| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2007 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile passing through given point |
| Difficulty | Standard +0.2 This is a straightforward projectile motion question requiring standard application of trajectory equations or parametric equations to find speed and time, followed by basic velocity component calculations. The given angle, horizontal and vertical displacements make this a routine two-equation problem with no novel insight required—slightly easier than average A-level mechanics. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(3 = 8\tan 35° - \frac{g \cdot 8^2}{2V^2\cos^2 35°}\) | M1 | For substituting \(\theta = 35°\), \(x=8\) and \(y=3\) into trajectory formula, or eliminating \(T\) from \(8 = V t\cos 35°\), \(3 = Vt\sin 35° - \frac{1}{2}gT^2\) |
| M1 | For solving for \(V\) | |
| Speed is \(13.5 \text{ ms}^{-1}\) | A1 | |
| OR: For eliminating \(VT\) from \(8 = Vt\cos 35°\), \(3 = Vt\sin 35° - \frac{1}{2}gT^2\) to find \(T\) (=0.721) | (M1) | |
| For back substituting to find \(V\) | (M1) | |
| Speed is \(13.5 \text{ ms}^{-1}\) | (A1) | |
| M1 | For substituting \(\theta=35°\), \(x=8\) and value of \(V\) into \(x = VT\cos\theta\) or stating value of \(T\) found in (i) (alternate method) | |
| \(T = 0.721\) | A1 | [5 marks] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For using \(v_x = V\cos 35°\) and \(v_y = V\sin 35° - gT\) | |
| M1 | For using \(\tan\alpha = \frac{v_y}{v_x}\) | |
| \(\tan\alpha = \frac{0.55(22)}{11(.09)}\) | A1 | May be implied by final answer |
| Direction \(2.85°\) to the horizontal | A1 | [4 marks] Accept 2.8 or 2.9 |
| OR: | (M1) | For differentiating trajectory equation w.r.t. \(x\) |
| \(y' = \tan 35° - \frac{gx}{V^2\cos^2 35°}\) | (A1) | |
| (M1) | For using \(\tan\alpha = y'(8)\) (0.0498) | |
| Direction \(2.85°\) to the horizontal | (A1) | [Total: 9] |
## Question 6:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3 = 8\tan 35° - \frac{g \cdot 8^2}{2V^2\cos^2 35°}$ | M1 | For substituting $\theta = 35°$, $x=8$ and $y=3$ into trajectory formula, or eliminating $T$ from $8 = V t\cos 35°$, $3 = Vt\sin 35° - \frac{1}{2}gT^2$ |
| | M1 | For solving for $V$ |
| Speed is $13.5 \text{ ms}^{-1}$ | A1 | |
| **OR:** For eliminating $VT$ from $8 = Vt\cos 35°$, $3 = Vt\sin 35° - \frac{1}{2}gT^2$ to find $T$ (=0.721) | (M1) | |
| For back substituting to find $V$ | (M1) | |
| Speed is $13.5 \text{ ms}^{-1}$ | (A1) | |
| | M1 | For substituting $\theta=35°$, $x=8$ and value of $V$ into $x = VT\cos\theta$ or stating value of $T$ found in (i) (alternate method) |
| $T = 0.721$ | A1 | **[5 marks]** |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For using $v_x = V\cos 35°$ and $v_y = V\sin 35° - gT$ |
| | M1 | For using $\tan\alpha = \frac{v_y}{v_x}$ |
| $\tan\alpha = \frac{0.55(22)}{11(.09)}$ | A1 | May be implied by final answer |
| Direction $2.85°$ to the horizontal | A1 | **[4 marks]** Accept 2.8 or 2.9 |
| **OR:** | (M1) | For differentiating trajectory equation w.r.t. $x$ |
| $y' = \tan 35° - \frac{gx}{V^2\cos^2 35°}$ | (A1) | |
| | (M1) | For using $\tan\alpha = y'(8)$ (0.0498) |
| Direction $2.85°$ to the horizontal | (A1) | **[Total: 9]** |
---6 A particle is projected from a point $O$ at an angle of $35 ^ { \circ }$ above the horizontal. At time $T$ s later the particle passes through a point $A$ whose horizontal and vertically upward displacements from $O$ are 8 m and 3 m respectively.\\
(i) By using the equation of the particle's trajectory, or otherwise, find (in either order) the speed of projection of the particle from $O$ and the value of $T$.\\
(ii) Find the angle between the direction of motion of the particle at $A$ and the horizontal.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b9080e9f-2c23-43ce-b171-bd68648dc56b-5_476_895_269_625}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
Fig. 1 shows the cross-section of a uniform solid. The cross-section has the shape and dimensions shown. The centre of mass $C$ of the solid lies in the plane of this cross-section. The distance of $C$ from $D E$ is $y \mathrm {~cm}$.\\
(i) Find the value of $y$.
The solid is placed on a rough plane. The coefficient of friction between the solid and the plane is $\mu$. The plane is tilted so that $E F$ lies along a line of greatest slope.\\
(ii)
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b9080e9f-2c23-43ce-b171-bd68648dc56b-5_375_431_1366_897}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
The solid is placed so that $F$ is higher up the plane than $E$ (see Fig. 2). When the angle of inclination is sufficiently great the solid starts to topple (without sliding). Show that $\mu > \frac { 1 } { 2 }$. [3]\\
(iii)
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b9080e9f-2c23-43ce-b171-bd68648dc56b-5_376_428_2069_900}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}
The solid is now placed so that $E$ is higher up the plane than $F$ (see Fig. 3). When the angle of inclination is sufficiently great the solid starts to slide (without toppling). Show that $\mu < \frac { 5 } { 6 }$. [3]
\hfill \mbox{\textit{CAIE M2 2007 Q6 [9]}}