| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2023 |
| Session | March |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Two particles: same start time, different heights |
| Difficulty | Moderate -0.8 Part (a) is a direct application of v² = u² + 2as with standard values. Part (b) requires setting up two SUVAT equations and solving simultaneously, but the setup is straightforward with no geometric complexity or novel insight needed. This is a routine two-particle kinematics problem, slightly easier than average due to the vertical-only motion and standard numerical values. |
| Spec | 3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| 2(a) | Use constant acceleration in an attempt to find v or v2 | |
| [v2 = 152 – 2g × 10] | M1 | e.g. v2 = u2 + 2as with a=g. |
| Speed = 5 m s-1 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 2(b) | (s =) 15t− 1 gt2 , ( s = ) 1 gt2 | |
| P 2 Q 2 | *B1 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| P Q | DM1 | Allow s +s =18. |
| Answer | Marks |
|---|---|
| So height = 10.8 m | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 15t | *B1 | Use of relative velocity (no acceleration). |
| Use 15t=18 and solve for t | DM1 | Allow 15t=18. |
| So height = 10.8 m | A1 | Not from t=−1.2 made positive without justification. |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 2:
--- 2(a) ---
2(a) | Use constant acceleration in an attempt to find v or v2
[v2 = 152 – 2g × 10] | M1 | e.g. v2 = u2 + 2as with a=g.
Speed = 5 m s-1 | A1
2
Question | Answer | Marks | Guidance
--- 2(b) ---
2(b) | (s =) 15t− 1 gt2 , ( s = ) 1 gt2
P 2 Q 2 | *B1 | 1
Use of s=ut+ at2 for either.
2
Allow if a not substituted, need both expressions with opposite
sign of t2 term and the same a.
Use s +s =18 and solve for t
P Q | DM1 | Allow s +s =18.
P Q
Must have s and s of the correct form.
P Q
So height = 10.8 m | A1
Alternative method for Question 2(b): Using relative velocity
15t | *B1 | Use of relative velocity (no acceleration).
Use 15t=18 and solve for t | DM1 | Allow 15t=18.
So height = 10.8 m | A1 | Not from t=−1.2 made positive without justification.
3
Question | Answer | Marks | GuidanceA particle $P$ is projected vertically upwards from horizontal ground with speed $15\,\text{m}\,\text{s}^{-1}$.
\begin{enumerate}[label=(\alph*)]
\item Find the speed of $P$ when it is 10 m above the ground. [2]
At the same instant that $P$ is projected, a second particle $Q$ is dropped from a height of 18 m above the ground in the same vertical line as $P$.
\item Find the height above the ground at which the two particles collide. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2023 Q2 [5]}}