| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2018 |
| Session | January |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Horizontal projection from height |
| Difficulty | Standard +0.3 This is a standard M2 projectiles question with routine derivations. Part (a) is a bookwork 'show that' using standard SUVAT equations. Part (b) requires simple application of Pythagoras to velocity components. Part (c) connects the two scenarios using given results. All steps are methodical applications of standard techniques 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 |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Horizontal distance in terms of \(U\), \(t\) and \(\alpha\) | M1 | |
| \(x = Ut\cos\alpha\) | A1 | Correct unsimplified equation |
| Vertical distance in terms of \(U\), \(t\) and \(\alpha\) | M1 | Condone sign error |
| \(y = Ut\sin\alpha - \frac{1}{2}gt^2\) | A1 | Correct unsimplified equation |
| \(y = U\sin\alpha \cdot \frac{x}{U\cos\alpha} - \frac{1}{2}g\left(\frac{x}{U\cos\alpha}\right)^2\) | DM1 | Substitute for \(t\). Dependent on the first 2 M marks |
| \(y = x\tan\alpha - \frac{gx^2\sec^2\alpha}{2U^2}\) | DM1 | Simplify the trig. and use Pythagoras. Dependent on the first 2 M marks |
| \(y = x\tan\alpha - \frac{gx^2(1+\tan^2\alpha)}{2U^2}\) given answer | A1 | Obtain given answer from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((\rightarrow)\ v_H = U\) | B1 | Horizontal component in \(U\), \(g\), \(T\) |
| \((\downarrow)\ v_V = gT\) | B1 | Vertical component in \(U\), \(g\), \(T\). Accept \(\pm\) |
| Use of Pythagoras | M1 | |
| \(v = \sqrt{U^2 + g^2T^2}\) | A1 | Or equivalent. Allow \(t\) for \(T\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(-h = d\tan 0 - \frac{gd^2}{2U^2}(1+\tan^2 0)\) | B1 | \(\left(h = \frac{gd^2}{2U^2}\right)\) |
| \(d = UT \Rightarrow h = \frac{gT^2}{2}\) | B1 | |
| \(\frac{1}{2}mv^2 - \frac{1}{2}mU^2 = mgh\) | M1 | Energy equation |
| \(v^2 = U^2 + 2gh = U^2 + g^2T^2,\ v = \sqrt{U^2 + g^2T^2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(d = UT\) | B1 | Horizontal distance |
| \(-h = d\tan\alpha - \frac{gd^2(1+\tan^2\alpha)}{2U^2}\) | M1 | Substitute for \(x\) and \(y\) in given equation. Condone sign error |
| \(h = \frac{1}{2}gT^2\) | B1 | Vertical distance |
| \(-\frac{1}{2}gT^2 = d\tan\alpha - \frac{g(UT)^2(1+\tan^2\alpha)}{2U^2}\) | M1 | Substitute to eliminate \(U\) from the equation |
| \(0 = d\tan\alpha - \frac{gT^2}{2}\tan^2\alpha\) | A1 | Correct equation in \(T\) and \(d\) |
| \(d = \frac{1}{2}gT^2\tan\alpha\) given answer | A1 | Obtain given answer from correct working |
## Question 7(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Horizontal distance in terms of $U$, $t$ and $\alpha$ | M1 | |
| $x = Ut\cos\alpha$ | A1 | Correct unsimplified equation |
| Vertical distance in terms of $U$, $t$ and $\alpha$ | M1 | Condone sign error |
| $y = Ut\sin\alpha - \frac{1}{2}gt^2$ | A1 | Correct unsimplified equation |
| $y = U\sin\alpha \cdot \frac{x}{U\cos\alpha} - \frac{1}{2}g\left(\frac{x}{U\cos\alpha}\right)^2$ | DM1 | Substitute for $t$. Dependent on the first 2 M marks |
| $y = x\tan\alpha - \frac{gx^2\sec^2\alpha}{2U^2}$ | DM1 | Simplify the trig. and use Pythagoras. Dependent on the first 2 M marks |
| $y = x\tan\alpha - \frac{gx^2(1+\tan^2\alpha)}{2U^2}$ **given answer** | A1 | Obtain given answer from correct working |
## Question 7(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(\rightarrow)\ v_H = U$ | B1 | Horizontal component in $U$, $g$, $T$ |
| $(\downarrow)\ v_V = gT$ | B1 | Vertical component in $U$, $g$, $T$. Accept $\pm$ |
| Use of Pythagoras | M1 | |
| $v = \sqrt{U^2 + g^2T^2}$ | A1 | Or equivalent. Allow $t$ for $T$ |
## Question 7(b) alt:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $-h = d\tan 0 - \frac{gd^2}{2U^2}(1+\tan^2 0)$ | B1 | $\left(h = \frac{gd^2}{2U^2}\right)$ |
| $d = UT \Rightarrow h = \frac{gT^2}{2}$ | B1 | |
| $\frac{1}{2}mv^2 - \frac{1}{2}mU^2 = mgh$ | M1 | Energy equation |
| $v^2 = U^2 + 2gh = U^2 + g^2T^2,\ v = \sqrt{U^2 + g^2T^2}$ | A1 | |
## Question 7(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $d = UT$ | B1 | Horizontal distance |
| $-h = d\tan\alpha - \frac{gd^2(1+\tan^2\alpha)}{2U^2}$ | M1 | Substitute for $x$ and $y$ in given equation. Condone sign error |
| $h = \frac{1}{2}gT^2$ | B1 | Vertical distance |
| $-\frac{1}{2}gT^2 = d\tan\alpha - \frac{g(UT)^2(1+\tan^2\alpha)}{2U^2}$ | M1 | Substitute to eliminate $U$ from the equation |
| $0 = d\tan\alpha - \frac{gT^2}{2}\tan^2\alpha$ | A1 | Correct equation in $T$ and $d$ |
| $d = \frac{1}{2}gT^2\tan\alpha$ **given answer** | A1 | Obtain given answer from correct working |7. A particle is projected from a point $O$ with speed $U$ at an angle of elevation $\alpha$ to the horizontal and moves freely under gravity. When the particle has moved a horizontal distance $x$, its height above $O$ is $y$.
\begin{enumerate}[label=(\alph*)]
\item Show that
$$y = x \tan \alpha - \frac { g x ^ { 2 } \left( 1 + \tan ^ { 2 } \alpha \right) } { 2 U ^ { 2 } }$$
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{54112b4a-3727-4e5b-97e5-4291e7172438-22_330_857_632_548}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A small stone is projected horizontally with speed $U$ from a point $C$ at the top of a vertical cliff $A C$ so as to hit a fixed target $B$ on the horizontal ground. The point $C$ is a height $h$ above the ground, as shown in Figure 3. The time of flight of the stone from $C$ to $B$ is $T$, and the stone is modelled as a particle moving freely under gravity.
\item Find, in terms of $U , g$ and $T$, the speed of the stone as it hits the target at $B$.
It is found that, using the same initial speed $U$, the target can also be hit by projecting the stone from $C$ at an angle $\alpha$ above the horizontal. The stone is again modelled as a particle moving freely under gravity and the distance $A B = d$.
\item Using the result in part (a), or otherwise, show that
$$d = \frac { 1 } { 2 } g T ^ { 2 } \tan \alpha$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2018 Q7 [17]}}