| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2018 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Deriving trajectory equation |
| Difficulty | Moderate -0.3 This is a standard projectile motion question requiring routine application of kinematic equations and trajectory derivation. Part (i) involves straightforward substitution of initial conditions into standard formulae and eliminating the parameter t. Part (ii) requires solving a quadratic equation from the trajectory. While it has multiple steps (7 marks total), all techniques are textbook exercises with no novel problem-solving or geometric insight required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| 4(i) | x = (20cos30)t or 10 3t | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | B1 | Use vertical motion |
| y = (20sin30)[x/(20cos30)] – 5[x/(20cos30)]2 | M1 | Attempt to eliminate t |
| y = x/ 3 – x2/60 or 0.577x – 0.0167x2 | A1 | |
| Total: | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| 4(ii) | x/ 3 – x2/60 = (x+15)/ 3 – ( x+15 )2 /60 | M1 |
| x = 9.821 | A1 | |
| y = 4.06(25) m | A1 | |
| Total: | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| 0.577x – 0.0167x2 = 0.577(x+15) – 0.0167 ( x+15 )2 | M1 | |
| x = 9.775 | A1 | |
| y = 4.044 | A1 | |
| Total: | 3 | |
| Question | Answer | Marks |
Question 4:
--- 4(i) ---
4(i) | x = (20cos30)t or 10 3t | B1 | Use horizontal motion
1
y = (20sin30)t – gt2 or 10t – 5t2
2 | B1 | Use vertical motion
y = (20sin30)[x/(20cos30)] – 5[x/(20cos30)]2 | M1 | Attempt to eliminate t
y = x/ 3 – x2/60 or 0.577x – 0.0167x2 | A1
Total: | 4
--- 4(ii) ---
4(ii) | x/ 3 – x2/60 = (x+15)/ 3 – ( x+15 )2 /60 | M1 | Simplifies to 0 = 15/ 3 – (30x+225)/60
x = 9.821 | A1
y = 4.06(25) m | A1
Total: | 3
Alternative method
0.577x – 0.0167x2 = 0.577(x+15) – 0.0167 ( x+15 )2 | M1
x = 9.775 | A1
y = 4.044 | A1
Total: | 3
Question | Answer | Marks | GuidanceA particle $P$ is projected from a point $O$ on horizontal ground with initial speed 20 m s$^{-1}$ and angle of projection 30°. At time $t$ s after projection, the horizontal and vertically upwards displacements of $P$ from $O$ are $x$ m and $y$ m respectively.
\begin{enumerate}[label=(\roman*)]
\item Express $x$ and $y$ in terms of $t$ and hence find the equation of the trajectory of $P$. [4]
\item Calculate this height. [3]
\end{enumerate}
$P$ is at the same height above the ground at two points which are a horizontal distance apart of 15 m.
\hfill \mbox{\textit{CAIE M2 2018 Q4 [7]}}