AQA Paper 2 Specimen — Question 15 11 marks

Exam BoardAQA
ModulePaper 2 (Paper 2)
SessionSpecimen
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeVector form projectile motion
DifficultyStandard +0.8 This is a multi-part mechanics question requiring integration of vector functions, solving transcendental equations (finding t when horizontal displacement = 100m involves ln), and deducing g from limiting behavior of exponential functions. Part (c) requires conceptual understanding that as t→∞, acceleration approaches -g. While each technique is A-level standard, the combination of vector calculus, exponential functions, and physical interpretation across 11 marks makes this moderately challenging, above average difficulty.
Spec1.06a Exponential function: a^x and e^x graphs and properties1.07j Differentiate exponentials: e^(kx) and a^(kx)1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.10h Vectors in kinematics: uniform acceleration in vector form3.02f Non-uniform acceleration: using differentiation and integration
At time \(t = 0\), a parachutist jumps out of an airplane that is travelling horizontally. The velocity, \(\mathbf{v}\) m s\(^{-1}\), of the parachutist at time \(t\) seconds is given by: $$\mathbf{v} = (40e^{-0.2t})\mathbf{i} + 50(e^{-0.2t} - 1)\mathbf{j}$$ The unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are horizontal and vertical respectively. Assume that the parachutist is at the origin when \(t = 0\) Model the parachutist as a particle.
  1. Find an expression for the position vector of the parachutist at time \(t\). [4 marks]
  2. The parachutist opens her parachute when she has travelled 100 metres horizontally. Find the vertical displacement of the parachutist from the origin when she opens her parachute. [4 marks]
  3. Carefully, explaining the steps that you take, deduce the value of \(g\) used in the formulation of this model. [3 marks]
Question 15:
AnswerMarks Guidance
15(a)Integrates both components
with at least one correctAO1.1a M1
(200e0.2t c)i(250e0.2t 50td)j
t 0,r 0i0jc 200, d 250
OR
t t
r = 40e0.2t dti50(e0.2t 1)dtj
0 0
t
 200e0.2ti(250e0.2t 50t)j
 
0
r 200(1e0.2t)i(250250e0.2t 50t)j
Obtains correct terms.
AnswerMarks Guidance
(condone missing constants)AO1.1b A1
Evaluates both constants (or
uses definite integration)
AnswerMarks Guidance
using ‘their’ expression for rAO3.4 M1
Obtains correct expressionAO1.1b A1
(b)Forms equation to find t
based on horizontal
AnswerMarks Guidance
componentAO3.4 M1
t 5ln23.4657
y250250e0.2×5ln2505ln2
48.3
The parachutist has a vertical displacement
of 50 m below the origin
AnswerMarks Guidance
Obtains correct timeAO1.1b A1
Substitutes ‘their’ time into
AnswerMarks Guidance
vertical componentAO1.1a M1
Obtains correct
displacement for ‘their’ time
(accept 1 sf, must have
metres)
FT only if both M1 marks
AnswerMarks Guidance
have been awardedAO3.2a A1F
(c)Identifies vertical component
of the velocity as first stepAO2.4 M1
(50(e0.2t 1))10e0.2t
dt
As there is no initial vertical component of
velocity (and hence no air resistance) the
initial acceleration is only due to gravity
Hence g is taken as 10 m s–2
Differentiates vertical
component of velocity
AnswerMarks Guidance
correctlyAO1.1b A1
Considers implication of
initial motion and reaches
AnswerMarks Guidance
correct conclusionAO2.2a R1
Total11
QMarking Instructions AO 
Question 15:
--- 15(a) ---
15(a) | Integrates both components
with at least one correct | AO1.1a | M1 | r 40e0.2t dti50(e0.2t 1)dtj
(200e0.2t c)i(250e0.2t 50td)j
t 0,r 0i0jc 200, d 250
OR
t t
r = 40e0.2t dti50(e0.2t 1)dtj
0 0
t
 200e0.2ti(250e0.2t 50t)j
 
0
r 200(1e0.2t)i(250250e0.2t 50t)j
Obtains correct terms.
(condone missing constants) | AO1.1b | A1
Evaluates both constants (or
uses definite integration)
using ‘their’ expression for r | AO3.4 | M1
Obtains correct expression | AO1.1b | A1
(b) | Forms equation to find t
based on horizontal
component | AO3.4 | M1 | 200(1e0.2t)100
t 5ln23.4657
y250250e0.2×5ln2505ln2
48.3
The parachutist has a vertical displacement
of 50 m below the origin
Obtains correct time | AO1.1b | A1
Substitutes ‘their’ time into
vertical component | AO1.1a | M1
Obtains correct
displacement for ‘their’ time
(accept 1 sf, must have
metres)
FT only if both M1 marks
have been awarded | AO3.2a | A1F
(c) | Identifies vertical component
of the velocity as first step | AO2.4 | M1 | d
(50(e0.2t 1))10e0.2t
dt
As there is no initial vertical component of
velocity (and hence no air resistance) the
initial acceleration is only due to gravity
Hence g is taken as 10 m s–2
Differentiates vertical
component of velocity
correctly | AO1.1b | A1
Considers implication of
initial motion and reaches
correct conclusion | AO2.2a | R1
Total | 11
Q | Marking Instructions | AO | Marks | Typical Solution
At time $t = 0$, a parachutist jumps out of an airplane that is travelling horizontally.

The velocity, $\mathbf{v}$ m s$^{-1}$, of the parachutist at time $t$ seconds is given by:
$$\mathbf{v} = (40e^{-0.2t})\mathbf{i} + 50(e^{-0.2t} - 1)\mathbf{j}$$

The unit vectors $\mathbf{i}$ and $\mathbf{j}$ are horizontal and vertical respectively.

Assume that the parachutist is at the origin when $t = 0$

Model the parachutist as a particle.

\begin{enumerate}[label=(\alph*)]
\item Find an expression for the position vector of the parachutist at time $t$.
[4 marks]

\item The parachutist opens her parachute when she has travelled 100 metres horizontally.

Find the vertical displacement of the parachutist from the origin when she opens her parachute.
[4 marks]

\item Carefully, explaining the steps that you take, deduce the value of $g$ used in the formulation of this model.
[3 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 2  Q15 [11]}}
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