CAIE Further Paper 3 2021 June — Question 1 5 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2021
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable acceleration (1D)
TypeAcceleration as function of velocity (separation of variables)
DifficultyChallenging +1.2 This is a variable acceleration problem requiring separation of variables and integration. While it involves a non-standard resistance force with both v and t dependence, the separation is straightforward (√v and (t+1)^2 separate cleanly), and the integrations are routine. The 5-mark allocation and Further Maths context indicate above-average difficulty, but the mathematical techniques are standard calculus without requiring novel insight.
Spec6.06a Variable force: dv/dt or v*dv/dx methods

A particle \(P\) of mass 1 kg is moving along a straight line against a resistive force of magnitude \(\frac{10\sqrt{v}}{(t+1)^2}\) N, where \(v\) ms\(^{-1}\) is the speed of \(P\) at time \(t\)s. When \(t = 0\), \(v = 25\). Find an expression for \(v\) in terms of \(t\). [5]

Question 1:
AnswerMarks
1dv 10dt
=−
AnswerMarks Guidance
v ( t+1 )2M1 Separate variables.
10
2 v = + A
AnswerMarks Guidance
t+1M1 A1 Attempt to integrate.
t=0, v=25, A=0M1 Use correct initial condition.
25
v=
AnswerMarks Guidance
( t+1 )2A1 CAO
5
AnswerMarks Guidance
QuestionAnswer Marks
Question 1:
1 | dv 10dt
=−
v ( t+1 )2 | M1 | Separate variables.
10
2 v = + A
t+1 | M1 A1 | Attempt to integrate.
t=0, v=25, A=0 | M1 | Use correct initial condition.
25
v=
( t+1 )2 | A1 | CAO
5
Question | Answer | Marks | Guidance
A particle $P$ of mass 1 kg is moving along a straight line against a resistive force of magnitude $\frac{10\sqrt{v}}{(t+1)^2}$ N, where $v$ ms$^{-1}$ is the speed of $P$ at time $t$s. When $t = 0$, $v = 25$.

Find an expression for $v$ in terms of $t$. [5]

\hfill \mbox{\textit{CAIE Further Paper 3 2021 Q1 [5]}}