CAIE M2 2010 November — Question 7 10 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2010
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeVelocity direction at specific time/point
DifficultyStandard +0.3 This is a standard projectiles question requiring routine application of kinematic equations and trajectory derivation. Part (i) involves direct substitution into standard formulae, part (ii) uses the geometric constraint tan(30°) = y/x with the trajectory equation, and part (iii) requires differentiation of the trajectory to find the gradient. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average.
Spec3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model

\includegraphics{figure_7} A particle \(P\) is projected from a point \(O\) with initial speed \(10 \text{ m s}^{-1}\) at an angle of \(45°\) above the horizontal. \(P\) subsequently passes through the point \(A\) which is at an angle of elevation of \(30°\) from \(O\) (see diagram). At time \(t \text{ s}\) after projection the horizontal and vertically upward displacements of \(P\) from \(O\) are \(x \text{ m}\) and \(y \text{ m}\) respectively.
  1. Write down expressions for \(x\) and \(y\) in terms of \(t\), and hence obtain the equation of the trajectory of \(P\). [3]
  2. Calculate the value of \(x\) when \(P\) is at \(A\). [3]
  3. Find the angle the trajectory makes with the horizontal when \(P\) is at \(A\). [4]

Question 7:
AnswerMarks
7(i) x = (10cos45°)t and
2
AnswerMarks
y = (10sin45°)t – gt /2B1
2
AnswerMarks
y = (10sin45° / 10cos45°)x – 10(x/10cos45°) /2M1
2
AnswerMarks
y = x – x /10A1
[3]
AnswerMarks Guidance
(ii) y/x = tan30°M1
1 – x/10 = tan30°A1
x = 4.23A1
[3]4.2264…
(iii) dy/dx = 1 – 2x/10M1 4.2264 = (10cos45°)t
θ
AnswerMarks Guidance
tan = dy/dxB1 t = 0.5977
θ
AnswerMarks Guidance
tan =1 – 2x4.23/10(= 0.15472..)M1 10sin45°−10x0.5977
θ
tan =
10cos45°
AnswerMarks
θ = 8.79°A1
[4]
Question 7:
7 | (i) x = (10cos45°)t and
2
y = (10sin45°)t – gt /2 | B1
2
y = (10sin45° / 10cos45°)x – 10(x/10cos45°) /2 | M1
2
y = x – x /10 | A1
[3]
(ii) y/x = tan30° | M1
1 – x/10 = tan30° | A1
x = 4.23 | A1
[3] | 4.2264…
(iii) dy/dx = 1 – 2x/10 | M1 | 4.2264 = (10cos45°)t
θ
tan = dy/dx | B1 | t = 0.5977
θ
tan =1 – 2x4.23/10(= 0.15472..) | M1 | 10sin45°−10x0.5977
θ
tan =
10cos45°
θ = 8.79° | A1
[4]
\includegraphics{figure_7}

A particle $P$ is projected from a point $O$ with initial speed $10 \text{ m s}^{-1}$ at an angle of $45°$ above the horizontal. $P$ subsequently passes through the point $A$ which is at an angle of elevation of $30°$ from $O$ (see diagram). At time $t \text{ s}$ after projection the horizontal and vertically upward displacements of $P$ from $O$ are $x \text{ m}$ and $y \text{ m}$ respectively.

\begin{enumerate}[label=(\roman*)]
\item Write down expressions for $x$ and $y$ in terms of $t$, and hence obtain the equation of the trajectory of $P$. [3]
\item Calculate the value of $x$ when $P$ is at $A$. [3]
\item Find the angle the trajectory makes with the horizontal when $P$ is at $A$. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE M2 2010 Q7 [10]}}