| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2021 |
| Session | October |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Solving quadratics and applications |
| Type | Finding quadratic constants from real-world trajectory or context |
| Difficulty | Moderate -0.3 This is a straightforward quadratic modelling question requiring students to form a quadratic from three given conditions (using vertex form is natural here), then find the maximum and a root. The context is accessible, the algebra is routine, and part (c) requires only a simple written response about model limitations. Slightly easier than average due to the clear structure and standard techniques involved. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H = ax^2 + bx + c\) and \(x=0,\ H=3 \Rightarrow H = ax^2 + bx + 3\) | M1 | Translates problem into suitable model; uses \(H=3\) when \(x=0\) to establish \(c=3\); condone \(a=\pm1\) |
| \(H = ax^2 + bx + 3\) and \(x=120,\ H=27 \Rightarrow 27 = 14400a + 120b + 3\) | M1 | Correct attempt using one of the two other pieces of information within a quadratic model |
| \(\frac{dH}{dx} = 2ax + b = 0\) when \(x=90 \Rightarrow 180a + b = 0\) | A1 | At least one correct equation connecting \(a\) and \(b\) |
| Both equations used simultaneously \(\Rightarrow a = ...,\ b = ...\) | dM1 | Fully correct strategy using \(H = ax^2 + bx + 3\) with both other pieces of information to find both \(a\) and \(b\) |
| \(H = -\frac{1}{300}x^2 + \frac{3}{5}x + 3\) | A1 | Correct equation; award if seen in part (b) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x=90 \Rightarrow H\left(-\frac{1}{300}(90)^2 + \frac{3}{5}(90) + 3\right) = 30\text{ m}\) | B1 | Correct height including units; CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H = 0 \Rightarrow -\frac{1}{300}x^2 + \frac{3}{5}x + 3 = 0 \Rightarrow x = ...\) | M1 | Uses \(H=0\) and attempts to solve for \(x\); usual rules for quadratics |
| \(x = (-4.868...)\ 184.868...\) \(\Rightarrow x = 185\text{ (m)}\) | A1 | Discards negative solution; identifies awrt 185 m; condone lack of units |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| E.g. The ground is unlikely to be horizontal; the ball is not a particle so has dimensions/size; the ball is unlikely to travel in a vertical plane (as it will spin); \(H\) is not likely to be a quadratic function in \(x\) | B1 | Must focus on why model may not be appropriate or give values/situations where model breaks down; do not accept answers referring to situation after ball hits ground; do not accept single word vague answers |
# Question 12:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H = ax^2 + bx + c$ and $x=0,\ H=3 \Rightarrow H = ax^2 + bx + 3$ | M1 | Translates problem into suitable model; uses $H=3$ when $x=0$ to establish $c=3$; condone $a=\pm1$ |
| $H = ax^2 + bx + 3$ and $x=120,\ H=27 \Rightarrow 27 = 14400a + 120b + 3$ | M1 | Correct attempt using one of the two other pieces of information within a quadratic model |
| $\frac{dH}{dx} = 2ax + b = 0$ when $x=90 \Rightarrow 180a + b = 0$ | A1 | At least one correct equation connecting $a$ and $b$ |
| Both equations used simultaneously $\Rightarrow a = ...,\ b = ...$ | dM1 | Fully correct strategy using $H = ax^2 + bx + 3$ with both other pieces of information to find both $a$ and $b$ |
| $H = -\frac{1}{300}x^2 + \frac{3}{5}x + 3$ | A1 | Correct equation; award if seen in part (b) |
## Part (b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=90 \Rightarrow H\left(-\frac{1}{300}(90)^2 + \frac{3}{5}(90) + 3\right) = 30\text{ m}$ | B1 | Correct height including units; CAO |
## Part (b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H = 0 \Rightarrow -\frac{1}{300}x^2 + \frac{3}{5}x + 3 = 0 \Rightarrow x = ...$ | M1 | Uses $H=0$ and attempts to solve for $x$; usual rules for quadratics |
| $x = (-4.868...)\ 184.868...$ $\Rightarrow x = 185\text{ (m)}$ | A1 | Discards negative solution; identifies awrt 185 m; condone lack of units |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| E.g. The ground is unlikely to be horizontal; the ball is not a particle so has dimensions/size; the ball is unlikely to travel in a vertical plane (as it will spin); $H$ is not likely to be a quadratic function in $x$ | B1 | Must focus on why model may not be appropriate or give values/situations where model breaks down; do not accept answers referring to situation after ball hits ground; do not accept single word vague answers |
---12.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{08ede5ea-85e9-44eb-be6a-5878096734e2-38_666_1189_244_440}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 is a graph of the trajectory of a golf ball after the ball has been hit until it first hits the ground.
The vertical height, $H$ metres, of the ball above the ground has been plotted against the horizontal distance travelled, $x$ metres, measured from where the ball was hit.
The ball is modelled as a particle travelling in a vertical plane above horizontal ground.\\
Given that the ball
\begin{itemize}
\item is hit from a point on the top of a platform of vertical height 3 m above the ground
\item reaches its maximum vertical height after travelling a horizontal distance of 90 m
\item is at a vertical height of 27 m above the ground after travelling a horizontal distance of 120 m
\end{itemize}
Given also that $H$ is modelled as a quadratic function in $x$
\begin{enumerate}[label=(\alph*)]
\item find $H$ in terms of $x$
\item Hence find, according to the model,
\begin{enumerate}[label=(\roman*)]
\item the maximum vertical height of the ball above the ground,
\item the horizontal distance travelled by the ball, from when it was hit to when it first hits the ground, giving your answer to the nearest metre.
\end{enumerate}\item The possible effects of wind or air resistance are two limitations of the model. Give one other limitation of this model.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 1 2021 Q12 [9]}}