| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2021 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Acceleration as function of velocity (separation of variables) |
| Difficulty | Challenging +1.2 This is a Further Maths mechanics question requiring separation of variables to solve a differential equation (a = dv/dt), then using an initial condition to find the constant. Part (b) applies F = ma with the derived expression. While it involves calculus and differential equations (making it harder than standard A-level), the separation and integration are straightforward, and the question follows a standard template for variable acceleration problems in Further Maths. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks |
|---|---|
| 2(a) | Separate variables and integrate: |
| Answer | Marks | Guidance |
|---|---|---|
| | M1 A1 | Must include logs. Condone missing modulus. |
| v = Ate −t2 , −v= Ate −t2 , v=−Ate −t2 | A1 | CAO. |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks |
|---|---|
| 2(b) | ( ) |
| Answer | Marks | Guidance |
|---|---|---|
| t | M1 | Substituting their answer to part (a) into given formula |
| t=1, a=5 ( A=5e ) | M1 | Use initial condition. |
| Answer | Marks | Guidance |
|---|---|---|
| Force = 5me1−t2 2t2 −1 | A1 | Use N2L, correct work only. |
| Answer | Marks | Guidance |
|---|---|---|
| t | M1 | Use initial condition. |
| Answer | Marks |
|---|---|
| Substituting in their answer to part (a) so ( A=5e ) | M1 |
| Answer | Marks |
|---|---|
| Force = 5me1−t2 2t2 −1 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 2:
--- 2(a) ---
2(a) | Separate variables and integrate:
dv 1−2t2
= dt so ln v =lnt−t2 +c
v t
| M1 A1 | Must include logs. Condone missing modulus.
v = Ate −t2 , −v= Ate −t2 , v=−Ate −t2 | A1 | CAO.
3
Question | Answer | Marks | Guidance
--- 2(b) ---
2(b) | ( )
−Ate −t2 1−2t2
( )
a= =−Ae −t2 1−2t2
t | M1 | Substituting their answer to part (a) into given formula
t=1, a=5 ( A=5e ) | M1 | Use initial condition.
( )
Force = 5me1−t2 2t2 −1 | A1 | Use N2L, correct work only.
Alternative method for question 2(b)
( )
v 1−2t2
a= substitute t=1, a=5 so v=−5
t | M1 | Use initial condition.
Use N2L, correct work only.
Substituting in their answer to part (a) so ( A=5e ) | M1
( )
Force = 5me1−t2 2t2 −1 | A1
3
Question | Answer | Marks | GuidanceA particle $P$ of mass $m$ kg moves along a horizontal straight line with acceleration $a$ ms$^{-2}$ given by
$$a = \frac{v(1-2t^2)}{t},$$
where $v$ ms$^{-1}$ is the velocity of $P$ at time $t$ s.
\begin{enumerate}[label=(\alph*)]
\item Find an expression for $v$ in terms of $t$ and an arbitrary constant. [3]
\item Given that $a = 5$ when $t = 1$, find an expression, in terms of $m$ and $t$, for the horizontal force acting on $P$ at time $t$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2021 Q2 [6]}}