| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2007 |
| Session | June |
| 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 with straightforward application of SUVAT equations and energy conservation. Part (a) uses standard maximum height formula, part (b) requires solving a quadratic from trajectory equations, and part (c) is direct energy conservation or velocity components. All techniques are routine for M2 students 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 model6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(0 = (35\sin\alpha)^2 - 2gh\) | M1 A1 | Use of \(v^2 = u^2 + 2as\), or possibly a 2-stage method. A1: correct expression; alternatives need complete method leading to equation in \(h\) only |
| \(h = 40\) m | A1 | 40(m); no more than 2sf due to use of \(g\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x = 168 \Rightarrow 168 = 35\cos\alpha \cdot t \Rightarrow t = 8\) s | M1 A1 | Use of \(x = u\cos\alpha \cdot t\) to find \(t\). A1: \(168 = 35 \times their\cos\alpha \times t\) |
| At \(t=8\): \(y = 35\sin\alpha \times t - \frac{1}{2}gt^2 = 28.8 - \frac{1}{2}g \cdot 8^2 = -89.6\) m | M1 A1 | Use of \(s = ut + \frac{1}{2}at^2\) to find vertical distance for their \(t\). A1: \(y = 35\sin\alpha \times t - \frac{1}{2}gt^2\) (\(u\), \(t\) consistent) |
| Hence height of \(A = 89.6\) m or 90 m | DM1 A1 | DM1: dependent on previous 2 M marks; complete method for AB; eliminate \(t\) and solve for \(s\). A1 cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{2}mv^2 = \frac{1}{2}m\cdot 35^2 + mg\cdot 89.6\) | M1 A1 | Conservation of energy: change in KE = change in GPE; all terms present; one side correct (follow their \(h\)) |
| \(\Rightarrow v = 54.6\) or \(55\) m s\(^{-1}\) | A1 | 54.6 or 55 (m/s). OR: M1 horizontal and vertical components found and combined using Pythagoras (\(v_x = 21\), \(v_y = 28 - 9.8\times8\)); A1 \(v_x\) and \(v_y\) expressions correct; A1 54.6 or 55. NB: penalty for inappropriate rounding after use of \(g\) only applies once per question |
## Question 6:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0 = (35\sin\alpha)^2 - 2gh$ | M1 A1 | Use of $v^2 = u^2 + 2as$, or possibly a 2-stage method. A1: correct expression; alternatives need complete method leading to equation in $h$ only |
| $h = 40$ m | A1 | 40(m); no more than 2sf due to use of $g$ |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = 168 \Rightarrow 168 = 35\cos\alpha \cdot t \Rightarrow t = 8$ s | M1 A1 | Use of $x = u\cos\alpha \cdot t$ to find $t$. A1: $168 = 35 \times their\cos\alpha \times t$ |
| At $t=8$: $y = 35\sin\alpha \times t - \frac{1}{2}gt^2 = 28.8 - \frac{1}{2}g \cdot 8^2 = -89.6$ m | M1 A1 | Use of $s = ut + \frac{1}{2}at^2$ to find vertical distance for their $t$. A1: $y = 35\sin\alpha \times t - \frac{1}{2}gt^2$ ($u$, $t$ consistent) |
| Hence height of $A = 89.6$ m or 90 m | DM1 A1 | DM1: dependent on previous 2 M marks; complete method for AB; eliminate $t$ and solve for $s$. A1 cso |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}mv^2 = \frac{1}{2}m\cdot 35^2 + mg\cdot 89.6$ | M1 A1 | Conservation of energy: change in KE = change in GPE; all terms present; one side correct (follow their $h$) |
| $\Rightarrow v = 54.6$ or $55$ m s$^{-1}$ | A1 | 54.6 or 55 (m/s). OR: M1 horizontal and vertical components found and combined using Pythagoras ($v_x = 21$, $v_y = 28 - 9.8\times8$); A1 $v_x$ and $v_y$ expressions correct; A1 54.6 or 55. NB: penalty for inappropriate rounding after use of $g$ only applies once per question |6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{778a0276-6738-40e6-90b2-a536ce5abe6a-10_447_908_205_516}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
A golf ball $P$ is projected with speed $35 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a point $A$ on a cliff above horizontal ground. The angle of projection is $\alpha$ to the horizontal, where $\tan \alpha = \frac { 4 } { 3 }$. The ball moves freely under gravity and hits the ground at the point $B$, as shown in Figure 4.
\begin{enumerate}[label=(\alph*)]
\item Find the greatest height of $P$ above the level of $A$.
The horizontal distance from $A$ to $B$ is 168 m .
\item Find the height of $A$ above the ground.
By considering energy, or otherwise,
\item find the speed of $P$ as it hits the ground at $B$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2007 Q6 [12]}}