CAIE Further Paper 3 2024 November — Question 5 3 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2024
SessionNovember
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable Force
TypeForce depends on time t
DifficultyModerate -0.5 This is a straightforward substitution question worth 3 marks, likely requiring evaluation of a given force function at t=3 and finding its magnitude. While it's Further Maths mechanics, this is a routine calculation with no problem-solving or conceptual challenge, making it slightly easier than average.
Spec3.03d Newton's second law: 2D vectors
  1. Find the magnitude of \(F\) when \(t = 3\). [3]
Question 5:
AnswerMarks
5(a)dx  4 
= −1dt
AnswerMarks Guidance
x t+1 M1 Separate variables, obtain RHS in integrable form.
ln x =4ln t+1−t+ AA1
t=0, x=5: A=ln5M1
x=5(t+1)4 e−tA1
4
AnswerMarks
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
AnswerMarks
dtM1
5e−t(t+1)2(5−t)(1−t)
AnswerMarks Guidance
Acceleration =AEF
F=2acceleration, so at F =10e−t(t+1)2(5−t)(1−t)M1
At t=3, magnitude of force is 640e−3 NA1 31.9 N
3
AnswerMarks Guidance
QuestionAnswer Marks 
Question 5:
--- 5(a) ---
5(a) | dx  4 
= −1dt
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=2acceleration, 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 | Guidance
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the magnitude of $F$ when $t = 3$. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2024 Q5 [3]}}
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Q1 5 Q2 5 Q3 6 Q3 2 Q4 4 Q4 3 Q5 4 Q5 3 Q6 3 Q6 3 Q6 2 Q7 4 Q7 6