AQA Further Paper 1 2021 June — Question 8 6 marks

Exam BoardAQA
ModuleFurther Paper 1 (Further Paper 1)
Year2021
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNumerical integration
TypeTrapezium rule with stated number of strips
DifficultyChallenging +1.2 This is a straightforward application of Euler's method with clearly stated forces and initial conditions. While it requires setting up Newton's second law (F=ma), differentiating to get dv/dt, and performing iterative calculations, the method is entirely procedural with no conceptual challenges. The 6 marks reflect computational work rather than problem-solving insight. Slightly above average difficulty due to being Further Maths and requiring careful arithmetic over multiple steps.
Spec1.09e Iterative method failure: convergence conditions6.06a Variable force: dv/dt or v*dv/dx methods

A particle of mass 4 kg moves horizontally in a straight line. At time \(t\) seconds the velocity of the particle is \(v\) m s\(^{-1}\) The following horizontal forces act on the particle: β€’ a constant driving force of magnitude 1.8 newtons β€’ another driving force of magnitude \(30\sqrt{t}\) newtons β€’ a resistive force of magnitude \(0.08v^2\) newtons When \(t = 70\), \(v = 54\) Use Euler's method with a step length of 0.5 to estimate the velocity of the particle after 71 seconds. Give your answer to four significant figures. [6 marks]

Question 8:
AnswerMarks
8Uses Newton’s second law to
form a four term differential
equation.
Must have correct terms,
condone wrong signs
AnswerMarks Guidance
PI3.3 M1
d𝑣𝑣 2 2
4 = 1.8+30𝑑𝑑 βˆ’0.08𝑣𝑣
d𝑑𝑑
1
d𝑣𝑣 2 2
= 0.45+7.5𝑑𝑑 βˆ’0.02𝑣𝑣
d𝑑𝑑
)
𝑣𝑣70.5 β‰… 54+0.5(1𝑣𝑣̇70)
2 2
= 54+0.5(0.45+7.5Γ—70 βˆ’0.02Γ—54
= 54+0.5(4.8795…)
= 56.439751
𝑣𝑣71 β‰… 56.439751+0 .5(𝑣𝑣̇70.5)
Vel𝑣𝑣o 7 c 1 ityβ‰… a5ft6e.r4 37917 s5e1c+on0d.s5 (=βˆ’ 506.2.3805 7m… s)-1
= 56.2969
Obtains correct expression
for or
d𝑣𝑣 d𝑣𝑣
AnswerMarks Guidance
PI1.1b A1
d𝑑𝑑 4d𝑑𝑑
Substitutes correct values
AnswerMarks Guidance
into their first Euler equation1.1a M1
Obtains value for which
rounds to 56.4
AnswerMarks Guidance
𝑣𝑣70.51.1b A1
Uses Euler’s method exactly
AnswerMarks Guidance
twice3.1a M1
Obtains correct answer to
required degree of accuracy
AnswerMarks Guidance
Condone lack of units3.2a A1
Total6
QMarking Instructions AO
Question 8:
8 | Uses Newton’s second law to
form a four term differential
equation.
Must have correct terms,
condone wrong signs
PI | 3.3 | M1 | 1
d𝑣𝑣 2 2
4 = 1.8+30𝑑𝑑 βˆ’0.08𝑣𝑣
d𝑑𝑑
1
d𝑣𝑣 2 2
= 0.45+7.5𝑑𝑑 βˆ’0.02𝑣𝑣
d𝑑𝑑
)
𝑣𝑣70.5 β‰… 54+0.5(1𝑣𝑣̇70)
2 2
= 54+0.5(0.45+7.5Γ—70 βˆ’0.02Γ—54
= 54+0.5(4.8795…)
= 56.439751
𝑣𝑣71 β‰… 56.439751+0 .5(𝑣𝑣̇70.5)
Vel𝑣𝑣o 7 c 1 ityβ‰… a5ft6e.r4 37917 s5e1c+on0d.s5 (=βˆ’ 506.2.3805 7m… s)-1
= 56.2969
Obtains correct expression
for or
d𝑣𝑣 d𝑣𝑣
PI | 1.1b | A1
d𝑑𝑑 4d𝑑𝑑
Substitutes correct values
into their first Euler equation | 1.1a | M1
Obtains value for which
rounds to 56.4
𝑣𝑣70.5 | 1.1b | A1
Uses Euler’s method exactly
twice | 3.1a | M1
Obtains correct answer to
required degree of accuracy
Condone lack of units | 3.2a | A1
Total | 6
Q | Marking Instructions | AO | Marks | Typical solution
A particle of mass 4 kg moves horizontally in a straight line.

At time $t$ seconds the velocity of the particle is $v$ m s$^{-1}$

The following horizontal forces act on the particle:

β€’ a constant driving force of magnitude 1.8 newtons

β€’ another driving force of magnitude $30\sqrt{t}$ newtons

β€’ a resistive force of magnitude $0.08v^2$ newtons

When $t = 70$, $v = 54$

Use Euler's method with a step length of 0.5 to estimate the velocity of the particle after 71 seconds.

Give your answer to four significant figures.
[6 marks]

\hfill \mbox{\textit{AQA Further Paper 1 2021 Q8 [6]}}