| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile with bounce or impact |
| Difficulty | Challenging +1.8 This is a challenging M2 projectile question requiring resolution in directions parallel/perpendicular to an inclined wall, applying impact conditions, and linking two separate projectile motions. It demands careful geometric reasoning about velocity components at 60° to horizontal, plus multi-stage calculations across two bounces. Significantly harder than routine projectile questions but uses standard M2 techniques without requiring novel mathematical insight. |
| Spec | 3.02i Projectile motion: constant acceleration model6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(U\cos\theta = 18\cos30\ (= 9\sqrt{3} = 15.588...)\) | B1 | |
| \(U\sin\theta - 2g = -18\sin30\) | B1 | \(U\sin\theta = 11\) |
| \(U^2 = 15.588^2 + 11^2\) | M1 | Pythagoras or \(\tan\theta = 11/15.588\) |
| \(U = 19.1\) | A1 | |
| \(\theta = 35.2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(X = 0.8V\cos30\) | B1 | Horizontal displacement |
| \(Y = -0.8V\sin30 + g(0.8)^2/2\) | B1 | Vertical displacement |
| \((3.2 - 0.4V)/(0.8V\cos30) = \tan60\) | M1 | Or \(0.8V\cos30/(3.2 - 0.4V) = \tan30\) |
| \(V = 2\) | A1 | |
| OR working perpendicular to the wall | B1* | |
| \(a = g\cos60\) | DB1* | |
| \(0 = 0.8V - g\cos60(0.8)^2/2\) | M1 | Uses \(s = 0\) |
| \(V = 2\) | A1 |
## Question 6:
### Part (i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $U\cos\theta = 18\cos30\ (= 9\sqrt{3} = 15.588...)$ | B1 | |
| $U\sin\theta - 2g = -18\sin30$ | B1 | $U\sin\theta = 11$ |
| $U^2 = 15.588^2 + 11^2$ | M1 | Pythagoras or $\tan\theta = 11/15.588$ |
| $U = 19.1$ | A1 | |
| $\theta = 35.2$ | A1 | |
**Total: 5 marks**
### Part (ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $X = 0.8V\cos30$ | B1 | Horizontal displacement |
| $Y = -0.8V\sin30 + g(0.8)^2/2$ | B1 | Vertical displacement |
| $(3.2 - 0.4V)/(0.8V\cos30) = \tan60$ | M1 | Or $0.8V\cos30/(3.2 - 0.4V) = \tan30$ |
| $V = 2$ | A1 | |
| OR working perpendicular to the wall | B1* | |
| $a = g\cos60$ | DB1* | |
| $0 = 0.8V - g\cos60(0.8)^2/2$ | M1 | Uses $s = 0$ |
| $V = 2$ | A1 | |
**Total: 4 marks**
---6
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a03ad6c1-b4a3-4007-8d3b-ce289a998a55-4_520_582_264_440}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a03ad6c1-b4a3-4007-8d3b-ce289a998a55-4_497_300_287_1411}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
A small ball $B$ is projected with speed $U \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $\theta ^ { \circ }$ above the horizontal from a point $O$. At time 2 s after the instant of projection, $B$ strikes a smooth wall which slopes at $60 ^ { \circ }$ to the horizontal. The speed of $B$ is $18 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and its direction of motion is perpendicular to the wall at the instant of impact (see Fig. 1). $B$ bounces off the wall with speed $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in a direction perpendicular to the wall. At time 0.8 s after $B$ bounces off the wall, $B$ strikes the wall again at a lower point $A$ (see Fig. 2).\\
(i) Find $U$ and $\theta$.\\
(ii) By considering the motion of $B$ after it bounces off the wall, calculate $V$.
\hfill \mbox{\textit{CAIE M2 2015 Q6 [9]}}