| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Session | Specimen |
| 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 projectiles question requiring resolution of initial velocity using tan α, then applying SUVAT equations to find U from given horizontal/vertical displacements. Part (b) uses energy conservation or trajectory equations. Part (c) is routine modelling critique. Slightly easier than average due to straightforward setup and nice numbers (tan α = 3/4 gives clean components), though multi-step calculation required. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(36 = Ut\cos\alpha\) | M1, A1 | Using \(s = ut\) horizontally |
| \(-18 = Ut\sin\alpha - \frac{1}{2}gt^2\) | M1, A1 | Using \(s = ut + \frac{1}{2}at^2\) vertically |
| Correct strategy solving two equations in \(t\) and \(U\) for \(U\) | M1 | Need both equations |
| \(U = 15\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(U\cos\alpha\) used as horizontal velocity component \((= 12)\) | B1 | |
| \(v^2 = (U\sin\alpha)^2 + 2(-10)(-7.2)\) | M1 | Attempt to find vertical component |
| \(v = 15\) | A1 | |
| Correct strategy: find both components, combine using Pythagoras: Speed \(= \sqrt{12^2 + 15^2}\) | M1 | |
| \(\sqrt{369} = 19 \text{ m s}^{-1}\) (2sf) | A1ft | Following through on incorrect component(s) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Any valid improvement e.g. include air resistance | B1 | |
| Any second valid improvement e.g. use more accurate value of \(g\); include wind effects; include dimensions of stone | B1 |
## Question 10:
### Part (a)
| Working/Answer | Mark | Guidance |
|---|---|---|
| $36 = Ut\cos\alpha$ | M1, A1 | Using $s = ut$ horizontally |
| $-18 = Ut\sin\alpha - \frac{1}{2}gt^2$ | M1, A1 | Using $s = ut + \frac{1}{2}at^2$ vertically |
| Correct strategy solving two equations in $t$ and $U$ for $U$ | M1 | Need both equations |
| $U = 15$ | A1 | |
**(6 marks)**
### Part (b)
| Working/Answer | Mark | Guidance |
|---|---|---|
| $U\cos\alpha$ used as horizontal velocity component $(= 12)$ | B1 | |
| $v^2 = (U\sin\alpha)^2 + 2(-10)(-7.2)$ | M1 | Attempt to find vertical component |
| $v = 15$ | A1 | |
| Correct strategy: find both components, combine using Pythagoras: Speed $= \sqrt{12^2 + 15^2}$ | M1 | |
| $\sqrt{369} = 19 \text{ m s}^{-1}$ (2sf) | A1ft | Following through on incorrect component(s) |
**(5 marks)**
### Part (c)
| Working/Answer | Mark | Guidance |
|---|---|---|
| Any valid improvement e.g. include air resistance | B1 | |
| Any second valid improvement e.g. use more accurate value of $g$; include wind effects; include dimensions of stone | B1 | |
**(2 marks)**
**(Total: 13 marks)**10.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e678bf51-6dca-4ad7-808b-dfa31b04dc63-22_719_1333_246_365}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A boy throws a stone with speed $U \mathrm {~ms} ^ { - 1 }$ from a point $O$ at the top of a vertical cliff. The point $O$ is 18 m above sea level.\\
The stone is thrown at an angle $\alpha$ above the horizontal, where $\tan \alpha = \frac { 3 } { 4 }$.\\
The stone hits the sea at the point $S$ which is at a horizontal distance of 36 m from the foot of the cliff, as shown in Figure 2.\\
The stone is modelled as a particle moving freely under gravity with $g = 10 \mathrm {~m} \mathrm {~s} ^ { - 2 }$\\
Find
\begin{enumerate}[label=(\alph*)]
\item the value of $U$,
\item the speed of the stone when it is 10.8 m above sea level, giving your answer to 2 significant figures.
\item Suggest two improvements that could be made to the model.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 3 Q10 [13]}}