| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2018 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Solving quadratics and applications |
| Type | Quadratic trajectory/projectile model |
| Difficulty | Moderate -0.8 This is a straightforward application of quadratic functions in context. Part (a) requires forming a quadratic from given vertex (20, 12) and root (40, 0), which is routine. Part (b) involves solving a quadratic equation. Part (c) is a standard modelling critique. All steps are textbook-standard with no novel problem-solving required. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02e Complete the square: quadratic polynomials and turning points1.02z Models in context: use functions in modelling |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H = Ax(40-x)\) | M1 | 3.3 — Translates situation into suitable equation |
| \(x=20, H=12 \Rightarrow 12=A(20)(20) \Rightarrow A=\dfrac{3}{100}\) | dM1 | 3.1b — Applies complete strategy with constraints |
| \(H=\dfrac{3}{100}x(40-x)\) or \(H=-\dfrac{3}{100}x(x-40)\) | A1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H=12-\lambda(x-20)^2\) | M1 | 3.3 |
| \(x=40, H=0 \Rightarrow 0=12-\lambda(20)^2 \Rightarrow \lambda=\dfrac{3}{100}\) | dM1 | 3.1b |
| \(H=12-\dfrac{3}{100}(x-20)^2\) | A1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H=ax^2+bx+c\); both \(x=0,H=0 \Rightarrow c=0\); and \(x=40,H=0\Rightarrow 0=1600a+40b\) or \(x=20,H=12\Rightarrow 12=400a+20b\); or \(\dfrac{-b}{2a}=20\) | M1 | 3.3 |
| \(b=-40a \Rightarrow 12=400a+20(-40a) \Rightarrow a=-0.03\), so \(b=1.2\) | dM1 | 3.1b |
| \(H=-0.03x^2+1.2x\) | A1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\{H=3\Rightarrow\}\ 3=\dfrac{3}{100}x(40-x)\Rightarrow x^2-40x+100=0\) or \(3=12-\dfrac{3}{100}(x-20)^2\Rightarrow(x-20)^2=300\) | M1 | 3.4 — Substitutes \(H=3\) and obtains 3TQ or form \((x\pm\alpha)^2=\beta\) |
| \(x=\dfrac{40\pm\sqrt{1600-400}}{2}\) or \(x=20\pm\sqrt{300}\) | dM1 | 1.1b — Correct method solving quadratic |
| Chooses \(20+\sqrt{300}\); greatest distance \(=\) awrt \(37.3\) m | A1 | 3.2a — Interprets solution, selects larger value, states correct units |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Gives a limitation of the model, e.g. ground is horizontal; ball kicked from ground; ball modelled as particle; horizontal bar modelled as a line; no wind/air resistance; no spin; no obstacles; trajectory is a perfect parabola | B1 | 3.5b — Do not accept: \(H\) negative when \(x>40\); bounce after hitting ground; model won't work for different ball; ball may not be kicked same way each time |
# Question 8:
## Part (a) Way 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H = Ax(40-x)$ | M1 | 3.3 — Translates situation into suitable equation |
| $x=20, H=12 \Rightarrow 12=A(20)(20) \Rightarrow A=\dfrac{3}{100}$ | dM1 | 3.1b — Applies complete strategy with constraints |
| $H=\dfrac{3}{100}x(40-x)$ or $H=-\dfrac{3}{100}x(x-40)$ | A1 | 1.1b |
## Part (a) Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H=12-\lambda(x-20)^2$ | M1 | 3.3 |
| $x=40, H=0 \Rightarrow 0=12-\lambda(20)^2 \Rightarrow \lambda=\dfrac{3}{100}$ | dM1 | 3.1b |
| $H=12-\dfrac{3}{100}(x-20)^2$ | A1 | 1.1b |
## Part (a) Way 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H=ax^2+bx+c$; both $x=0,H=0 \Rightarrow c=0$; and $x=40,H=0\Rightarrow 0=1600a+40b$ or $x=20,H=12\Rightarrow 12=400a+20b$; or $\dfrac{-b}{2a}=20$ | M1 | 3.3 |
| $b=-40a \Rightarrow 12=400a+20(-40a) \Rightarrow a=-0.03$, so $b=1.2$ | dM1 | 3.1b |
| $H=-0.03x^2+1.2x$ | A1 | 1.1b |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{H=3\Rightarrow\}\ 3=\dfrac{3}{100}x(40-x)\Rightarrow x^2-40x+100=0$ or $3=12-\dfrac{3}{100}(x-20)^2\Rightarrow(x-20)^2=300$ | M1 | 3.4 — Substitutes $H=3$ and obtains 3TQ or form $(x\pm\alpha)^2=\beta$ |
| $x=\dfrac{40\pm\sqrt{1600-400}}{2}$ or $x=20\pm\sqrt{300}$ | dM1 | 1.1b — Correct method solving quadratic |
| Chooses $20+\sqrt{300}$; greatest distance $=$ awrt $37.3$ m | A1 | 3.2a — Interprets solution, selects larger value, states correct units |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Gives a limitation of the model, e.g. ground is horizontal; ball kicked from ground; ball modelled as particle; horizontal bar modelled as a line; no wind/air resistance; no spin; no obstacles; trajectory is a perfect parabola | B1 | 3.5b — Do not accept: $H$ negative when $x>40$; bounce after hitting ground; model won't work for different ball; ball may not be kicked same way each time |
---8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{580fc9b9-d78c-4a86-91fc-22638cb5186d-20_540_1465_294_301}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 is a graph showing the trajectory of a rugby ball.
The height of the ball above the ground, $H$ metres, has been plotted against the horizontal distance, $x$ metres, measured from the point where the ball was kicked.
The ball travels in a vertical plane.
The ball reaches a maximum height of 12 metres and hits the ground at a point 40 metres from where it was kicked.
\begin{enumerate}[label=(\alph*)]
\item Find a quadratic equation linking $H$ with $x$ that models this situation.
The ball passes over the horizontal bar of a set of rugby posts that is perpendicular to the path of the ball. The bar is 3 metres above the ground.
\item Use your equation to find the greatest horizontal distance of the bar from $O$.
\item Give one limitation of the model.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 2 2018 Q8 [7]}}