| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Basic trajectory calculations |
| Difficulty | Moderate -0.3 This is a standard M2 projectiles question requiring routine application of SUVAT equations and trajectory formulas. Part (a) is straightforward modelling discussion, parts (b) and (c) involve standard range and velocity calculations with given initial conditions. The calculations are multi-step but follow well-practiced procedures with no novel insight required, making it slightly easier than average. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Particle moving freely under gravity | B2 | |
| (b) Vert. disp. \(= 0\) \(\therefore u\sin\alpha - \frac{1}{2}gt) = 0\) | M1 | |
| \(t = 0\) at O, we require \(49\sin 30° - 4.9t = 0\) \(\therefore t = 5\) | M1 A1 | |
| Horiz. disp. \(= u\cos\alpha = 49(5)\cos 30° = 212.17\) i.e. \(212 - 170\) beyond hole \(= 42.2 \text{ m}\) (3sf) | M1 A1 | |
| (c) When horiz. disp. \(= 170\), \(u\cos\alpha = 170\) \(\therefore t = 4.006\) | M1 | |
| Horiz. vel. \(= u\cos\alpha = 42.44\) Vert. vel. \(= u\sin\alpha - gt = -14.76\) | A2 | |
| Mag. of vel. \(= \sqrt{(42.44)^2 + (-14.76)^2} = 44.9 \text{ ms}^{-1}\) (3sf) | M1 A1 | |
| Req'd angle \(= \tan^{-1}\frac{14.76}{42.44} = 19.2°\) below horizontal (3sf) | M1 A1 | (15 marks) |
**(a)** Particle moving freely under gravity | B2 |
**(b)** Vert. disp. $= 0$ $\therefore u\sin\alpha - \frac{1}{2}gt) = 0$ | M1 |
$t = 0$ at O, we require $49\sin 30° - 4.9t = 0$ $\therefore t = 5$ | M1 A1 |
Horiz. disp. $= u\cos\alpha = 49(5)\cos 30° = 212.17$ i.e. $212 - 170$ beyond hole $= 42.2 \text{ m}$ (3sf) | M1 A1 |
**(c)** When horiz. disp. $= 170$, $u\cos\alpha = 170$ $\therefore t = 4.006$ | M1 |
Horiz. vel. $= u\cos\alpha = 42.44$ Vert. vel. $= u\sin\alpha - gt = -14.76$ | A2 |
Mag. of vel. $= \sqrt{(42.44)^2 + (-14.76)^2} = 44.9 \text{ ms}^{-1}$ (3sf) | M1 A1 |
Req'd angle $= \tan^{-1}\frac{14.76}{42.44} = 19.2°$ below horizontal (3sf) | M1 A1 | (15 marks)
**Total: (75 marks)**7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f0e751be-f095-4a56-8ee9-8433cc4873e9-4_236_942_1101_479}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}
Figure 3 shows the path of a golf ball which is hit from the point $O$ with speed $49 \mathrm {~ms} ^ { - 1 }$ at an angle of $30 ^ { \circ }$ to the horizontal. The path of the ball is in a vertical plane containing $O$ and the hole at which the ball is aimed.
The hole is 170 m from $O$ and on the same horizontal level as $O$.
\begin{enumerate}[label=(\alph*)]
\item Suggest a suitable model for the motion of the golf ball.
Find, correct to 3 significant figures,
\item the distance beyond the hole at which the ball hits the ground,
\item the magnitude and direction of the velocity of the ball when it is directly above the hole.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q7 [15]}}