| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Finding angle given constraints |
| Difficulty | Standard +0.3 This is a standard M2 projectiles question requiring application of kinematic equations with constant acceleration. Part (i) uses velocity components and the speed formula to find the angle—straightforward but requires careful algebra. Part (ii) involves finding vertical displacement, which is routine. The question tests standard techniques without requiring novel insight or complex problem-solving, making it slightly easier than average for M2 level. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| \(15\cos\theta = v_H\) and \(15\sin\theta - 4g = v_V\) | B1 | Use horizontal and vertical motion |
| \((15\cos\theta)^2 + (15\sin\theta - 4g)^2 = 30^2\) | M1 | Use Pythagoras's theorem |
| \([225 - 1200\sin\theta + 1600 = 900]\) | M1 | Attempt to solve for \(\theta\) |
| \(\theta = 50.4°\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(h = (15\sin\theta)\times 4 - \frac{g(4)^2}{2}\) | B1 | |
| \(\frac{m(15)^2}{2} = \frac{m(30)^2}{2} + mgh\) | M1 | Allow \(h\) not replaced |
| M1 | Attempt to eliminate \(h\) and attempt to solve for \(\theta\) | |
| \(\theta = 50.4°\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(s = 15\sin 50.4 \times 4 - \frac{1}{2} \times g \times 4^2\) | M1 | Use vertical motion. Allow *their* \(\theta\) for first M1 |
| \(s = 33.75\) m | A1 | AG |
| \(\cos\alpha = \frac{15\cos 50.4}{30}\) | M1 | Use trigonometry of a right angled triangle |
| \(\alpha = 71.4°\) below the horizontal | A1 | |
| If \(g = 9.8\) or \(9.81\) used then M1A0M1A0 |
## Question 6(i):
| $15\cos\theta = v_H$ and $15\sin\theta - 4g = v_V$ | B1 | Use horizontal and vertical motion |
| $(15\cos\theta)^2 + (15\sin\theta - 4g)^2 = 30^2$ | M1 | Use Pythagoras's theorem |
| $[225 - 1200\sin\theta + 1600 = 900]$ | M1 | Attempt to solve for $\theta$ |
| $\theta = 50.4°$ | A1 | |
**Alternative Method:**
| $h = (15\sin\theta)\times 4 - \frac{g(4)^2}{2}$ | B1 | |
| $\frac{m(15)^2}{2} = \frac{m(30)^2}{2} + mgh$ | M1 | Allow $h$ not replaced |
| | M1 | Attempt to eliminate $h$ and attempt to solve for $\theta$ |
| $\theta = 50.4°$ | A1 | |
## Question 6(ii):
| $s = 15\sin 50.4 \times 4 - \frac{1}{2} \times g \times 4^2$ | M1 | Use vertical motion. Allow *their* $\theta$ for first M1 |
| $s = 33.75$ m | A1 | AG |
| $\cos\alpha = \frac{15\cos 50.4}{30}$ | M1 | Use trigonometry of a right angled triangle |
| $\alpha = 71.4°$ below the horizontal | A1 | |
| | | If $g = 9.8$ or $9.81$ used then M1A0M1A0 |
---6 A particle is projected with speed $15 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $\theta ^ { \circ }$ above the horizontal. At the instant 4 s after projection the speed of the particle is $30 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) Find $\theta$.\\
(ii) Show that at the instant 4 s after projection the particle is 33.75 m below the level of the point of projection and find the direction of motion at this instant.\\
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f3a35846-075d-4e03-ba6b-82774ef0e4f8-12_259_609_255_769}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
Fig. 1 shows an object made from a uniform wire of length 0.8 m . The object consists of a straight part $A B$, and a semicircular part $B C$ such that $A , B$ and $C$ lie in the same straight line. The radius of the semicircle is $r \mathrm {~m}$ and the centre of mass of the object is 0.1 m from line $A B C$.\\
(i) Show that $r = 0.2$.\\
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f3a35846-075d-4e03-ba6b-82774ef0e4f8-13_615_383_260_881}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
The object is freely suspended at $A$ and a horizontal force of magnitude 7 N is applied to the object at $C$ so that the object is in equilibrium with $A B C$ vertical (see Fig. 2).\\
(ii) Calculate the weight of the object.\\
The 7 N force is removed and the object hangs in equilibrium with $A B C$ at an angle of $\theta ^ { \circ }$ with the vertical.\\
(iii) Find $\theta$.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.\\
\hfill \mbox{\textit{CAIE M2 2019 Q6 [8]}}