CAIE M2 2002 November — Question 3 6 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2002
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeHorizontal projection from height
DifficultyModerate -0.3 This is a standard two-part projectiles question requiring kinematic equations for horizontal projection and basic trajectory analysis. Part (i) involves routine application of s=ut and s=ut+½at² to find x and y, then eliminating t. Part (ii) requires finding time to hit sea level and calculating the angle using velocity components. All steps are textbook-standard with no novel problem-solving required, making it slightly easier than average.
Spec3.02i Projectile motion: constant acceleration model

3 \includegraphics[max width=\textwidth, alt={}, center]{fcf239a6-6558-43ec-b404-70aa349af6a9-2_502_789_1742_680} A stone is projected horizontally, with speed \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), from the top of a vertical cliff of height 45 m above sea level (see diagram). At time \(t \mathrm {~s}\) after projection the horizontal and vertically upward displacements of the stone from the top of the cliff are \(x \mathrm {~m}\) and \(y \mathrm {~m}\) respectively.
  1. Write down expressions for \(x\) and \(y\) in terms of \(t\), and hence obtain the equation of the stone's trajectory.
  2. Find the angle the trajectory makes with the horizontal at the point where the stone reaches sea level.

(i)
AnswerMarks Guidance
\(x = 10t, y = -5t^2\)B1
Eliminates \(t\) to find an equation in \(x\) and \(y\)M1
Obtains \(y = -x^2/20\)A1 3 marks
(ii)
AnswerMarks Guidance
Uses \(\tan\theta = dy/dx\) or \(\tan\theta = y/x\)M1
Obtains \(x = 30\) when \(y = -45\) or \(t = 3\) when \(y = -45\) or \(x \geq 1\sigma\) and \(y = (∓)30\)A1
Obtains angle as 108.4° (108.435) or 71.6° (71.565)A1 3 marks
**(i)**
$x = 10t, y = -5t^2$ | B1 |
Eliminates $t$ to find an equation in $x$ and $y$ | M1 |
Obtains $y = -x^2/20$ | A1 | 3 marks |

**(ii)**
Uses $\tan\theta = dy/dx$ or $\tan\theta = y/x$ | M1 |
Obtains $x = 30$ when $y = -45$ or $t = 3$ when $y = -45$ or $x \geq 1\sigma$ and $y = (∓)30$ | A1 |
Obtains angle as 108.4° (108.435) or 71.6° (71.565) | A1 | 3 marks |
3\\
\includegraphics[max width=\textwidth, alt={}, center]{fcf239a6-6558-43ec-b404-70aa349af6a9-2_502_789_1742_680}

A stone is projected horizontally, with speed $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, from the top of a vertical cliff of height 45 m above sea level (see diagram). At time $t \mathrm {~s}$ after projection the horizontal and vertically upward displacements of the stone from the top of the cliff are $x \mathrm {~m}$ and $y \mathrm {~m}$ respectively.\\
(i) Write down expressions for $x$ and $y$ in terms of $t$, and hence obtain the equation of the stone's trajectory.\\
(ii) Find the angle the trajectory makes with the horizontal at the point where the stone reaches sea level.

\hfill \mbox{\textit{CAIE M2 2002 Q3 [6]}}