| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | January |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Vector form projectile motion |
| Difficulty | Standard +0.3 This is a standard M2 projectiles question requiring derivation of trajectory equation (bookwork), finding range and max height using standard methods, and applying perpendicular velocity condition. All techniques are routine for M2 students with no novel insight required, making it slightly easier than average. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10h Vectors in kinematics: uniform acceleration in vector form3.02f Non-uniform acceleration: using differentiation and integration3.02i Projectile motion: constant acceleration model |
| Answer | Marks |
|---|---|
| \(x = ut\) | B1 |
| \(y = cut - 4.9t^2\) | M1 A1 |
| eliminating \(t\) and simplifying to give \(y = cx - \frac{4.9x^2}{u^2}\) \*\* | DM1 A1 (5) |
| Answer | Marks |
|---|---|
| \(0 = cx - \frac{4.9x^2}{u^2}\) | M1 |
| \(0 = x(c - \frac{4.9x}{u^2}) \Rightarrow R = \frac{u^2c}{4.9} = 10c\) | M1 A1 |
| Answer | Marks |
|---|---|
| When \(x = 5c\), \(y = H\) | M1 |
| \(= 5c^2 - \frac{(5c)^2}{10} = 2.5c^2\) | M1 A1 (6) |
| Answer | Marks |
|---|---|
| \(\frac{dy}{dx} = c - \frac{9.8x}{u^2} = c - \frac{x}{5}\) | M1 A1 |
| When \(x = 0\), \(\frac{dy}{dx} = c\) | B1 |
| So, \(c - \frac{x}{5} = -1 \Rightarrow c\) | DM1 A1 |
| \(x = 5(c + \frac{1}{c})\) | A1 (6) |
| Answer | Marks |
|---|---|
| \(\tan\theta = \frac{u}{cu} = \frac{1}{c} = \frac{v}{u}\) | B1 |
| \(\Rightarrow v = \frac{u}{c} = \frac{7}{c}\) | M1 A1 |
| \(v = u + at \; ; -\frac{7}{c} = 7c - 9.8t\) | M1 |
| \(t = \frac{7}{9.8}(c + \frac{1}{c})\) | A1 |
| \(x = ut = 7t \; ; \quad x = 5(c + \frac{1}{c})\) | A1 |
## (a)
$x = ut$ | B1 |
$y = cut - 4.9t^2$ | M1 A1 |
eliminating $t$ and simplifying to give $y = cx - \frac{4.9x^2}{u^2}$ **\*\*** | DM1 A1 (5) |
## (b)(i)
$0 = cx - \frac{4.9x^2}{u^2}$ | M1 |
$0 = x(c - \frac{4.9x}{u^2}) \Rightarrow R = \frac{u^2c}{4.9} = 10c$ | M1 A1 |
## (b)(ii)
When $x = 5c$, $y = H$ | M1 |
$= 5c^2 - \frac{(5c)^2}{10} = 2.5c^2$ | M1 A1 (6) |
## (c)
$\frac{dy}{dx} = c - \frac{9.8x}{u^2} = c - \frac{x}{5}$ | M1 A1 |
When $x = 0$, $\frac{dy}{dx} = c$ | B1 |
So, $c - \frac{x}{5} = -1 \Rightarrow c$ | DM1 A1 |
$x = 5(c + \frac{1}{c})$ | A1 (6) |
**Total: [17]**
### Alternative to 8(c)
$\tan\theta = \frac{u}{cu} = \frac{1}{c} = \frac{v}{u}$ | B1 |
$\Rightarrow v = \frac{u}{c} = \frac{7}{c}$ | M1 A1 |
$v = u + at \; ; -\frac{7}{c} = 7c - 9.8t$ | M1 |
$t = \frac{7}{9.8}(c + \frac{1}{c})$ | A1 |
$x = ut = 7t \; ; \quad x = 5(c + \frac{1}{c})$ | A1 |[In this question $\mathbf{i}$ and $\mathbf{j}$ are unit vectors in a horizontal and upward vertical direction respectively]
A particle $P$ is projected from a fixed point $O$ on horizontal ground with velocity $u(\mathbf{i} + c\mathbf{j}) \text{ ms}^{-1}$, where $c$ and $u$ are positive constants. The particle moves freely under gravity until it strikes the ground at $A$, where it immediately comes to rest. Relative to $O$, the position vector of a point on the path of $P$ is $(x\mathbf{i} + y\mathbf{j})$ m.
\begin{enumerate}[label=(\alph*)]
\item Show that
$$y = cx - \frac{4.9x^2}{u^2}.$$
[5]
\end{enumerate}
Given that $u = 7$, $OA = R$ m and the maximum vertical height of $P$ above the ground is $H$ m,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item using the result in part (a), or otherwise, find, in terms of $c$,
\begin{enumerate}[label=(\roman*)]
\item $R$
\item $H$.
\end{enumerate}
[6]
\end{enumerate}
Given also that when $P$ is at the point $Q$, the velocity of $P$ is at right angles to its initial velocity,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item find, in terms of $c$, the value of $x$ at $Q$.
[6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2010 Q8 [17]}}