| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2024 |
| Session | November |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Given velocity function find force |
| Difficulty | Challenging +1.2 This is a Further Maths mechanics problem requiring separation of variables and integration. The given relationship v/x = (3-t)/(1+t) leads to a separable differential equation dx/x = [(3-t)/(1+t)]dt. While it involves Further Maths content (variable force, differential equations), the actual mathematical technique is standard: separate variables, integrate both sides (requiring partial fractions or substitution for the right side), and apply initial conditions. The setup is clear and the method is routine for Further Maths students, making it slightly above average difficulty overall but not requiring novel insight. |
| Spec | 3.03d Newton's second law: 2D vectors6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks |
|---|---|
| 5(a) | dx 4 |
| Answer | Marks | Guidance |
|---|---|---|
| x t+1 | M1 | Separate variables, obtain RHS in integrable form. |
| ln x =4ln t+1−t+ A | A1 | |
| t=0, x=5: A=ln5 | M1 | |
| x=5(t+1)4 e−t | A1 |
| Answer | Marks |
|---|---|
| 5(b) | v=(3−t)5(t+1)3 e−t |
| Answer | Marks |
|---|---|
| dt | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Acceleration = | AEF | |
| F=2acceleration, so at F =10e−t(t+1)2(5−t)(1−t) | M1 | |
| At t=3, magnitude of force is 640e−3 N | A1 | 31.9 N |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 5:
--- 5(a) ---
5(a) | dx 4
= −1dt
x t+1 | M1 | Separate variables, obtain RHS in integrable form.
ln x =4ln t+1−t+ A | A1
t=0, x=5: A=ln5 | M1
x=5(t+1)4 e−t | A1
4
--- 5(b) ---
5(b) | v=(3−t)5(t+1)3 e−t
dv ( )
Acceleration = =5e−t −(t+1)3+(3−t)3(t+1)2 −(3−t)(t+1)3
dt | M1
5e−t(t+1)2(5−t)(1−t)
Acceleration = | AEF
F=2acceleration, so at F =10e−t(t+1)2(5−t)(1−t) | M1
At t=3, magnitude of force is 640e−3 N | A1 | 31.9 N
3
Question | Answer | Marks | GuidanceA particle $P$ of mass $2\text{kg}$ moving on a horizontal straight line has displacement $x\text{m}$ from a fixed point $O$ on the line and velocity $v\text{ms}^{-1}$ at time $t$. The only horizontal force acting on $P$ is a variable force $F\text{N}$ which can be expressed as a function of $t$. It is given that
$$\frac{v}{x} = \frac{3-t}{1+t}$$
and when $t = 0$, $x = 5$.
\begin{enumerate}[label=(\alph*)]
\item Find an expression for $x$ in terms of $t$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2024 Q5 [4]}}