CAIE M2 2014 November — Question 6 9 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2014
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable Force
TypeAir resistance with other powers
DifficultyChallenging +1.2 This is a multi-part variable force mechanics question requiring integration of a differential equation with fractional powers. Part (i) is straightforward application of F=ma and the chain rule. Part (ii) requires separating variables and integrating v^(1/2), which is standard but slightly beyond routine. Part (iii) needs another integration using the v-x relationship. The fractional power adds mild algebraic complexity, but the overall structure follows a predictable pattern for M2 variable force questions, making it moderately above average difficulty.
Spec3.02f Non-uniform acceleration: using differentiation and integration6.06a Variable force: dv/dt or v*dv/dx methods

\(O\), \(A\) and \(B\) are three points in a straight line on a smooth horizontal surface. A particle \(P\) of mass \(0.6\) kg moves along the line. At time \(t\) s the particle has displacement \(x\) m from \(O\) and speed \(v\) m s\(^{-1}\). The only horizontal force acting on \(P\) has magnitude \(0.4v^{\frac{1}{2}}\) N and acts in the direction \(OA\). Initially the particle is at \(A\), where \(x = 1\) and \(v = 1\).
  1. Show that \(3v^{\frac{1}{2}}\frac{dv}{dx} = 2\). [2]
  2. Express \(v\) in terms of \(x\). [4]
  3. Given that \(AB = 7\) m, find the value of \(t\) when \(P\) passes through \(B\). [3]

Question 6:

(ii) ---
6 (i)
AnswerMarks
(ii)dv
1/2
0.6v = 0.4v
dx
dv
1/2
3v = 2 AG
dx
1
3∫v2dv =2∫dx
3
3v2
=2x (+c)
3
2
3 2
3 × 12 × = 2 +c
3
2
AnswerMarks
v= x3M1
A1
[2]
M1
A1
M1
A1
AnswerMarks
[4]dv
Newton’s 2nd law, a = v
dx
Integrates
Accept omission of +c
Evaluates c (=0)
AnswerMarks Guidance
Page 6Mark Scheme Syllabus
Cambridge International A Level – October/November 20149709 52
(iii)−2
∫x3dx=∫dt
8
 
1
x3 
  =t
1
 
 
3 1
AnswerMarks
t = 3M1
A1
A1
AnswerMarks
[3]dx
Integrates using v =
dt
Question 6:
--- 6 (i)
(ii) ---
6 (i)
(ii) | dv
1/2
0.6v = 0.4v
dx
dv
1/2
3v = 2 AG
dx
1
3∫v2dv =2∫dx
3
3v2
=2x (+c)
3
2
3 2
3 × 12 × = 2 +c
3
2
v= x3 | M1
A1
[2]
M1
A1
M1
A1
[4] | dv
Newton’s 2nd law, a = v
dx
Integrates
Accept omission of +c
Evaluates c (=0)
Page 6 | Mark Scheme | Syllabus | Paper
Cambridge International A Level – October/November 2014 | 9709 | 52
(iii) | −2
∫x3dx=∫dt
8
 
1
x3 
  =t
1
 
 
3 1
t = 3 | M1
A1
A1
[3] | dx
Integrates using v =
dt
$O$, $A$ and $B$ are three points in a straight line on a smooth horizontal surface. A particle $P$ of mass $0.6$ kg moves along the line. At time $t$ s the particle has displacement $x$ m from $O$ and speed $v$ m s$^{-1}$. The only horizontal force acting on $P$ has magnitude $0.4v^{\frac{1}{2}}$ N and acts in the direction $OA$. Initially the particle is at $A$, where $x = 1$ and $v = 1$.

\begin{enumerate}[label=(\roman*)]
\item Show that $3v^{\frac{1}{2}}\frac{dv}{dx} = 2$. [2]
\item Express $v$ in terms of $x$. [4]
\item Given that $AB = 7$ m, find the value of $t$ when $P$ passes through $B$. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE M2 2014 Q6 [9]}}