| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projection from elevated point - angle above horizontal |
| Difficulty | Standard +0.3 This is a standard M2 projectile question with projection from an elevated point. It requires resolving initial velocity, applying SUVAT equations for vertical motion to find time, then calculating horizontal range and final speed using standard formulas. The calculations are straightforward with no conceptual challenges beyond routine application of projectile motion principles, making it slightly easier than average. |
| Spec | 3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((\uparrow):\ -52.5 = 14t - \frac{1}{2} \times 9.8t^2\) | M1 A2 | |
| \(7t^2 - 20t - 75 = 0\) | ||
| \((7t+15)(t-5) = 0\) | M1 A1 A1 | |
| \(t = 5\ \left(\text{or } t = -\frac{15}{7}\right)\) | M1 | |
| \((\rightarrow):\ S = 28\cos 30° \times 5\) | A1 | (8) |
| \(= 70\sqrt{3} = 121\ \text{m (3 s.f.)}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(v_{\text{horizontal}}:\ 28\cos 30° = 14\sqrt{3}\) | B1 | |
| \(v_{\text{vertical}}:\ 28\sin 30° - 5g = -35\) | M1 A1 | |
| \(\therefore \text{speed} = \sqrt{(14\sqrt{3})^2 + 35^2} = \sqrt{1813} = 42.6\ \text{m s}^{-1}\) | M1 A1 | (5) (13) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| KE gain = PE loss | ||
| \(\frac{1}{2}m(v^2 - 28^2) = mg \times 52.5\) | M1 A2 | |
| \(\Rightarrow v = \sqrt{1813} = 42.6\ \text{m s}^{-1}\) | M1 A1 | (5) |
# Question 7:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(\uparrow):\ -52.5 = 14t - \frac{1}{2} \times 9.8t^2$ | M1 A2 | |
| $7t^2 - 20t - 75 = 0$ | | |
| $(7t+15)(t-5) = 0$ | M1 A1 A1 | |
| $t = 5\ \left(\text{or } t = -\frac{15}{7}\right)$ | M1 | |
| $(\rightarrow):\ S = 28\cos 30° \times 5$ | A1 | (8) |
| $= 70\sqrt{3} = 121\ \text{m (3 s.f.)}$ | | |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $v_{\text{horizontal}}:\ 28\cos 30° = 14\sqrt{3}$ | B1 | |
| $v_{\text{vertical}}:\ 28\sin 30° - 5g = -35$ | M1 A1 | |
| $\therefore \text{speed} = \sqrt{(14\sqrt{3})^2 + 35^2} = \sqrt{1813} = 42.6\ \text{m s}^{-1}$ | M1 A1 | (5) **(13)** |
## Alternative for Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| KE gain = PE loss | | |
| $\frac{1}{2}m(v^2 - 28^2) = mg \times 52.5$ | M1 A2 | |
| $\Rightarrow v = \sqrt{1813} = 42.6\ \text{m s}^{-1}$ | M1 A1 | (5) |
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9126ebb1-eaa7-4a40-953f-5dc819c9f479-6_675_1243_392_415}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}
A shot is projected upwards from the top of a cliff with a velocity of $28 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $30 ^ { \circ }$ above the horizontal. It strikes the ground 52.5 m vertically below the level of the point of projection, as shown in Fig. 3. The motion of the shot is modelled as that of a particle moving freely under gravity.
Find, to 3 significant figures,
\begin{enumerate}[label=(\alph*)]
\item the horizontal distance from the point of projection at which the shot strikes the ground,
\item the speed of the shot as it strikes the ground.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q7 [13]}}