| Exam Board | Edexcel |
|---|---|
| Module | CP1 (Core Pure 1) |
| Year | 2021 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Modeling context with interpretation |
| Difficulty | Standard +0.3 This is a standard second-order differential equation problem with damped oscillation. Part (a) uses initial conditions to find k (straightforward substitution), part (b) requires solving a second-order linear ODE with given initial conditions (standard CP1 technique), part (c) is simple evaluation, and part (d) asks for a modeling limitation (routine). While it involves multiple steps and careful algebra, all techniques are core syllabus material with no novel insight required, making it slightly easier than average. |
| Spec | 4.10d Second order homogeneous: auxiliary equation method4.10f Simple harmonic motion: x'' = -omega^2 x |
| Answer | Marks | Guidance |
|---|---|---|
| 6(a) | 5 k ( 1 3 .6 ) + 2 k ( 0 ) + 1 7 ( − 2 0 ) = 0 k = ... | M1 |
| k = 5 | A1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | Solves their 2 5 m 2 + 1 0 m + 1 7 = 0 m = ... | M1 |
| m = − 0 .2 0 .8 i | A1 | 1.1b |
| x = e − 0 .2 t ( A c o s 0 .8 t + B s i n 0 .8 t ) | A1ft | 1.1b |
| t = 0 , x = − 2 0 A = . . . ( = − 2 0 ) | M1 | 3.4 |
| Answer | Marks | Guidance |
|---|---|---|
| + e − 0 .2 t ( − 0 . 8 A s i n 0 . 8 t + 0 . 8 B c o s 0 . 8 t ) | M1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| d t | dM1 | 3.4 |
| x = e − 0 .2 t ( − 2 0 c o s 0 .8 t − 5 s i n 0 .8 t ) o.e. | A1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| (c) | Vertical height =30+e−0.215(−20cos(0.815)−5sin(0.815) ) | |
| | M1 | 3.4 |
| Vertical height = awrt 29.3 m | A1 | 2.2b |
| Answer | Marks |
|---|---|
| (d) | For example |
| Answer | Marks | Guidance |
|---|---|---|
| The tourist is modelled as a particle | B1 | 3.5b |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Scheme | Marks |
Question 6:
--- 6(a) ---
6(a) | 5 k ( 1 3 .6 ) + 2 k ( 0 ) + 1 7 ( − 2 0 ) = 0 k = ... | M1 | 3.3
k = 5 | A1 | 1.1b
(2)
(b) | Solves their 2 5 m 2 + 1 0 m + 1 7 = 0 m = ... | M1 | 3.1b
m = − 0 .2 0 .8 i | A1 | 1.1b
x = e − 0 .2 t ( A c o s 0 .8 t + B s i n 0 .8 t ) | A1ft | 1.1b
t = 0 , x = − 2 0 A = . . . ( = − 2 0 ) | M1 | 3.4
d x
= − 0 . 2 e − 0 .2 t ( A c o s 0 . 8 t + B s i n 0 . 8 ) t
d t
+ e − 0 .2 t ( − 0 . 8 A s i n 0 . 8 t + 0 . 8 B c o s 0 . 8 t ) | M1 | 1.1b
d x
t = 0 = 0 − 0 . 2 A + 0 . 8 B = 0 B = . . . ( = − 5 )
d t | dM1 | 3.4
x = e − 0 .2 t ( − 2 0 c o s 0 .8 t − 5 s i n 0 .8 t ) o.e. | A1 | 1.1b
(7)
(c) | Vertical height =30+e−0.215(−20cos(0.815)−5sin(0.815) )
| M1 | 3.4
Vertical height = awrt 29.3 m | A1 | 2.2b
(2)
(d) | For example
It is unlikely that the rope will remain taut
The model predicts the tourist will continue to move up and down,
(but in fact they will lose momentum)
The tourist is modelled as a particle | B1 | 3.5b
(1)
(12 marks)
Notes:
(a)
d 2 x
M1: Substitutes = 1 3 . 6 dx =0 and x =−20into the differential equation to find a value for k.
d t 2 dt
Allow if there are sign slips but must be attempting the values in the correct places.
A1: Correct value k = 5
(b)
M1: Forms and solves the auxiliary equation.
A1: Correct solution to the auxiliary equation (not follow through).
A1ft: Correct complementary function for their solutions to their auxiliary equation. (Follow
through on distinct real, repeated or complex roots.)
M1: Uses the information from the model t = 0 x = − 2 0 to find a constant or equation linking two
constants in their equation.
M1: Differentiates an expression of the form e kt ( A c o s t B s i n t ) + using the product rule to find
1 2
an expression for the velocity.
d x
dM1: Uses the information from the model, t = 0 = 0 to find and solve another equation for the
d t
constants.
A1: Correct equation for displacement.
(c)
M1: Finds the height above the river by finding the displacement after 15 seconds and adding 30
A1: Vertical height = awrt 29.3 m
(d)
B1: Any suitable comment relating to the given model or the outcomes of it. See scheme for
examples. Do not accept just “air resistance has not been considered” as the question does not say
this was ignored. However, if a valid consequence of what including air resistance would mean to
the model, then the mark may be awarded.
Question | Scheme | Marks | AOsA tourist decides to do a bungee jump from a bridge over a river. One end of an elastic rope is attached to the bridge and the other end of the elastic rope is attached to the tourist. The tourist jumps off the bridge.
At time $t$ seconds after the tourist reaches their lowest point, their vertical displacement is $x$ metres above a fixed point 30 metres vertically above the river.
When $t = 0$
\begin{itemize}
\item $x = -20$
\item the velocity of the tourist is $0\text{ms}^{-1}$
\item the acceleration of the tourist is $13.6\text{ms}^{-2}$
\end{itemize}
In the subsequent motion, the elastic rope is assumed to remain taut so that the vertical displacement of the tourist can be modelled by the differential equation
$$5k\frac{d^2x}{dt^2} + 2k\frac{dx}{dt} + 17x = 0 \quad t \geq 0$$
where $k$ is a positive constant.
\begin{enumerate}[label=(\alph*)]
\item Determine the value of $k$
[2]
\item Determine the particular solution to the differential equation.
[7]
\item Hence find, according to the model, the vertical height of the tourist above the river 15 seconds after they have reached their lowest point.
[2]
\item Give a limitation of the model.
[1]
\end{enumerate}
\hfill \mbox{\textit{Edexcel CP1 2021 Q6 [12]}}