| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2020 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Vertical motion under gravity |
| Difficulty | Moderate -0.8 This is a straightforward kinematics problem using standard SUVAT equations. Part (a) requires finding maximum height using v²=u²+2as, and part (b) involves solving a quadratic equation for times when the particle is at a specific height. Both parts are routine applications of well-practiced techniques with no conceptual challenges or novel problem-solving required. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| 3(a) | 0 = 52 – 2gs | M1 |
| s = 1.25 | A1 | |
| [Height above ground =] 4.05 m | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 3(b) | Use of s = ut + ½ at2 | M1 |
| 0.8 = 5t – 5t2 | A1 | |
| t = 0.2 or 0.8 | M1 | |
| Length of time = 0.6 s | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 3:
--- 3(a) ---
3(a) | 0 = 52 – 2gs | M1
s = 1.25 | A1
[Height above ground =] 4.05 m | A1
3
--- 3(b) ---
3(b) | Use of s = ut + ½ at2 | M1
0.8 = 5t – 5t2 | A1
t = 0.2 or 0.8 | M1
Length of time = 0.6 s | A1
4
Question | Answer | MarksA particle $P$ is projected vertically upwards with speed $5\text{ m s}^{-1}$ from a point $A$ which is $2.8\text{ m}$ above horizontal ground.
\begin{enumerate}[label=(\alph*)]
\item Find the greatest height above the ground reached by $P$. [3]
\item Find the length of time for which $P$ is at a height of more than $3.6\text{ m}$ above the ground. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2020 Q3 [7]}}