| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2001 |
| Session | June |
| Marks | 16 |
| 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 combined with basic kinematics. Part (a) uses standard SUVAT equations with given angle/speed. Parts (b)-(d) involve straightforward integration of acceleration to find velocity and position, then comparing distances. All techniques are routine for M2 with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02f Non-uniform acceleration: using differentiation and integration3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(U_y = 23.75 \sin \alpha (= 19)\) | B1 | |
| Complete method to find time, e.g. \(-2.4 = 23.75 \sin \alpha \cdot t - \frac{1}{2}gt^2\) | M1A1 | |
| Solving to find t; t = 4 | M1A1 | (5) |
| (b) \(\frac{dv}{dt} = -\frac{1}{4}t^2\) | M1A1 | |
| \(\Rightarrow v = -\frac{1}{12}t^3 + c\) | M1A1 | |
| \(t = 0, v = 18\) \(\Rightarrow\) \(v = 18 - \frac{1}{12}t^3\) | A1 | (3) |
| (c) Putting \(v = 0\) expression in (b) | M1 | |
| Solving equation [dependent on previous M1 and M1 in (b)] | M1 | |
| Finding \(T = 6\), with no wrong working seen [allow verification] | A1 cso | (3) |
| (d) Distance \(\rightarrow\) travelled by package = \(23.75 \cos \alpha \times 4c_s = 57\) m | M1A1√ | |
| [\(\sqrt{}\) only on 14.25 × 4\(c_s\)] | ||
| For lorry \(s = 18t - \frac{1}{48}t^4\) | M1;A1√ | |
| Showing \(s = 66\frac{2}{3}\) for lorry, and distance them between is just under 10m | A1 cso | (5) |
| [If lorry moving in direction CA, allow final answer of just under 124m] |
**(a)** $U_y = 23.75 \sin \alpha (= 19)$ | B1 |
Complete method to find time, e.g. $-2.4 = 23.75 \sin \alpha \cdot t - \frac{1}{2}gt^2$ | M1A1 |
Solving to find t; **t = 4** | M1A1 | (5)
**(b)** $\frac{dv}{dt} = -\frac{1}{4}t^2$ | M1A1 |
$\Rightarrow v = -\frac{1}{12}t^3 + c$ | M1A1 |
$t = 0, v = 18$ $\Rightarrow$ $v = 18 - \frac{1}{12}t^3$ | A1 | (3)
**(c)** Putting $v = 0$ expression in (b) | M1 |
Solving equation [dependent on previous M1 and M1 in (b)] | M1 |
Finding $T = 6$, with no wrong working seen [allow verification] | A1 cso | (3)
**(d)** Distance $\rightarrow$ travelled by package = $23.75 \cos \alpha \times 4c_s = 57$ m | M1A1√ |
[$\sqrt{}$ only on 14.25 × 4$c_s$] | |
For lorry $s = 18t - \frac{1}{48}t^4$ | M1;A1√ |
Showing $s = 66\frac{2}{3}$ for lorry, and distance them between is just under 10m | A1 cso | (5)
[If lorry moving in direction CA, allow final answer of just under 124m] | |\includegraphics{figure_2}
At time $t = 0$ a small package is projected from a point $B$ which is 2.4 m above a point $A$ on horizontal ground. The package is projected with speed 23.75 m s$^{-1}$ at an angle $\alpha$ to the horizontal, where $\tan \alpha = \frac{4}{3}$. The package strikes the ground at the point $C$, as shown in Fig. 2. The package is modelled as a particle moving freely under gravity.
\begin{enumerate}[label=(\alph*)]
\item Find the time taken for the package to reach $C$.
[5]
\end{enumerate}
A lorry moves along the line $AC$, approaching $A$ with constant speed 18 m s$^{-1}$. At time $t = 0$ the rear of the lorry passes $A$ and the lorry starts to slow down. It comes to rest $T$ seconds later. The acceleration, $a$ m s$^{-2}$ of the lorry at time $t$ seconds is given by
$$a = -\frac{1}{4}t^2, \quad 0 \leq t \leq T.$$
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the speed of the lorry at time $t$ seconds.
[3]
\item Hence show that $T = 6$.
[3]
\item Show that when the package reaches $C$ it is just under 10 m behind the rear of the moving lorry.
[5]
\end{enumerate}
END
\hfill \mbox{\textit{Edexcel M2 2001 Q7 [16]}}