| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2002 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Variable power or two power scenarios |
| Difficulty | Standard +0.8 This M4 variable force question requires setting up F=ma with power and resistance, then solving a separable differential equation with substitution. While the setup is guided by part (a), finding v(t) requires integration techniques beyond standard A-level including partial fractions with the given limiting speed condition. Moderately challenging for Further Maths. |
| Spec | 6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(800\frac{dv}{dt} = \frac{32000}{v} - kv\) | M1 A1 | Equation of motion with two force terms |
| \(\Rightarrow 800v\frac{dv}{dt} = 32000 - kv^2\) (*) | A1 | (3) Given answer on paper |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(v=40, \frac{dv}{dt}=0 \Rightarrow 32000 = k \times 40^2\) | M1 | |
| \(\Rightarrow k = 20\) | A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(\int dt = 800\int\frac{v\,dv}{32000-20v^2} = \int\frac{40v\,dv}{1600-v^2}\) | M1 | |
| \(t = -20\ln(1600-v^2)(+C)\) | M1 A1 ft | |
| \(t=0, v=0 \Rightarrow C = 20\ln 1600\) (or use of limits) | ||
| \(t = 20\ln 1600 - 20\ln(1600-v^2)\) | M1 A1 ft | |
| \(\Rightarrow t = 20\ln\left(\frac{1600}{1600-v^2}\right)\) | ||
| \(\frac{1600}{1600-v^2} = e^{\frac{t}{20}}\) | M1 | |
| \(1600e^{-\frac{t}{20}} = 1600 - v^2\) | ||
| \(v = 40\sqrt{\left(1-e^{-\frac{t}{20}}\right)}\) | A1 | (7) |
# Question 3:
## Part (a):
| Working/Answer | Marks | Notes |
|---|---|---|
| $800\frac{dv}{dt} = \frac{32000}{v} - kv$ | M1 A1 | Equation of motion with two force terms |
| $\Rightarrow 800v\frac{dv}{dt} = 32000 - kv^2$ (*) | A1 | (3) Given answer on paper |
## Part (b):
| Working/Answer | Marks | Notes |
|---|---|---|
| $v=40, \frac{dv}{dt}=0 \Rightarrow 32000 = k \times 40^2$ | M1 | |
| $\Rightarrow k = 20$ | A1 | (2) |
## Part (c):
| Working/Answer | Marks | Notes |
|---|---|---|
| $\int dt = 800\int\frac{v\,dv}{32000-20v^2} = \int\frac{40v\,dv}{1600-v^2}$ | M1 | |
| $t = -20\ln(1600-v^2)(+C)$ | M1 A1 ft | |
| $t=0, v=0 \Rightarrow C = 20\ln 1600$ (or use of limits) | | |
| $t = 20\ln 1600 - 20\ln(1600-v^2)$ | M1 A1 ft | |
| $\Rightarrow t = 20\ln\left(\frac{1600}{1600-v^2}\right)$ | | |
| $\frac{1600}{1600-v^2} = e^{\frac{t}{20}}$ | M1 | |
| $1600e^{-\frac{t}{20}} = 1600 - v^2$ | | |
| $v = 40\sqrt{\left(1-e^{-\frac{t}{20}}\right)}$ | A1 | (7) |
---3. The engine of a car of mass 800 kg works at a constant rate of 32 kW . The car travels along a straight horizontal road and the resistance to motion of the car is proportional to the speed of the car. The car starts from rest and $t$ seconds later it has a speed of $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that
$$800 v \frac { \mathrm {~d} v } { \mathrm {~d} t } = 32000 - k v ^ { 2 } , \text { where } k \text { is a positive constant. }$$
Given that the limiting speed of the car is $40 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, find
\item the value of $k$,
\item $v$ in terms of $t$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2002 Q3 [12]}}