| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2020 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Inverse power force - non-gravitational context |
| Difficulty | Standard +0.8 This M3 variable force question requires setting up F=ma with inverse cube force, integrating to find v² (using chain rule dv/dt = v dv/dx), then separating variables and integrating again for time. While methodical, it demands fluency with multiple integration techniques and careful algebraic manipulation across two connected parts, making it moderately challenging but within standard M3 scope. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| VIXV SIHIANI III IM IONOO | VIAV SIHI NI JYHAM ION OO | VI4V SIHI NI JLIYM ION OO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0.5v\dfrac{dv}{dx} = -\dfrac{2}{x^3}\) | M1 | Dimensionally correct equation of motion. Acceleration must be in form \(v\dfrac{dv}{dx}\). Condone missing minus sign. |
| \(\int v \, dv = \int -\dfrac{4}{x^3} \, dx\) | DM1 | Separate variables and attempt integration. Condone missing minus sign. |
| \(\dfrac{v^2}{2} = \dfrac{2}{x^2} \quad (+k)\) | A1 | Correct integrals. Condone missing constant. |
| \(v = 3, x = 1 \Rightarrow 2k = 5\) | DM1 | Use \(v=3, x=1\) to find constant of integration. Dependent on both previous M marks. |
| \(v^2 = \dfrac{4}{x^2} + 5\) | A1* | Reach given result with no errors. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2}{\times}2v^2 - \frac{1}{2}{\times}2{\times}3^2 = \int(\pm)\dfrac{2}{x^3}\,dx\) | M1 | Equate change in KE to integral for WD. Integration not needed for this mark. Condone inconsistent signs. |
| Attempt integration | DM1 | Condone inconsistent signs. |
| Correct integration with consistent signs for KE and WD | A1 | |
| Substitute in limits | DM1 | |
| Reach given result with no errors | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{dx}{dt} = \sqrt{\dfrac{4}{x^2}+5}\) | M1 | Use \(v = \dfrac{dx}{dt}\) to form differential equation in \(x\) and \(t\). |
| \(\int \dfrac{x}{\sqrt{5x^2+4}} \, dx = \int dt\) | M1, A1 | Separate variables and produce functions ready to be integrated. A1: correct integrands written in integrable form. |
| \(\dfrac{1}{5}\sqrt{5x^2+4} = t \quad (+k_1)\) | DM1, A1 | Valid attempt to integrate. A1: correct integration, condone missing constant. |
| \(x=1, t=1 \Rightarrow k_1 = -\dfrac{2}{5}\) | DM1 | Use \(x=1, t=1\) to find constant. Dependent on all 3 M marks. |
| \(t = \dfrac{2+\sqrt{5x^2+4}}{5}\) | A1 | Correct result (cso). |
# Question 5(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.5v\dfrac{dv}{dx} = -\dfrac{2}{x^3}$ | M1 | Dimensionally correct equation of motion. Acceleration must be in form $v\dfrac{dv}{dx}$. Condone missing minus sign. |
| $\int v \, dv = \int -\dfrac{4}{x^3} \, dx$ | DM1 | Separate variables and attempt integration. Condone missing minus sign. |
| $\dfrac{v^2}{2} = \dfrac{2}{x^2} \quad (+k)$ | A1 | Correct integrals. Condone missing constant. |
| $v = 3, x = 1 \Rightarrow 2k = 5$ | DM1 | Use $v=3, x=1$ to find constant of integration. Dependent on both previous M marks. |
| $v^2 = \dfrac{4}{x^2} + 5$ | A1* | Reach given result with no errors. |
**Alt using Energy/Work:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}{\times}2v^2 - \frac{1}{2}{\times}2{\times}3^2 = \int(\pm)\dfrac{2}{x^3}\,dx$ | M1 | Equate change in KE to integral for WD. Integration not needed for this mark. Condone inconsistent signs. |
| Attempt integration | DM1 | Condone inconsistent signs. |
| Correct integration with consistent signs for KE and WD | A1 | |
| Substitute in limits | DM1 | |
| Reach given result with no errors | A1 | |
---
# Question 5(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{dx}{dt} = \sqrt{\dfrac{4}{x^2}+5}$ | M1 | Use $v = \dfrac{dx}{dt}$ to form differential equation in $x$ and $t$. |
| $\int \dfrac{x}{\sqrt{5x^2+4}} \, dx = \int dt$ | M1, A1 | Separate variables and produce functions ready to be integrated. A1: correct integrands written in integrable form. |
| $\dfrac{1}{5}\sqrt{5x^2+4} = t \quad (+k_1)$ | DM1, A1 | Valid attempt to integrate. A1: correct integration, condone missing constant. |
| $x=1, t=1 \Rightarrow k_1 = -\dfrac{2}{5}$ | DM1 | Use $x=1, t=1$ to find constant. Dependent on all 3 M marks. |
| $t = \dfrac{2+\sqrt{5x^2+4}}{5}$ | A1 | Correct result (cso). |
---5. A particle $P$ of mass 0.5 kg moves along the positive $x$-axis in the positive $x$ direction. At time $t$ seconds, $t \geqslant 1 , P$ is $x$ metres from the origin $O$ and is moving with speed $v \mathrm {~ms} ^ { - 1 }$. The resultant force acting on $P$ has magnitude $\frac { 2 } { x ^ { 3 } } \mathrm {~N}$ and is directed towards $O$.
When $t = 1 , x = 1$ and $v = 3$\\
Show that
\begin{enumerate}[label=(\alph*)]
\item $v ^ { 2 } = \frac { 4 } { x ^ { 2 } } + 5$
\item $t = \frac { a + \sqrt { b x ^ { 2 } + c } } { d }$, where $a , b , c$ and $d$ are integers to be found.\\
\includegraphics[max width=\textwidth, alt={}, center]{ace84823-db30-463e-b24b-f0cd7df73746-13_2255_50_314_34}\\
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VIXV SIHIANI III IM IONOO & VIAV SIHI NI JYHAM ION OO & VI4V SIHI NI JLIYM ION OO \\
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\hfill \mbox{\textit{Edexcel M3 2020 Q5 [12]}}