CAIE M2 2014 November — Question 6 12 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2014
SessionNovember
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicForces, equilibrium and resultants
TypeForces in vector form: kinematics extension
DifficultyStandard +0.3 This is a straightforward application of Newton's second law with vector calculus. Students integrate the force to find velocity (using F=ma), then integrate velocity for position. The calculations are routine with no conceptual challenges—slightly easier than average due to the mechanical nature of the integration and standard kinetic energy formula application.
Spec1.10h Vectors in kinematics: uniform acceleration in vector form3.02f Non-uniform acceleration: using differentiation and integration6.02e Calculate KE and PE: using formulae

A particle of mass \(2\) kg moves under the action of a variable force. At time \(t\) seconds the force is \((6t - 3)\mathbf{i} + 4\mathbf{j}\) newtons, where \(\mathbf{i}\) and \(\mathbf{j}\) are perpendicular unit vectors. When \(t = 0\), the particle is at rest at the origin.
  1. Find the velocity of the particle when \(t = 4\). [4]
  2. Find the kinetic energy of the particle when \(t = 4\). [2]
  3. Find the distance of the particle from the origin when \(t = 2\). [6]

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 51
(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 | 51
(iii) | −2
∫x3dx=∫dt
8
 
1
x3 
  =t
1
 
 
3 1
t = 3 | M1
A1
A1
[3] | dx
Integrates using v =
dt
A particle of mass $2$ kg moves under the action of a variable force. At time $t$ seconds the force is $(6t - 3)\mathbf{i} + 4\mathbf{j}$ newtons, where $\mathbf{i}$ and $\mathbf{j}$ are perpendicular unit vectors. When $t = 0$, the particle is at rest at the origin.

\begin{enumerate}[label=(\roman*)]
\item Find the velocity of the particle when $t = 4$. [4]
\item Find the kinetic energy of the particle when $t = 4$. [2]
\item Find the distance of the particle from the origin when $t = 2$. [6]
\end{enumerate}

\hfill \mbox{\textit{CAIE M2 2014 Q6 [12]}}