| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Session | Specimen |
| 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 applying standard SUVAT equations and projectile formulas across multiple parts, but follows a predictable structure: finding angle from maximum height, then range, then final speed. The calculations are straightforward with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| Vertical motion using \(v^2 = u^2 + 2as\) | M1 | |
| \((40\sin\theta)^2 = 2 \times g \times 12\) | A1 | |
| \((\sin\theta)^2 = \dfrac{2 \times g \times 12}{40^2}\) | ||
| \(\theta = 22.54 = 22.5°\) (accept 23) | A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| Vertical motion \(P \to R\): \(s = ut + \frac{1}{2}at^2\) | ||
| \(-36 = 40\sin\theta t - \dfrac{g}{2}t^2\) | M1 | |
| \(\dfrac{g}{2}t^2 - 40\sin\theta t - 36 = 0\) | A1 A1 | |
| \(t = \dfrac{40\sin 22.54 \pm \sqrt{(40\sin 22.54)^2 + 4 \times 4.9 \times 36}}{9.8}\) | ||
| \(t = 4.694...\) | A1 | |
| Horizontal \(P\) to \(R\): \(s = 40\cos\theta t\) | M1 | |
| \(= 173\) m (or 170 m) | A1 | (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| Using Energy: \(\dfrac{1}{2}mv^2 - \dfrac{1}{2}m \times 40^2 = m \times g \times 36\) | M1 A1 | |
| \(v^2 = 2(9.8 \times 36 + \frac{1}{2} \times 40^2)\) | ||
| \(v = 48.0....\) | ||
| \(v = 48\) m s\(^{-1}\) (accept 48.0) | A1 | (3) [12] |
# Question 7:
## Part (a):
| Working | Marks | Notes |
|---------|-------|-------|
| Vertical motion using $v^2 = u^2 + 2as$ | M1 | |
| $(40\sin\theta)^2 = 2 \times g \times 12$ | A1 | |
| $(\sin\theta)^2 = \dfrac{2 \times g \times 12}{40^2}$ | | |
| $\theta = 22.54 = 22.5°$ (accept 23) | A1 | (3) |
## Part (b):
| Working | Marks | Notes |
|---------|-------|-------|
| Vertical motion $P \to R$: $s = ut + \frac{1}{2}at^2$ | | |
| $-36 = 40\sin\theta t - \dfrac{g}{2}t^2$ | M1 | |
| $\dfrac{g}{2}t^2 - 40\sin\theta t - 36 = 0$ | A1 A1 | |
| $t = \dfrac{40\sin 22.54 \pm \sqrt{(40\sin 22.54)^2 + 4 \times 4.9 \times 36}}{9.8}$ | | |
| $t = 4.694...$ | A1 | |
| Horizontal $P$ to $R$: $s = 40\cos\theta t$ | M1 | |
| $= 173$ m (or 170 m) | A1 | (6) |
## Part (c):
| Working | Marks | Notes |
|---------|-------|-------|
| Using Energy: $\dfrac{1}{2}mv^2 - \dfrac{1}{2}m \times 40^2 = m \times g \times 36$ | M1 A1 | |
| $v^2 = 2(9.8 \times 36 + \frac{1}{2} \times 40^2)$ | | |
| $v = 48.0....$ | | |
| $v = 48$ m s$^{-1}$ (accept 48.0) | A1 | (3) **[12]** |
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{0a4e4cdd-bec4-4059-b9f7-9ce00bc34b71-24_629_1029_251_461}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A ball is projected with speed $40 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a point $P$ on a cliff above horizontal ground. The point O on the ground is vertically below P and OP is 36 m . The ball is projected at an angle $\theta ^ { \circ }$ to the horizontal. The point Q is the highest point of the path of the ball and is 12 m above the level of P. The ball moves freely under gravity and hits the ground at the point R , as shown in Figure 3. Find
\begin{enumerate}[label=(\alph*)]
\item the value of $\theta$,
\item the distance OR ,
\item the speed of the ball as it hits the ground at R.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q7 [12]}}