| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2002 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile with bounce or impact |
| Difficulty | Standard +0.3 This is a standard M2 projectiles question with momentum conservation. Parts (a)-(b) use routine projectile formulas (vertical component becomes zero at maximum height). Parts (c)-(d) apply conservation of momentum in an explosion scenario—a common M2 topic. All steps are methodical applications of standard techniques with no novel insight required, making it slightly easier than the average A-level question. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model6.03b Conservation of momentum: 1D two particles |
| Answer | Marks | Guidance |
|---|---|---|
| - Height is 260 m (Accept 265) | B1, B1, M1, B1 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| - \(t = 7.1\) (s) | N1 (Accept 7.07) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| - \(v = (-)20\) (cso) | B1, M1, A1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| - \(264.9 = \frac{1}{2} \times 4.8 \times t^2 \Rightarrow t \approx 7.35\) (avrt) | M1, B1 | M1, B1 |
| - \(CN = 20 \times 7.35 \approx 147\) (avrt) | M1, B1 | |
| - \(OC = 140\) (m) (accept 136) | A1 | 6 |
## (a)
**Answer/Working:**
- $u_y = 80\sin60°$, $v_y = 0$
- $0^2 = (80\sin60°)^2 - 2 \times 9.8 \times s$
- $s \approx 244.9$ m
- Height is 260 m (Accept 265) | B1, B1, M1, B1 | A1 | 4
## (b)
**Answer/Working:**
- $0 = 90\sin60° - 4.8t$
- $t = 7.1$ (s) | N1 (Accept 7.07) | A1 | 2
## (c)
**Answer/Working:**
- $u_x = 90\cos60° (= 40)$
- L/M: $100s = 40 \times v + 60 \times 80$
- $v = (-)20$ (cso) | B1, M1, A1 | 3
## (d)
**Answer/Working:**
- Let N be point on ground vertically below B
- $ON = 90\cos60° \times$ their (b) $(= 212.7)$
- $264.9 = \frac{1}{2} \times 4.8 \times t^2 \Rightarrow t \approx 7.35$ (avrt) | M1, B1 | M1, B1 |
- $CN = 20 \times 7.35 \approx 147$ (avrt) | M1, B1 |
- $OC = 140$ (m) (accept 136) | A1 | 6\includegraphics{figure_3}
A rocket $R$ of mass 100 kg is projected from a point $A$ with speed 80 m s$^{-1}$ at an angle of elevation of 60°, as shown in Fig. 3. The point $A$ is 20 m vertically above a point $O$ which is on horizontal ground. The rocket $R$ moves freely under gravity. At $B$ the velocity of $R$ is horizontal. By modelling $R$ as a particle, find
\begin{enumerate}[label=(\alph*)]
\item the height in m of $B$ above the ground,
[4]
\item the time taken for $R$ to reach $B$ from $A$.
[2]
\end{enumerate}
When $R$ is at $B$, there is an internal explosion and $R$ breaks into two parts $P$ and $Q$ of masses 60 kg and 40 kg respectively. Immediately after the explosion the velocity of $P$ is 80 m s$^{-1}$ horizontally away from $A$. After the explosion the paths of $P$ and $Q$ remain in the plane $OAB$. Part $Q$ strikes the ground at $C$. By modelling $P$ and $Q$ as particles,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item show that the speed of $Q$ immediately after the explosion is 20 m s$^{-1}$,
[3]
\item find the distance $OC$.
[6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2002 Q7 [15]}}