CAIE M2 2012 June — Question 4 7 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2012
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable Force
TypeForce depends on time t
DifficultyStandard +0.3 This is a straightforward variable force problem requiring integration of F=ma with a simple linear force function. Students apply Newton's second law to get dv/dt = -12t, integrate once to find initial velocity using the boundary condition, then integrate again to find distance. All steps are standard M2 techniques with no conceptual surprises.
Spec6.06a Variable force: dv/dt or v*dv/dx methods

4 A particle \(P\) of mass 0.25 kg moves in a straight line on a smooth horizontal surface. At time \(t \mathrm {~s}\) the velocity of \(P\) is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). A variable force of magnitude \(3 t \mathrm {~N}\) opposes the motion of \(P\).
  1. Given that \(P\) comes to rest when \(t = 3\), find \(v\) when \(t = 0\).
  2. Calculate the distance travelled by \(P\) in the interval \(0 \leqslant t \leqslant 3\).

(i)
AnswerMarks Guidance
\(0.25v\frac{dv}{dt} = -3t\)M1 Newton's Second Law, – sign essential
\(v = -12t^2/(2 + c)\)A1 Accept uncancelled form
\(0 = 12 \times 3^2/2 + c\)M1 Appropriate use of \(v = 0\), \(t = 3\)
Initial speed \(= 54\,\mathrm{ms}^{-1}\)A1 [4]
(ii)
AnswerMarks Guidance
\(\int dx = \int [54 - 6t^2]\,dt\)M1 Separates variables, integrates \(v\)
\(x = [54t - 6t^3/3]_0^3\)A1 \(\unicode{x2713}\) candidates value [\(v\) in (i)]
\(x = 108\,\mathrm{m}\)A1 [3]
**(i)**

$0.25v\frac{dv}{dt} = -3t$ | M1 | Newton's Second Law, – sign essential
$v = -12t^2/(2 + c)$ | A1 | Accept uncancelled form
$0 = 12 \times 3^2/2 + c$ | M1 | Appropriate use of $v = 0$, $t = 3$
Initial speed $= 54\,\mathrm{ms}^{-1}$ | A1 | [4] | Goes beyond $c = 54$

**(ii)**

$\int dx = \int [54 - 6t^2]\,dt$ | M1 | Separates variables, integrates $v$
$x = [54t - 6t^3/3]_0^3$ | A1 | $\unicode{x2713}$ candidates value [$v$ in (i)]
$x = 108\,\mathrm{m}$ | A1 | [3] | [7]
4 A particle $P$ of mass 0.25 kg moves in a straight line on a smooth horizontal surface. At time $t \mathrm {~s}$ the velocity of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. A variable force of magnitude $3 t \mathrm {~N}$ opposes the motion of $P$.\\
(i) Given that $P$ comes to rest when $t = 3$, find $v$ when $t = 0$.\\
(ii) Calculate the distance travelled by $P$ in the interval $0 \leqslant t \leqslant 3$.

\hfill \mbox{\textit{CAIE M2 2012 Q4 [7]}}