CAIE M2 2012 November — Question 5 7 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2012
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeSpeed at specific time or position
DifficultyStandard +0.3 This is a standard projectiles question requiring application of energy conservation and kinematic equations. Part (i) is straightforward using v² = u² - 2as or energy methods. Part (ii) requires finding time and position components, but follows routine procedures. The 'speed increasing' condition guides students to the correct phase of motion. Slightly above average due to the two-part structure and coordinate geometry in part (ii), but all techniques are standard M2 material.
Spec3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model

A particle \(P\) is projected with speed \(30\) m s\(^{-1}\) at an angle of \(60°\) above the horizontal from a point \(O\) on horizontal ground. For the instant when the speed of \(P\) is \(17\) m s\(^{-1}\) and increasing,
  1. show that the vertical component of the velocity of \(P\) is \(8\) m s\(^{-1}\) downwards, [2]
  2. calculate the distance of \(P\) from \(O\). [5]

(i)
AnswerMarks Guidance
\(v^2 = 17^2 - (30 \cos60)^2\)M1 Finds vertical speed
\(v = -8\)A1 [2] – may be implied by later work
(ii)
AnswerMarks Guidance
\(-8 = 30 \sin60 - gt\)M1 Finds relevant time
\(t = 3.4\)A1 3.398
\(y = [(30 \sin60)^2 - 8^2]/(2g) = (30.55)\)B1 Or \(y = (30 \sin60) \times 3.4 - g \cdot 3.4^2/2 = (30.53)\)
\(OP^2 = (30 \cos60 \times 3.4)^2 + 30.55^2\)M1 Use of Pythagoras
\(OP = 59.4 \, \text{m}\)A1 [5] Accept 59.5
## (i)

$v^2 = 17^2 - (30 \cos60)^2$ | M1 | Finds vertical speed
$v = -8$ | A1 [2] | – may be implied by later work

## (ii)

$-8 = 30 \sin60 - gt$ | M1 | Finds relevant time
$t = 3.4$ | A1 | 3.398
$y = [(30 \sin60)^2 - 8^2]/(2g) = (30.55)$ | B1 | Or $y = (30 \sin60) \times 3.4 - g \cdot 3.4^2/2 = (30.53)$
$OP^2 = (30 \cos60 \times 3.4)^2 + 30.55^2$ | M1 | Use of Pythagoras
$OP = 59.4 \, \text{m}$ | A1 [5] | Accept 59.5
A particle $P$ is projected with speed $30$ m s$^{-1}$ at an angle of $60°$ above the horizontal from a point $O$ on horizontal ground. For the instant when the speed of $P$ is $17$ m s$^{-1}$ and increasing,

\begin{enumerate}[label=(\roman*)]
\item show that the vertical component of the velocity of $P$ is $8$ m s$^{-1}$ downwards, [2]
\item calculate the distance of $P$ from $O$. [5]
\end{enumerate}

\hfill \mbox{\textit{CAIE M2 2012 Q5 [7]}}