| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2015 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Two projectiles meeting - 2D flight |
| Difficulty | Challenging +1.2 This is a two-projectile collision problem requiring simultaneous equations for horizontal and vertical positions. While it involves multiple steps and algebraic manipulation (including solving a quadratic to find t), the setup is methodical and the techniques are standard M2 fare. The given answer for part (i) guides students through the hardest step, making parts (ii) and (iii) straightforward applications of projectile formulae. Moderately above average difficulty due to the multi-stage reasoning and algebraic complexity, but not requiring novel insight. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model |
| Answer | Marks |
|---|---|
| 7 (i) | (x = ) Vcos45t = Vcos60(t+1) |
| t = 2.414 AG | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| A1 | 3 | Equates horizontal distances |
| Answer | Marks |
|---|---|
| (iii) | gt2 |
| Answer | Marks |
|---|---|
| Falls 9.72 m | M1 |
| Answer | Marks |
|---|---|
| A1 | 4 |
| 4 | Equates vertical distances |
Question 7:
--- 7 (i) ---
7 (i) | (x = ) Vcos45t = Vcos60(t+1)
t = 2.414 AG | M1
A1
A1 | 3 | Equates horizontal distances
Terms correct
(ii)
(iii) | gt2
(y = ) Vsin45t – =
2
g ( t+ 1)2
Vsin60(t + 1) –
2
V{sin60(3.414) – sin45(2.414)} =
2 2
5{(3.414) – (2.414) }
V = 23.3
23.322sin260
Greatest H =
(2g )
(3.414)2
g
h = 23.3sin60(3.414) –
2
h = 10.67
Falls 9.72 m | M1
A1
M1
A1
B1
M1
A1
A1 | 4
4 | Equates vertical distances
Terms correct
Gathers terms correctly
23.32...
2
20.39, ft cv(23.3) ×3/80A particle $P$ is projected with speed $V\text{ m s}^{-1}$ at an angle of $60°$ above the horizontal from a point $O$. At the instant $1\text{ s}$ later a particle $Q$ is projected from $O$ with the same initial speed at an angle of $45°$ above the horizontal. The two particles collide when $Q$ has been in motion for $t\text{ s}$.
\begin{enumerate}[label=(\roman*)]
\item Show that $t = 2.414$, correct to 3 decimal places. [3]
\item Find the value of $V$. [4]
\end{enumerate}
The collision occurs after $P$ has passed through the highest point of its trajectory.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Calculate the vertical distance of $P$ below its greatest height when $P$ and $Q$ collide. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2015 Q7 [11]}}