| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2023 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Air resistance kv² - falling from rest or projected downward |
| Difficulty | Challenging +1.2 This is a standard variable force mechanics problem requiring separation of variables to solve a first-order ODE with resistance proportional to v². While it involves Further Maths content (differential equations in mechanics), the setup is straightforward with clear forces, and the method is a direct application of standard techniques. The 6-mark allocation and routine 'large t' follow-up confirm this is a typical textbook-style question, placing it moderately above average difficulty. |
| 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 |
|---|---|
| 2(a) | dv |
| Answer | Marks |
|---|---|
| dt | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 0.1 | M1 | Integrate to a ln term of the correct form. |
| Answer | Marks |
|---|---|
| 20 10−v | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 20 | M1 | Use initial condition. |
| Answer | Marks | Guidance |
|---|---|---|
| 3(10−v) 3(10−v) | M1 | Rearrange, removing ln. |
| Answer | Marks | Guidance |
|---|---|---|
| 3+e−2t | A1 | AEF |
| Answer | Marks | Guidance |
|---|---|---|
| 2(b) | v→10 | B1FT |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 2:
--- 2(a) ---
2(a) | dv
2 =2g−0.2v2
dt | B1
dv
Separate variables and attempt to integrate =dt
( 100−v2)
0.1 | M1 | Integrate to a ln term of the correct form.
1 10+v
ln =0.1t+c
20 10−v | A1
1
t=0, v=5, c= ln3
20 | M1 | Use initial condition.
10+v 10+v
2t =ln , e2t =
3(10−v) 3(10−v) | M1 | Rearrange, removing ln.
30−10e−2t
v=
3+e−2t | A1 | AEF
6
--- 2(b) ---
2(b) | v→10 | B1FT | FT from expression of correct form.
1
Question | Answer | Marks | GuidanceA ball of mass $2$ kg is projected vertically downwards with speed $5\text{ ms}^{-1}$ through a liquid. At time $t$ s after projection, the velocity of the ball is $v\text{ ms}^{-1}$ and its displacement from its starting point is $x$ m. The forces acting on the ball are its weight and a resistive force of magnitude $0.2v^2$ N.
\begin{enumerate}[label=(\alph*)]
\item Find an expression for $v$ in terms of $t$. [6]
\item Deduce what happens to $v$ for large values of $t$. [1]
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2023 Q2 [7]}}