| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2018 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Force depends on time t |
| Difficulty | Standard +0.3 This is a standard M3 variable force question requiring integration of F=ma with force as a function of time, followed by a second integration for displacement. The steps are routine: set up the differential equation, integrate with given initial conditions, then integrate v to find x. The algebra is straightforward with simple fractions and logarithms, making this slightly easier than average for M3 content. |
| Spec | 1.08a Fundamental theorem of calculus: integration as reverse of differentiation3.03c Newton's second law: F=ma one dimension6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(0.4\frac{dv}{dt} = -\frac{8}{(t+4)^2}\) | M1 | Form equation of motion with acceleration in form shown (minus sign may be missing); 0.4 or \(m\) for mass for first 4 marks |
| \(v = \int -\frac{20}{(t+4)^2}\,dt\) | ||
| \(v = \frac{20}{(t+4)} (+c)\) | DM1A1 | DM1: attempt integration; A1: correct integration including correct sign, constant not needed |
| \(t=0, v=10 \Rightarrow 10 = 5 + c,\; c = 5\) | DM1 | Use \(t=0\), \(v=10\) to obtain value for \(c\) |
| \(v = \frac{20}{(t+4)} + 5\) | A1cso | Correct answer only, no errors seen |
| Part (a) Total: (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x = \int \frac{20}{(t+4)} + 5\; dt\) | ||
| \(x = 20\ln(t+4) + 5t + c\) | M1A1 | M1: integrate given expression for \(v\); A1: correct integration, constant not needed |
| \(t=0, x=0 \Rightarrow c = -20\ln 4\) | A1 | Obtain correct constant |
| \(v=6\): \(\frac{20}{t+4} = 1 \Rightarrow t = 16\) | B1 | Correct value of \(t\) when \(v=6\) |
| \(x = 20\ln 20 + 80 - 20\ln 4 = 80 + 20\ln 5\) | ||
| \(a = 80,\; b = 20\) | A1cao | Correct values for \(a\) and \(b\); need not be shown explicitly |
| Part (b) Total: (5) [10] |
## Question 3(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0.4\frac{dv}{dt} = -\frac{8}{(t+4)^2}$ | M1 | Form equation of motion with acceleration in form shown (minus sign may be missing); 0.4 or $m$ for mass for first 4 marks |
| $v = \int -\frac{20}{(t+4)^2}\,dt$ | | |
| $v = \frac{20}{(t+4)} (+c)$ | DM1A1 | DM1: attempt integration; A1: correct integration including correct sign, constant not needed |
| $t=0, v=10 \Rightarrow 10 = 5 + c,\; c = 5$ | DM1 | Use $t=0$, $v=10$ to obtain value for $c$ |
| $v = \frac{20}{(t+4)} + 5$ | A1cso | Correct answer only, no errors seen |
| **Part (a) Total: (5)** | | |
## Question 3(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = \int \frac{20}{(t+4)} + 5\; dt$ | | |
| $x = 20\ln(t+4) + 5t + c$ | M1A1 | M1: integrate given expression for $v$; A1: correct integration, constant not needed |
| $t=0, x=0 \Rightarrow c = -20\ln 4$ | A1 | Obtain correct constant |
| $v=6$: $\frac{20}{t+4} = 1 \Rightarrow t = 16$ | B1 | Correct value of $t$ when $v=6$ |
| $x = 20\ln 20 + 80 - 20\ln 4 = 80 + 20\ln 5$ | | |
| $a = 80,\; b = 20$ | A1cao | Correct values for $a$ and $b$; need not be shown explicitly |
| **Part (b) Total: (5) [10]** | | |
---\begin{enumerate}
\item A particle $P$ of mass 0.4 kg moves along the $x$-axis in the positive direction. At time $t = 0 , P$ passes through the origin $O$ with speed $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. At time $t$ seconds $P$ is $x$ metres from $O$ and the speed of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The resultant force acting on $P$ has magnitude $\frac { 8 } { ( t + 4 ) ^ { 2 } } \mathrm {~N}$ and is directed towards $O$.\\
(a) Show that $v = \frac { 20 } { t + 4 } + 5$
\end{enumerate}
When $v = 6 , x = a + b \ln 5$, where $a$ and $b$ are integers.\\
(b) Using algebraic integration, find the value of $a$ and the value of $b$.\\
\begin{center}
\end{center}
\begin{center}
\end{center}
\hfill \mbox{\textit{Edexcel M3 2018 Q3 [10]}}