CAIE M2 2012 June — Question 1 4 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2012
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable Force
TypeVariable force (position x) - find velocity
DifficultyStandard +0.8 This is a variable force problem requiring the v dv/dx method with integration. Students must set up F = ma as 0.3x = 0.6v dv/dx, separate variables, and integrate to find v when x = 8. While the integration itself is straightforward, recognizing the appropriate method and correctly applying it with initial conditions requires solid understanding beyond routine exercises. This is moderately challenging for M2 level.
Spec6.06a Variable force: dv/dt or v*dv/dx methods

1 A particle \(P\) of mass 0.6 kg is projected horizontally with velocity \(2 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) from a point \(O\) on a smooth horizontal surface. A horizontal force of magnitude \(0.3 x \mathrm {~N}\) acts on \(P\) in the direction \(O P\), where \(x \mathrm {~m}\) is the distance of \(P\) from \(O\). Calculate the velocity of \(P\) when \(x = 8\).

AnswerMarks Guidance
\(0.6y\frac{dy}{dx} = 0.3x\)M1 Newton's Second Law with \(a = v\frac{dv}{dx}\)
\(0.6v^2/2 = 0.3x^2/(2 + c)\)A1 From \(\int 0.6v\,dv = \int 0.3x\,dx\)
\([x^2/2]_0^8 = [v^2]_2^5\)M1 Uses limits of finds constant
\(v = 6\,\mathrm{ms}^{-1}\)A1 [4]
$0.6y\frac{dy}{dx} = 0.3x$ | M1 | Newton's Second Law with $a = v\frac{dv}{dx}$
$0.6v^2/2 = 0.3x^2/(2 + c)$ | A1 | From $\int 0.6v\,dv = \int 0.3x\,dx$
$[x^2/2]_0^8 = [v^2]_2^5$ | M1 | Uses limits of finds constant
$v = 6\,\mathrm{ms}^{-1}$ | A1 | [4]
1 A particle $P$ of mass 0.6 kg is projected horizontally with velocity $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a point $O$ on a smooth horizontal surface. A horizontal force of magnitude $0.3 x \mathrm {~N}$ acts on $P$ in the direction $O P$, where $x \mathrm {~m}$ is the distance of $P$ from $O$. Calculate the velocity of $P$ when $x = 8$.

\hfill \mbox{\textit{CAIE M2 2012 Q1 [4]}}