Correct reasoning: $U^2 - 2gR = 0$, or $U^2 \to 2gR$, or $U^2 \geq 2gR$ | dM1 | Dependent on previous M. Correct reasoning leading to correct equation or inequality |
| $U_{\min} = \sqrt{2gR}$ | A1 | cso |
---\begin{enumerate}
\item The centre of the Earth is the point $O$ and the Earth is modelled as a fixed sphere of radius $R$.\\
At time $t = 0$, a particle $P$ is projected vertically upwards with speed $U$ from a point $A$ on the surface of the Earth.
\end{enumerate}
At time $t$ seconds, where $t \geqslant 0$
\begin{itemize}
\item $\quad P$ is a distance $x$ from $O$
\item $P$ is moving with speed $v$
\item $P$ has acceleration of magnitude $\frac { g R ^ { 2 } } { x ^ { 2 } }$ directed towards $O$
\end{itemize}
Air resistance is modelled as being negligible.\\
(a) Show that $v ^ { 2 } = \frac { 2 g R ^ { 2 } } { x } + U ^ { 2 } - 2 g R$
Particle $P$ is first moving with speed $\frac { 1 } { 2 } \sqrt { g R }$ at the point $B$.\\
(b) Given that $U = \sqrt { g R }$ find, in terms of $R$, the distance $A B$.\\
(c) Find, in terms of $g$ and $R$, the smallest value of $U$ that would ensure that $P$ never comes to rest, explaining your reasoning.
\hfill \mbox{\textit{Edexcel M3 2023 Q5 [12]}}