| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2024 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Inverse-square gravitational force |
| Difficulty | Standard +0.3 This is a standard M3 gravitational mechanics question requiring integration of F=ma with variable force. Part (a) is a straightforward 'show that' using separation of variables (v dv/dx = acceleration), and part (b) substitutes given values to find C then evaluates at x=R. The setup is clearly guided with all necessary information provided, making it slightly easier than average for A-level. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration6.02c Work by variable force: using integration6.02d Mechanical energy: KE and PE concepts6.02i Conservation of energy: mechanical energy principle6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(-\dfrac{mgR^2}{2x^2} = mv\dfrac{\mathrm{d}v}{\mathrm{d}x}\) | M1 | Form differential equation in \(v\) and \(x\) only. Need to see \(\dfrac{\mathrm{d}v}{\mathrm{d}x}\) or \(v\dfrac{\mathrm{d}v}{\mathrm{d}x}\). Cannot get this mark using \(t\). Allow with both \(m\)'s cancelled. Condone sign error. |
| \(-\dfrac{gR^2}{2}\int\dfrac{1}{x^2}\mathrm{d}x = \int v\mathrm{d}v\) | M1 | Separate variables correctly and integrate at least one side. Cannot get this mark using \(t\). Condone sign error. |
| \(v^2 = \dfrac{gR^2}{x} + C\) * | A1* | Obtain given answer from correct work. Must include at least one line of working between integral and final answer. Correct signs seen throughout working. Condone \(\dfrac{v^2}{2} = \dfrac{gR^2}{2x} + C\) followed by \(v^2 = \dfrac{gR^2}{x} + C\). Note: If first line of working is \(\dfrac{1}{2}v^2 = -\int\dfrac{gR^2}{2x^2}\,\mathrm{d}x\) followed by integration of RHS, this scores M0M1A0* |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{1}{2}mu^2 - \dfrac{1}{2}mv^2 = \int\dfrac{gmR^2}{x^2}\mathrm{d}x\) | M1 | Form an energy equation with 2 KE terms and the integral of the variable force. Condone sign errors. |
| \(\dfrac{1}{2}mu^2 - \dfrac{1}{2}mv^2 = -\dfrac{gmR^2}{x} + A\) | M1 | Integrate the force wrt \(x\). Condone sign errors. |
| \(v^2 = \dfrac{gR^2}{x} + C\) * | A1* | Obtain given answer from correct work. Must include at least one line of working and correct signs seen throughout working. |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = 3R,\, v^2 = 3gR\) | M1 | Use initial conditions to evaluate \(C\) in the given answer. |
| \(\Rightarrow C = 3gR - \dfrac{gR^2}{3R}\left(= \dfrac{8gR}{3}\right)\) | A1 | Or equivalent |
| \(x = R \Rightarrow v = \sqrt{\dfrac{11gR}{3}}\) | A1 | Accept \(\dfrac{\sqrt{33gR}}{3}\). Answer must be in terms of \(g\) and \(R\). |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left[v^2\right]_{\sqrt{3gR}}^{v} = \left[\dfrac{gR^2}{x}\right]_{3R}^{R}\) | M1 | Use initial conditions in a definite integral. |
| \(v^2 - 3gR = \dfrac{gR^2}{R} - \dfrac{gR^2}{3R}\) | A1 | Or equivalent |
| \(v = \sqrt{\dfrac{11gR}{3}}\) | A1 | Accept \(\dfrac{\sqrt{33gR}}{3}\). Answer must be in terms of \(g\) and \(R\). |
| [6] total |
# Question 1:
## Part 1a:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $-\dfrac{mgR^2}{2x^2} = mv\dfrac{\mathrm{d}v}{\mathrm{d}x}$ | M1 | Form differential equation in $v$ and $x$ only. Need to see $\dfrac{\mathrm{d}v}{\mathrm{d}x}$ or $v\dfrac{\mathrm{d}v}{\mathrm{d}x}$. Cannot get this mark using $t$. Allow with both $m$'s cancelled. Condone sign error. |
| $-\dfrac{gR^2}{2}\int\dfrac{1}{x^2}\mathrm{d}x = \int v\mathrm{d}v$ | M1 | Separate variables correctly and integrate at least one side. Cannot get this mark using $t$. Condone sign error. |
| $v^2 = \dfrac{gR^2}{x} + C$ * | A1* | Obtain given answer from correct work. Must include at least one line of working between integral and final answer. Correct signs seen throughout working. Condone $\dfrac{v^2}{2} = \dfrac{gR^2}{2x} + C$ followed by $v^2 = \dfrac{gR^2}{x} + C$. Note: If first line of working is $\dfrac{1}{2}v^2 = -\int\dfrac{gR^2}{2x^2}\,\mathrm{d}x$ followed by integration of RHS, this scores M0M1A0* |
**ALT1(a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{1}{2}mu^2 - \dfrac{1}{2}mv^2 = \int\dfrac{gmR^2}{x^2}\mathrm{d}x$ | M1 | Form an energy equation with 2 KE terms and the integral of the variable force. Condone sign errors. |
| $\dfrac{1}{2}mu^2 - \dfrac{1}{2}mv^2 = -\dfrac{gmR^2}{x} + A$ | M1 | Integrate the force wrt $x$. Condone sign errors. |
| $v^2 = \dfrac{gR^2}{x} + C$ * | A1* | Obtain given answer from correct work. Must include at least one line of working and correct signs seen throughout working. |
| | [3] | |
## Part 1b:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = 3R,\, v^2 = 3gR$ | M1 | Use initial conditions to evaluate $C$ in the given answer. |
| $\Rightarrow C = 3gR - \dfrac{gR^2}{3R}\left(= \dfrac{8gR}{3}\right)$ | A1 | Or equivalent |
| $x = R \Rightarrow v = \sqrt{\dfrac{11gR}{3}}$ | A1 | Accept $\dfrac{\sqrt{33gR}}{3}$. Answer must be in terms of $g$ and $R$. |
| | [3] | |
**ALT1(b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left[v^2\right]_{\sqrt{3gR}}^{v} = \left[\dfrac{gR^2}{x}\right]_{3R}^{R}$ | M1 | Use initial conditions in a definite integral. |
| $v^2 - 3gR = \dfrac{gR^2}{R} - \dfrac{gR^2}{3R}$ | A1 | Or equivalent |
| $v = \sqrt{\dfrac{11gR}{3}}$ | A1 | Accept $\dfrac{\sqrt{33gR}}{3}$. Answer must be in terms of $g$ and $R$. |
| | [6] total | |
---\begin{enumerate}
\item A spacecraft $S$ of mass $m$ moves in a straight line towards the centre, $O$, of a planet.
\end{enumerate}
The planet is modelled as a fixed sphere of radius $R$.\\
The spacecraft $S$ is modelled as a particle.\\
The gravitational force of the planet is the only force acting on $S$.\\
When $S$ is a distance $x ( x \geqslant R )$ from $O$
\begin{itemize}
\item the gravitational force is directed towards $O$ and has magnitude $\frac { m g R ^ { 2 } } { 2 x ^ { 2 } }$
\item the speed of $S$ is $v$\\
(a) Show that
\end{itemize}
$$v ^ { 2 } = \frac { g R ^ { 2 } } { x } + C$$
where $C$ is a constant.
When $x = 3 R , v = \sqrt { 3 g R }$\\
(b) Find, in terms of $g$ and $R$, the speed of $S$ as it hits the surface of the planet.
\hfill \mbox{\textit{Edexcel M3 2024 Q1 [6]}}