| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Deriving trajectory equation |
| Difficulty | Standard +0.3 This is a standard M2 projectile motion question with straightforward application of SUVAT equations and trajectory derivation. Parts (a) and (d) are 'show that' questions with given answers, reducing problem-solving demand. The trigonometry is simple (3-4-5 triangle scaled), and all parts follow a predictable sequence. While it requires multiple steps and careful algebra, it involves no novel insight or challenging problem-solving—just methodical application of standard projectile formulae. |
| Spec | 1.02n Sketch curves: simple equations including polynomials3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model3.03g Gravitational acceleration |
| Answer | Marks |
|---|---|
| (a) \(y = (52\sin\theta)t - \frac{1}{2}gt^2 = 20t - 5t^2\) | M1 A1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Ball is coming down, so \(t = 3\) | M1 A1 A1 | |
| (c) \(x = (52\cos\theta)t = 52 \times \frac{12}{13}t = 48t\). When \(t = 3\), \(x = 144 \text{ m}\) | M1 A1 A1 | |
| (d) \(y = 20x \times \frac{1}{48} - 5x \times \left(\frac{1}{48}\right)^2 = \frac{5}{12}x - \frac{5}{2304}x^2\) | M1 M1 A1 | |
| (e) Have ignored air resistance, which would make answer larger | B1 B1 | Total: 15 marks |
(a) $y = (52\sin\theta)t - \frac{1}{2}gt^2 = 20t - 5t^2$ | M1 A1 A1 |
(b) Lands when $y = 15$: $t^2 - 4t + 3 = 0$, $(t-1)(t-3) = 0$
Ball is coming down, so $t = 3$ | M1 A1 A1 |
(c) $x = (52\cos\theta)t = 52 \times \frac{12}{13}t = 48t$. When $t = 3$, $x = 144 \text{ m}$ | M1 A1 A1 |
(d) $y = 20x \times \frac{1}{48} - 5x \times \left(\frac{1}{48}\right)^2 = \frac{5}{12}x - \frac{5}{2304}x^2$ | M1 M1 A1 |
(e) Have ignored air resistance, which would make answer larger | B1 B1 | Total: 15 marksTake $g = 10$ ms$^{-2}$ in this question.
\includegraphics{figure_6}
A golfer hits a ball from a point $T$ at an angle $\theta$ to the horizontal, where $\sin \theta = \frac{5}{13}$, giving it an initial speed of 52 ms$^{-1}$. The ball lands on top of a mound, 15 m above the level of $T$, as shown.
\begin{enumerate}[label=(\alph*)]
\item Show that the height, $y$ m, of the ball above $T$ at time $t$ seconds after it was hit is given by
$$y = 20t - 5t^2.$$ [3 marks]
\item Find the time for which the ball is in flight. [4 marks]
\item Find the horizontal distance travelled by the ball. [3 marks]
\item Show that, if the ball is $x$ m horizontally from $T$ at time $t$ seconds, then
$$y = \frac{5}{12}x - \frac{5}{2304}x^2.$$ [3 marks]
\item Name a force that has been ignored in your mathematical model and state whether the answer to part (b) would be larger or smaller if this force were taken into account. [2 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q6 [15]}}