| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2024 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Acceleration as function of velocity (separation of variables) |
| Difficulty | Standard +0.8 This is a standard variable acceleration problem requiring setting up and solving a differential equation (F=ma with resistance proportional to v), then integrating to find displacement. While it involves multiple steps and integration techniques, the method is well-practiced in Further Maths mechanics. The resistance being mv (not mv²) makes the differential equation separable and straightforward. More challenging than typical A-level due to the multi-step nature and being Further Maths content, but follows a standard template without requiring novel insight. |
| Spec | 4.10c Integrating factor: first order equations6.02a Work done: concept and definition |
| Answer | Marks |
|---|---|
| 7(a) | dv |
| Answer | Marks | Guidance |
|---|---|---|
| dt | B1 | No marks in this part if suvat used. |
| Answer | Marks | Guidance |
|---|---|---|
| ln10v t A or lnv10t A | *M1 A1 | Separate variables and integrate to obtain a ln |
| Answer | Marks | Guidance |
|---|---|---|
| Use t 0, v50: Aln 40 | DM1 | Find constant, dependent on previous M1. |
| Answer | Marks | Guidance |
|---|---|---|
| 0.1vet 4et | M1 | Remove all logs . |
| v1040et | A1 | Correct work only . |
| Answer | Marks | Guidance |
|---|---|---|
| 7(b) | x10t40et B | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Use t 0, x0: B40 | M1 | Use initial condition in their expression for x in |
| Answer | Marks |
|---|---|
| x10t40et 40 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks |
|---|---|
| 7(c) | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| 8 | M1 | No marks in this part if suvat used in part (a) part |
| Answer | Marks | Guidance |
|---|---|---|
| x 55.8 (metres) | A1 | Note 35 + 10 ln 8 scores A0. |
Question 7:
--- 7(a) ---
7(a) | dv
m m10v
dt | B1 | No marks in this part if suvat used.
Must have sight of m (for example in F = ma).
ln10v t A or lnv10t A | *M1 A1 | Separate variables and integrate to obtain a ln
term. Constant may be omitted.
Constant needed for A1
Use t 0, v50: Aln 40 | DM1 | Find constant, dependent on previous M1.
May use limits instead.
0.1vet 4et | M1 | Remove all logs .
v1040et | A1 | Correct work only .
6
--- 7(b) ---
7(b) | x10t40et B | M1 | No marks in this part if suvat used in part (a) or
part (b).
Integrate their answer to part (a).
Constant may be omitted.
Use t 0, x0: B40 | M1 | Use initial condition in their expression for x in
terms of t .
x10t40et 40 | A1
3
Question | Answer | Marks | Guidance
--- 7(c) ---
7(c) | 1
When v15, et , t 2.08 or ln 8
8 | M1 | No marks in this part if suvat used in part (a) part
(b) or part (c).
Find value of t from their answer to part (a).
x 55.8 (metres) | A1 | Note 35 + 10 ln 8 scores A0.
2
A parachutist of mass $m$ kg opens his parachute when he is moving vertically downwards with a speed of $50\text{ ms}^{-1}$. At time $t$ s after opening his parachute, he has fallen a distance $x$ m from the point where he opened his parachute, and his speed is $v\text{ ms}^{-1}$. The forces acting on him are his weight and a resistive force of magnitude $mv$ N.
\begin{enumerate}[label=(\alph*)]
\item Find an expression for $v$ in terms of $t$. [6]
\item Find an expression for $x$ in terms of $t$. [3]
\item Find the distance that the parachutist has fallen, since opening his parachute, when his speed is $15\text{ ms}^{-1}$. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2024 Q7 [11]}}