| 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-particle projectile collision problem requiring simultaneous equations for horizontal and vertical positions. While it involves multiple steps (setting up equations for both particles, accounting for the 1s delay, solving a quadratic), the techniques are standard M2 fare. The given answer for part (i) guides students through the hardest algebraic step, and parts (ii)-(iii) follow systematically. More challenging than basic single-projectile questions but doesn't require novel insight beyond careful bookkeeping of the time offset. |
| 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]}}