| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projection from elevated point - angle above horizontal |
| Difficulty | Standard +0.3 This is a standard M2 projectiles question involving projection from an elevated point. It requires routine application of SUVAT equations in vertical motion (parts a,b) and combining projectile motion with constant speed motion (part c). The given sin α = 7/8 simplifies calculations. While multi-part, each step follows standard procedures without requiring novel insight, making it slightly easier than average. |
| Spec | 3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| (a) at max. ht., \(v_y = 0\) \(\therefore\) \(0 = (22\sin\alpha)^2 - 2gs\) | M1 A1 | |
| \(s_y = \dfrac{(22\frac{7}{8})^2}{2g} = 18.91\) | M1 | |
| starts 1.6 m above \(P\) so max. ht. above ground \(= 20.5\) m (3sf) | A1 | |
| (b) \(s_y = -1.4\) \(\therefore\) \(ut\sin\alpha - \frac{1}{2}gt^2 = -1.4\) | ||
| \(\frac{77}{4}t - 4.9t^2 = -1.4\) | M1 A1 | |
| \(14t^2 - 55t - 4 = 0\) \(\therefore\) \((14t+1)(t-4) = 0\) | M1 | |
| \(t = 4\) in this case \(\therefore\) ball in flight for 4 seconds | A1 | |
| (c) \(s_x = ut\cos\alpha = 22 \times 4 \times \frac{\sqrt{15}}{8} = 11\sqrt{15} = 42.6\) | M1 A1 | |
| max. dist. fielder can run is \(4 \times 6 = 24\) m | A1 | |
| max. initial dist. between fielder and ball \(= 42.6 + 24 = 66.6\) m (3sf) | A1 | (12) |
## Question 5:
| Answer/Working | Marks | Notes |
|---|---|---|
| **(a)** at max. ht., $v_y = 0$ $\therefore$ $0 = (22\sin\alpha)^2 - 2gs$ | M1 A1 | |
| $s_y = \dfrac{(22\frac{7}{8})^2}{2g} = 18.91$ | M1 | |
| starts 1.6 m above $P$ so max. ht. above ground $= 20.5$ m (3sf) | A1 | |
| **(b)** $s_y = -1.4$ $\therefore$ $ut\sin\alpha - \frac{1}{2}gt^2 = -1.4$ | | |
| $\frac{77}{4}t - 4.9t^2 = -1.4$ | M1 A1 | |
| $14t^2 - 55t - 4 = 0$ $\therefore$ $(14t+1)(t-4) = 0$ | M1 | |
| $t = 4$ in this case $\therefore$ ball in flight for 4 seconds | A1 | |
| **(c)** $s_x = ut\cos\alpha = 22 \times 4 \times \frac{\sqrt{15}}{8} = 11\sqrt{15} = 42.6$ | M1 A1 | |
| max. dist. fielder can run is $4 \times 6 = 24$ m | A1 | |
| max. initial dist. between fielder and ball $= 42.6 + 24 = 66.6$ m (3sf) | A1 | **(12)** |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{0ea2267e-6c46-4a4f-9a38-c242de57901d-3_405_718_1169_555}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
During a cricket match, a batsman hits the ball giving it an initial velocity of $22 \mathrm {~ms} ^ { - 1 }$ at an angle $\alpha$ to the horizontal where $\sin \alpha = \frac { 7 } { 8 }$. When the batsman strikes the ball it is 1.6 metres above the ground, as shown in Figure 2, and it subsequently moves freely under gravity.
\begin{enumerate}[label=(\alph*)]
\item Find, correct to 3 significant figures, the maximum height above the ground reached by the ball.
The ball is caught by a fielder when it is 0.2 metres above the ground.
\item Find the length of time for which the ball is in the air.
Assuming that the fielder who caught the ball ran at a constant speed of $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$,
\item find, correct to 3 significant figures, the maximum distance that the fielder could have been from the ball when it was struck.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q5 [12]}}