| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2020 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Air resistance kv² - horizontal motion or engine power |
| Difficulty | Challenging +1.8 This is a challenging Further Maths mechanics problem requiring v dv/dx formulation for variable force, separation of variables, and integration with logarithms. Part (a) is a standard 'show that' requiring one differential equation setup and integration. Part (b) requires setting up a new equation with two forces, careful algebraic manipulation of the resulting integral, and pattern matching to the given form. The conceptual demand (choosing correct formulation, handling 1/v force term) and multi-step algebraic complexity place this above average difficulty, though the 'show that' structure provides guidance. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(mv\dfrac{dv}{dx} = -kmv^2\) | B1 | N2L, with \(m\) |
| \(\ln v = -kx + c\) | M1 | Separate variables and integrate |
| \(x=0, v=u: \quad c = \ln u\) | M1 | Use initial condition |
| \(v = \dfrac{1}{2}u: \quad \ln\dfrac{1}{2} = -kx,\) \(\quad x = \dfrac{1}{k}\ln 2\) | A1 | AG |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(mv\dfrac{dv}{dx} = -mkv^2 + \dfrac{5m}{v}\) | B1 | N2L (allow missing \(m\) in this part) |
| \(\dfrac{v^2\,dv}{5-kv^3} = dx \Rightarrow -\dfrac{1}{3k}\ln\!\left(5-kv^3\right) = x(+d)\) | M1A1 | Separate variables and integrate |
| Using (a): \(-\dfrac{1}{3k}\ln\!\left(5-kv^3\right) = x - \dfrac{1}{3k}\ln\!\left(5-ku^3\right) - \dfrac{1}{k}\ln 2\) | M1M1 | Use condition. M0 if \(v=\dfrac{1}{2}u,\, x=0\) used unless \(\dfrac{1}{k}\ln 2\) is added on later. Rearrange dependent on ln solution |
| \(x = \dfrac{1}{3k}\ln\!\left(\dfrac{40-ku^3}{5-ku^3}\right)\) | M1A1 | Use \(v = u\) |
| 7 |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $mv\dfrac{dv}{dx} = -kmv^2$ | **B1** | N2L, with $m$ |
| $\ln v = -kx + c$ | **M1** | Separate variables and integrate |
| $x=0, v=u: \quad c = \ln u$ | **M1** | Use initial condition |
| $v = \dfrac{1}{2}u: \quad \ln\dfrac{1}{2} = -kx,$ $\quad x = \dfrac{1}{k}\ln 2$ | **A1** | AG |
| | **4** | |
---
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $mv\dfrac{dv}{dx} = -mkv^2 + \dfrac{5m}{v}$ | **B1** | N2L (allow missing $m$ in this part) |
| $\dfrac{v^2\,dv}{5-kv^3} = dx \Rightarrow -\dfrac{1}{3k}\ln\!\left(5-kv^3\right) = x(+d)$ | **M1A1** | Separate variables and integrate |
| Using **(a):** $-\dfrac{1}{3k}\ln\!\left(5-kv^3\right) = x - \dfrac{1}{3k}\ln\!\left(5-ku^3\right) - \dfrac{1}{k}\ln 2$ | **M1M1** | Use condition. M0 if $v=\dfrac{1}{2}u,\, x=0$ used unless $\dfrac{1}{k}\ln 2$ is added on later. Rearrange dependent on ln solution |
| $x = \dfrac{1}{3k}\ln\!\left(\dfrac{40-ku^3}{5-ku^3}\right)$ | **M1A1** | Use $v = u$ |
| | **7** | |7 A particle $P$ of mass $m \mathrm {~kg}$ moves in a horizontal straight line against a resistive force of magnitude $\mathrm { mkv } ^ { 2 } \mathrm {~N}$, where $v \mathrm {~ms} ^ { - 1 }$ is the speed of $P$ after it has moved a distance $x \mathrm {~m}$ and $k$ is a positive constant. The initial speed of $P$ is $u \mathrm {~ms} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { x } = \frac { 1 } { \mathrm { k } } \ln 2$ when $\mathrm { v } = \frac { 1 } { 2 } \mathrm { u }$.\\
Beginning at the instant when the speed of $P$ is $\frac { 1 } { 2 } u$, an additional force acts on $P$. This force has magnitude $\frac { 5 \mathrm {~m} } { \mathrm { v } } \mathrm { N }$ and acts in the direction of increasing $x$.
\item Show that when the speed of $P$ has increased again to $u \mathrm {~ms} ^ { - 1 }$, the total distance travelled by $P$ is given by an expression of the form
$$\frac { 1 } { 3 k } \ln \left( \frac { A - k u ^ { 3 } } { B - k u ^ { 3 } } \right) ,$$
stating the values of the constants $A$ and $B$.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2020 Q7 [11]}}