| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2021 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Given velocity function find force |
| Difficulty | Standard +0.8 This M3 question requires using F=ma with position-dependent force, applying the chain rule (v dv/dx = a), integrating a trigonometric expression, then separating variables and integrating sec(x) to find time. While systematic, it demands fluency with multiple techniques (variable force, chain rule for acceleration, integration of trig functions including sec x) and careful algebraic manipulation across two connected parts, making it moderately challenging but within reach of well-prepared M3 students. |
| Spec | 1.05o Trigonometric equations: solve in given intervals6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(0.5v\dfrac{dv}{dx} = -\sin 2x\) | M1 | Equation of motion with acceleration in form \(v\dfrac{dv}{dx}\). Condone sign error. |
| \(\int 0.5v\, dv = \int -\sin 2x\, dx\) | DM1 | Separate variables to prepare for integration. Depends on M mark above. |
| \(0.25v^2 = \dfrac{1}{2}\cos 2x\ (+c)\) | A1 | Correct integration. Constant not needed. |
| \(x=0, v=2 \Rightarrow 4 = 2 + c\) | DM1 | Substitute \(x=0\), \(v=2\) to find constant. Depends on both M marks above. |
| \(v^2 = 2\cos 2x + 2\ (= 4\cos^2 x)\) | A1 | A correct result for \(v^2\) |
| \(v = 2\cos x\) ✱ | A1✱ | Given result reached through use of double angle formula. (Formula need not be shown.) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(0.5v\dfrac{dv}{dx} = -\sin 2x\) | M1 | Equation of motion with acceleration in form \(v\dfrac{dv}{dx}\). Condone sign error. |
| \(\int_2^v 0.5v\, dv = \int_0^x -\sin 2x\, dx\) | DM1 | Separate variables, prepare for integration. Limits not needed for this mark. Depends on M mark above. |
| \(\left[0.25v^2\right]_2^v = \left[\dfrac{1}{2}\cos 2x\right]_0^x\) | A1 | Correct integration — limits not needed |
| \(0.25(v^2 - 4) = \dfrac{1}{2}\cos 2x - \dfrac{1}{2}\) | DM1A1 | Correct substitution of correct limits. Limits must be "paired" correctly. Correct expression which can yield \(v^2\). |
| \(v = 2\cos x\) ✱ | A1✱ | Given result through use of double angle formula. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\dfrac{dx}{dt} = 2\cos x\) | M1 | Use of \(v = \dfrac{dx}{dt}\) |
| \(\int \sec x\, dx = \int 2\, dt\) | DM1 | Correct separation and attempt integration (integral in formula book). Depends on first M of (b). Modulus signs may be missing. |
| \(\ln | \sec x + \tan x | = 2t + k\) |
| \(t=0, x=0 \Rightarrow \ln 1 = 2(0) + k \Rightarrow k=0\) | ||
| \(t = \dfrac{1}{2}\ln | \sec x + \tan x | = \dfrac{1}{2}\ln\!\left(\sec\dfrac{\pi}{4} + \tan\dfrac{\pi}{4}\right)\) |
| \(t = \dfrac{1}{2}\ln(\sqrt{2}+1)\) ✱ | A1✱ | Given result reached from fully correct working. (Modulus signs may be missing throughout.) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\dfrac{dx}{dt} = 2\cos x\) | M1 | Use of \(v = \dfrac{dx}{dt}\) |
| \(\int_0^{\pi/4} \sec x\, dx = \int_0^t 2\, dt\) | DM1 | Correct separation and attempt integration. Limits not needed. Depends on first M of (b). Modulus signs may be missing. |
| \(\left[\ln | \sec x + \tan x | \right]_0^{\pi/4} = \left[2t\right]_0^t\) |
| \(2t = \ln\!\left(\sec\dfrac{\pi}{4} + \tan\dfrac{\pi}{4}\right)\) | DM1 | Substitute limits and solve for \(t\). Depends on both previous M marks in (b). |
| \(t = \dfrac{1}{2}\ln(\sqrt{2}+1)\) ✱ | A1✱ | Given result from fully correct working. (Modulus signs may be missing throughout.) |
