| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2023 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Two possible trajectories through point |
| Difficulty | Standard +0.3 This is a standard A-level mechanics projectile question requiring systematic application of SUVAT equations in two dimensions. Parts (a) and (b) are guided 'show that' proofs with clear targets, (c) requires finding maximum height from two possible angles, and (d) is a standard modelling assumption recall. While it involves multiple steps and algebraic manipulation, the approach is entirely routine for this topic with no novel insight required. |
| Spec | 3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(28\cos\alpha \times T = 40\) | M1 | 3.4 — Correct no. of terms, dim correct, condone sin/cos confusion and sign errors |
| \(T = \dfrac{10}{7\cos\alpha}\) * | A1* | 1.1b — Allow \(\frac{40}{28\cos\alpha} = \frac{10}{7\cos\alpha}\) or \(t\) instead of \(T\) |
| (2 marks total) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Vertical motion equation in \(T\) and \(\alpha\) only | M1 | 3.3 — Correct no. of terms, dim correct, condone sin/cos confusion and sign errors |
| \(20 = (28\sin\alpha)T - \frac{1}{2}gT^2\) | A1 | 1.1b |
| Eliminate \(T\): \(20 = (28\sin\alpha)\times\dfrac{10}{7\cos\alpha} - \frac{1}{2}g\left(\dfrac{10}{7\cos\alpha}\right)^2\) | M1 | 1.1b — Eliminate \(T\) using (a) or own \(T\) to give equation in \(\alpha\) only |
| Use \(\sec^2\alpha = 1+\tan^2\alpha\): \(20 = 40\tan\alpha - \frac{1}{2}g\times\dfrac{100}{49}(1+\tan^2\alpha)\) | M1 | 3.1b |
| \(\tan^2\alpha - 4\tan\alpha + 3 = 0\) * (allow \(0 = \tan^2\alpha - 4\tan\alpha + 3\)) | A1* | 2.2a — Must be \(\alpha\) (or \(a\)) in final answer |
| (5 marks total) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Solve and use \(\tan\alpha = 3\) or \(\sin\alpha = \dfrac{3}{\sqrt{10}}\) or \(\alpha = 71.565...°\) to find equation in \(H\) only | M1 | 3.1b — Select larger value of \(\tan\alpha\) |
| \(0 = (28\sin\alpha)^2 - 2gH\) where \(\tan\alpha = 3\) \((\alpha = 71.565...°)\) | M1 | 3.4 — Complete method for \(H\) only, using larger \(\alpha\), correct no. of terms, condone sin/cos confusion |
| \(H = 36\) or \(36.0\) (m) | A1 | 1.1b — Must be positive; rounded to 2 or 3 sf since \(g = 9.8\) |
| (3 marks total) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| e.g. spin of ball, wind, dimensions/shape of ball, ball modelled as particle, inaccurate value of \(g\), motion in 3D not 2D, \(g\) could be variable. B0 if mass/weight mentioned. B0 for "ground may not be horizontal" | B1 | 3.5b — B0 if any incorrect extras |
| (1 mark total) |
# Question 5:
## Part 5(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $28\cos\alpha \times T = 40$ | M1 | 3.4 — Correct no. of terms, dim correct, condone sin/cos confusion and sign errors |
| $T = \dfrac{10}{7\cos\alpha}$ * | A1* | 1.1b — Allow $\frac{40}{28\cos\alpha} = \frac{10}{7\cos\alpha}$ or $t$ instead of $T$ |
| **(2 marks total)** | | |
## Part 5(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Vertical motion equation in $T$ and $\alpha$ only | M1 | 3.3 — Correct no. of terms, dim correct, condone sin/cos confusion and sign errors |
| $20 = (28\sin\alpha)T - \frac{1}{2}gT^2$ | A1 | 1.1b |
| Eliminate $T$: $20 = (28\sin\alpha)\times\dfrac{10}{7\cos\alpha} - \frac{1}{2}g\left(\dfrac{10}{7\cos\alpha}\right)^2$ | M1 | 1.1b — Eliminate $T$ using (a) or own $T$ to give equation in $\alpha$ only |
| Use $\sec^2\alpha = 1+\tan^2\alpha$: $20 = 40\tan\alpha - \frac{1}{2}g\times\dfrac{100}{49}(1+\tan^2\alpha)$ | M1 | 3.1b |
| $\tan^2\alpha - 4\tan\alpha + 3 = 0$ * (allow $0 = \tan^2\alpha - 4\tan\alpha + 3$) | A1* | 2.2a — Must be $\alpha$ (or $a$) in final answer |
| **(5 marks total)** | | |
## Part 5(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Solve and use $\tan\alpha = 3$ or $\sin\alpha = \dfrac{3}{\sqrt{10}}$ or $\alpha = 71.565...°$ to find equation in $H$ only | M1 | 3.1b — Select larger value of $\tan\alpha$ |
| $0 = (28\sin\alpha)^2 - 2gH$ where $\tan\alpha = 3$ $(\alpha = 71.565...°)$ | M1 | 3.4 — Complete method for $H$ only, using larger $\alpha$, correct no. of terms, condone sin/cos confusion |
| $H = 36$ or $36.0$ (m) | A1 | 1.1b — Must be positive; rounded to 2 or 3 sf since $g = 9.8$ |
| **(3 marks total)** | | |
## Part 5(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| e.g. spin of ball, wind, dimensions/shape of ball, ball modelled as particle, inaccurate value of $g$, motion in 3D not 2D, $g$ could be variable. B0 if mass/weight mentioned. B0 for "ground may not be horizontal" | B1 | 3.5b — B0 if any incorrect extras |
| **(1 mark total)** | | |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f9dc8158-8ed8-4138-9c75-050cf52e6f7e-12_965_1226_244_422}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A small ball is projected with speed $28 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a point $O$ on horizontal ground. After moving for $T$ seconds, the ball passes through the point $A$.
The point $A$ is 40 m horizontally and 20 m vertically from the point $O$, as shown in Figure 2.
The motion of the ball from $O$ to $A$ is modelled as that of a particle moving freely under gravity.
Given that the ball is projected at an angle $\alpha$ to the ground, use the model to
\begin{enumerate}[label=(\alph*)]
\item show that $T = \frac { 10 } { 7 \cos \alpha }$
\item show that $\tan ^ { 2 } \alpha - 4 \tan \alpha + 3 = 0$
\item find the greatest possible height, in metres, of the ball above the ground as the ball moves from $O$ to $A$.
The model does not include air resistance.
\item State one other limitation of the model.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 3 2023 Q5 [11]}}