| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2019 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Two projectiles meeting - 2D flight |
| Difficulty | Challenging +1.2 This is a two-projectile collision problem requiring systematic application of projectile motion equations. While it involves multiple steps (finding velocity components, position matching, solving simultaneous equations), the mathematical techniques are standard for Further Maths mechanics. The collision condition provides clear constraints, making it more structured than problems requiring novel geometric insight. |
| Spec | 3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Horizontal speed \(= 20\cos 30°\) | B1 | 3.4 |
| Vertical velocity at \(t = 2\): \(= 20\sin 30° - 2g\) | M1, A1 | 3.4, 1.1b |
| \(\theta = \tan^{-1}\!\left(\pm\dfrac{9.6}{10\sqrt{3}}\right)\) | M1 | 1.1b |
| Speed \(= \sqrt{100 \times 3 + 9.6^2}\) or e.g. speed \(= \dfrac{9.6}{\sin\theta}\) | M1 | 1.1b |
| \(19.8\) or \(20\ (\text{m s}^{-1})\) at \(29.0°\) or \(29°\) to horizontal | A1 | 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| Using sum of horizontal distances \(= 50\) at \(t = 2\) | M1 | 3.3 |
| \((u\cos\theta) \times 2 + (20\cos 30°) \times 2 = 50 \Rightarrow u\cos\theta = 25 - 20\cos 30°\) | A1 | 1.1b |
| Vertical distances equal | M1 | 3.4 |
| \((20\sin 30°) \times 2 - \dfrac{g}{2} \times 4 = (u\sin\theta) \times 2 - \dfrac{g}{2} \times 4 \Rightarrow 20\sin 30° = u\sin\theta\) | A1 | 1.1b |
| Solving for both \(\theta\) and \(u\) | M1 | 3.1b |
| \(\theta = 52°\) or better \((52.47756849\ldots°)\); \(u = 13\) or better \((12.6085128\ldots)\) | A1 | 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| It does not take account of the fact that they are not particles (moving freely under gravity); does not account for size(s) of balls; spin of balls; wind; \(g\) is not exactly \(9.8\ \text{m s}^{-2}\). N.B. If they refer to mass or weight of balls give B0 | B1 | 3.5b |
# Question 5(a):
Horizontal speed $= 20\cos 30°$ | B1 | 3.4
Vertical velocity at $t = 2$: $= 20\sin 30° - 2g$ | M1, A1 | 3.4, 1.1b
$\theta = \tan^{-1}\!\left(\pm\dfrac{9.6}{10\sqrt{3}}\right)$ | M1 | 1.1b
Speed $= \sqrt{100 \times 3 + 9.6^2}$ or e.g. speed $= \dfrac{9.6}{\sin\theta}$ | M1 | 1.1b
$19.8$ or $20\ (\text{m s}^{-1})$ at $29.0°$ or $29°$ to horizontal | A1 | 2.2a
---
# Question 5(b):
Using sum of horizontal distances $= 50$ at $t = 2$ | M1 | 3.3
$(u\cos\theta) \times 2 + (20\cos 30°) \times 2 = 50 \Rightarrow u\cos\theta = 25 - 20\cos 30°$ | A1 | 1.1b
Vertical distances equal | M1 | 3.4
$(20\sin 30°) \times 2 - \dfrac{g}{2} \times 4 = (u\sin\theta) \times 2 - \dfrac{g}{2} \times 4 \Rightarrow 20\sin 30° = u\sin\theta$ | A1 | 1.1b
Solving for both $\theta$ and $u$ | M1 | 3.1b
$\theta = 52°$ or better $(52.47756849\ldots°)$; $u = 13$ or better $(12.6085128\ldots)$ | A1 | 2.2a
---
# Question 5(c):
It does not take account of the fact that they are not particles (moving freely under gravity); does not account for size(s) of balls; spin of balls; wind; $g$ is not exactly $9.8\ \text{m s}^{-2}$. **N.B.** If they refer to mass or weight of balls give B0 | B1 | 3.5b
**Examiner Notes for Q5:**
- 5a B1: Seen or implied, possibly on diagram
- 5a M1: Use of $v = u + at$ or any complete method using $t = 2$; condone sign errors and sin/cos confusion
- 5a A1: Correct unsimplified equation in $v$ or $v^2$
- 5a M1: Correct use of trig to find relevant angle; must have horizontal and vertical velocity components
- 5a M1: Use Pythagoras or trig to find magnitude; must have both components
- 5a A1: Need magnitude **and** direction stated or implied in diagram (0.506 or 0.51 rads)
- 5b M1: First equation in terms of $u$ and $\theta$ using horizontal motion with $t = 2$; condone sign errors/sin/cos confusion
- 5b A1: Correct unsimplified equation — any equivalent form
- 5b M1: Second equation in terms of $u$ and $\theta$ using vertical motion — equating distances or vertical velocity components; condone sign errors/sin/cos confusion
- 5b A1: Correct unsimplified equation — any equivalent form
- 5b M1: Complete strategy — all necessary equations formed and solve for $u$ and $\theta$; independent mark but can only be earned if 50 m used
- 5b A1: Both values correct (accept 2SF or better since $g$'s cancel); allow radians for $\theta$: 0.92 or better (0.915906…) rads
- 5c B1: Any factor **related to the model** as stated in question; penalise incorrect extras but ignore consequences; e.g. '$AB$ (or ground) is not horizontal' — penalised; 'they do not move in a vertical plane' — penalised
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\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8399dae8-1b9d-4564-a95b-7ab857368b86-14_223_855_239_605}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
The points $A$ and $B$ lie 50 m apart on horizontal ground.\\
At time $t = 0$ two small balls, $P$ and $Q$, are projected in the vertical plane containing $A B$.\\
Ball $P$ is projected from $A$ with speed $20 \mathrm {~ms} ^ { - 1 }$ at $30 ^ { \circ }$ to $A B$.\\
Ball $Q$ is projected from $B$ with speed $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at angle $\theta$ to $B A$, as shown in Figure 3.\\
At time $t = 2$ seconds, $P$ and $Q$ collide.\\
Until they collide, the balls are modelled as particles moving freely under gravity.
\begin{enumerate}[label=(\alph*)]
\item Find the velocity of $P$ at the instant before it collides with $Q$.
\item Find
\begin{enumerate}[label=(\roman*)]
\item the size of angle $\theta$,
\item the value of $u$.
\end{enumerate}\item State one limitation of the model, other than air resistance, that could affect the accuracy of your answers.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 3 2019 Q5 [13]}}