| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2024 |
| Session | November |
| Marks | 11 |
| 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 Further Maths mechanics problem involving variable resistive force. Part (a) requires separating variables and integrating dv/dt = g - 0.1v² (6 marks suggests standard integration with partial fractions). Part (b) uses v dv/dx = g - 0.1v² which separates more easily. While this requires familiarity with differential equations and resistance proportional to v², these are well-practiced techniques in FM mechanics with no novel insight required—moderately above average difficulty due to the algebraic manipulation involved. |
| Spec | 3.02h Motion under gravity: vector form3.03c Newton's second law: F=ma one dimension6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks |
|---|---|
| 7(a) | dv |
| Answer | Marks | Guidance |
|---|---|---|
| dt | B1 | Use of suvat means 0 marks in this part |
| Answer | Marks | Guidance |
|---|---|---|
| 10−v | M1* | Separate variables and integrate. May see partial |
| Answer | Marks |
|---|---|
| A1 | Must see modulus sign |
| Use t=0, v=0, A=0 | DM1 |
| Remove logs to obtain v in terms of t | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| e2t +1 | A1 | ( 1−e−2t) |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks |
|---|---|
| 7(b) | dv 1 ( 100−v2) |
| Answer | Marks | Guidance |
|---|---|---|
| 2 10 | M1* | Use of suvat means 0 marks in this part |
| Answer | Marks |
|---|---|
| A1 | For A1, allow missing modulus sign |
| Answer | Marks |
|---|---|
| 2 | DM1 |
| Remove logs to obtain v2 in terms of x | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| v2 =100(1−e 5) | A1 | AEF |
Question 7:
--- 7(a) ---
7(a) | dv
m =mg−0.1mv2
dt | B1 | Use of suvat means 0 marks in this part
Note that no mg term means 0 marks in this part.
dv
Must see m, may be cancelled before a = used
dt
dv =10−0.1v2 = 1 ( 100−v2)
dt 10
dv 1
= dt
100−v2 10
v+10
ln =2t+A
10−v | M1* | Separate variables and integrate. May see partial
fractions, but integral is on Formula sheet,
allow missing +A for M1 only
A1 | Must see modulus sign
Use t=0, v=0, A=0 | DM1
Remove logs to obtain v in terms of t | M1
10 ( e2t −1 )
v= aef
e2t +1 | A1 | ( 1−e−2t)
10
v=
1+e−2t
6
Question | Answer | Marks | Guidance
--- 7(b) ---
7(b) | dv 1 ( 100−v2)
v =
dx 10
1 ( ) 1
− ln 100−v2 = x+B
2 10 | M1* | Use of suvat means 0 marks in this part
Separate variables and integrate, allow missing
+A for M1 only.
A1 | For A1, allow missing modulus sign
1
Use x=0, v=0, B=− ln100
2 | DM1
Remove logs to obtain v2 in terms of x | M1
x
−
v2 =100(1−e 5) | A1 | AEF
Allow 10g instead of 100.
5A particle $P$ of mass $m$ kg is held at rest at a point $O$ and released so that it moves vertically under gravity against a resistive force of magnitude $0.1mv^2$ N, where $v$ m s$^{-1}$ is the velocity of $P$ at time $t$ s.
\begin{enumerate}[label=(\alph*)]
\item Find an expression for $v$ in terms of $t$. [6]
\item Find an expression for $v^2$ in terms of $x$. [5]
\end{enumerate}
The displacement of $P$ from $O$ at time $t$ s is $x$ m.
\hfill \mbox{\textit{CAIE Further Paper 3 2024 Q7 [11]}}