## Question 5:
### Part 5a:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0.5v\dfrac{dv}{dx} = -\sin 2x$ | M1 | Equation of motion with acceleration in form $v\dfrac{dv}{dx}$. Condone sign error. |
| $\int 0.5v\, dv = \int -\sin 2x\, dx$ | DM1 | Separate variables to prepare for integration. Depends on M mark above. |
| $0.25v^2 = \dfrac{1}{2}\cos 2x\ (+c)$ | A1 | Correct integration. Constant not needed. |
| $x=0, v=2 \Rightarrow 4 = 2 + c$ | DM1 | Substitute $x=0$, $v=2$ to find constant. Depends on both M marks above. |
| $v^2 = 2\cos 2x + 2\ (= 4\cos^2 x)$ | A1 | A correct result for $v^2$ |
| $v = 2\cos x$ ✱ | A1✱ | **Given** result reached through use of double angle formula. (Formula need not be shown.) |
**ALT (Definite integration):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0.5v\dfrac{dv}{dx} = -\sin 2x$ | M1 | Equation of motion with acceleration in form $v\dfrac{dv}{dx}$. Condone sign error. |
| $\int_2^v 0.5v\, dv = \int_0^x -\sin 2x\, dx$ | DM1 | Separate variables, prepare for integration. Limits not needed for this mark. Depends on M mark above. |
| $\left[0.25v^2\right]_2^v = \left[\dfrac{1}{2}\cos 2x\right]_0^x$ | A1 | Correct integration — limits not needed |
| $0.25(v^2 - 4) = \dfrac{1}{2}\cos 2x - \dfrac{1}{2}$ | DM1A1 | Correct substitution of correct limits. Limits must be "paired" correctly. Correct expression which can yield $v^2$. |
| $v = 2\cos x$ ✱ | A1✱ | **Given** result through use of double angle formula. |
**Total: (6)**
---
### Part 5b:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\dfrac{dx}{dt} = 2\cos x$ | M1 | Use of $v = \dfrac{dx}{dt}$ |
| $\int \sec x\, dx = \int 2\, dt$ | DM1 | Correct separation and attempt integration (integral in formula book). Depends on first M of (b). Modulus signs may be missing. |
| $\ln|\sec x + \tan x| = 2t + k$ | A1 | Correct integration and use of limits to find correct value for constant. |
| $t=0, x=0 \Rightarrow \ln 1 = 2(0) + k \Rightarrow k=0$ | | |
| $t = \dfrac{1}{2}\ln|\sec x + \tan x| = \dfrac{1}{2}\ln\!\left(\sec\dfrac{\pi}{4} + \tan\dfrac{\pi}{4}\right)$ | DM1 | Substitute $x = \dfrac{\pi}{4}$ and solve for $t$. Depends on both previous M marks in (b). |
| $t = \dfrac{1}{2}\ln(\sqrt{2}+1)$ ✱ | A1✱ | **Given** result reached from fully correct working. (Modulus signs may be missing throughout.) |
**ALT (Definite integration):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\dfrac{dx}{dt} = 2\cos x$ | M1 | Use of $v = \dfrac{dx}{dt}$ |
| $\int_0^{\pi/4} \sec x\, dx = \int_0^t 2\, dt$ | DM1 | Correct separation and attempt integration. Limits not needed. Depends on first M of (b). Modulus signs may be missing. |
| $\left[\ln|\sec x + \tan x|\right]_0^{\pi/4} = \left[2t\right]_0^t$ | A1 | Correct integration including correct limits. |
| $2t = \ln\!\left(\sec\dfrac{\pi}{4} + \tan\dfrac{\pi}{4}\right)$ | DM1 | Substitute limits and solve for $t$. Depends on both previous M marks in (b). |
| $t = \dfrac{1}{2}\ln(\sqrt{2}+1)$ ✱ | A1✱ | **Given** result from fully correct working. (Modulus signs may be missing throughout.) |
**Total: (5)**
**Overall Total: [11]**\begin{enumerate}
\item A particle $P$ of mass 0.5 kg moves on the $x$-axis under the action of a single force.
\end{enumerate}
At time $t$ seconds, $t \geqslant 0$
\begin{itemize}
\item $O P = x$ metres, $0 \leqslant x < \frac { \pi } { 2 }$
\item the force has magnitude $\sin 2 x \mathrm {~N}$ and is directed towards the origin $O$
\item $P$ is moving in the positive $x$ direction with speed $v \mathrm {~ms} ^ { - 1 }$
\end{itemize}
At time $t = 0 , P$ passes through the origin with speed $2 \mathrm {~ms} ^ { - 1 }$\\
(a) Show that $v = 2 \cos x$\\
(b) Show that $t = \frac { 1 } { 2 } \ln ( \sqrt { 2 } + 1 )$ when $x = \frac { \pi } { 4 }$
\hfill \mbox{\textit{Edexcel M3 2021 Q5 [11]}}