| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2023 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Multi-phase journey: find unknown speed or time |
| Difficulty | Standard +0.3 This is a straightforward multi-part M1 question combining basic SUVAT calculations with Newton's second law. Parts (a)-(b) require only routine kinematics (v = u + at, area under graph), while parts (c)-(d) involve standard force resolution with weight and tension. All steps are textbook-standard with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae3.03c Newton's second law: F=ma one dimension |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Greatest speed \(=2\ [\text{ms}^{-1}]\ [0.4\times5]\) | B1 | This can be seen on the graph and not stated explicitly. |
| Trapezium shape | B1 | Sitting on \(t\)-axis, starting at origin. |
| All correct including height of 2 and \(t\)-values of 5, 30, 40 on horizontal axis. Labels not needed. Does not need to be to scale. | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Distance \(=\frac{1}{2}(25+5+25+their\ 10)\times their\ 2\) or \(\frac{1}{2}\times5\times their\ 2+25\times their\ 2+\frac{1}{2}\times their\ 10\times their\ 2\) | M1 | Allow M1 for finding total area under their trapezium or appropriate 'suvat' in each phase. If presented as 3 areas, they do not need to be added for M1. Allow one wrong value but must represent all 3 phases of motion. |
| Distance \(=65\ [\text{m}]\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Attempt at Newton's second law | M1 | Must have correct number of terms (5). Allow sign errors. Allow \(g\) missing. Use of \(a=g\) is M0A0A0 but condone use of \(a=0.4\) (from wrong phase). |
| \(12250-1200g-mg=(1200+m)\times(-0.2)\) or \(1200g+mg-12250=(1200+m)\times0.2\) | A1 | Correct equation. Note that taking \(a=0.2\) and omitting \(mg\) gets M0A0A0. |
| \(m=50\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Realise that this is when accelerating and attempt Newton's second law for the crate only | M1 | Must have correct number of terms (3). Allow sign errors. Allow \(g\) missing. Must use \(a=\pm0.4\), M0A0A0 otherwise. |
| \(R-50g=50\times0.4\) or \(50g-R=50\times(-0.4)\) | A1FT | Correct equation using their 50. |
| Force \(R=520\ [\text{N}]\), upwards | A1 | Must include 'upwards' OE. |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Greatest speed $=2\ [\text{ms}^{-1}]\ [0.4\times5]$ | **B1** | This can be seen on the graph and not stated explicitly. |
| Trapezium shape | **B1** | Sitting on $t$-axis, starting at origin. |
| All correct including height of 2 and $t$-values of 5, 30, 40 on horizontal axis. Labels not needed. Does not need to be to scale. | **B1** | |
---
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Distance $=\frac{1}{2}(25+5+25+their\ 10)\times their\ 2$ or $\frac{1}{2}\times5\times their\ 2+25\times their\ 2+\frac{1}{2}\times their\ 10\times their\ 2$ | **M1** | Allow M1 for finding total area under their trapezium or appropriate 'suvat' in each phase. If presented as 3 areas, they do not need to be added for M1. Allow one wrong value but must represent all 3 phases of motion. |
| Distance $=65\ [\text{m}]$ | **A1** | |
---
## Question 6(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt at Newton's second law | **M1** | Must have correct number of terms (5). Allow sign errors. Allow $g$ missing. Use of $a=g$ is M0A0A0 but condone use of $a=0.4$ (from wrong phase). |
| $12250-1200g-mg=(1200+m)\times(-0.2)$ or $1200g+mg-12250=(1200+m)\times0.2$ | **A1** | Correct equation. Note that taking $a=0.2$ and omitting $mg$ gets M0A0A0. |
| $m=50$ | **A1** | |
---
## Question 6(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Realise that this is when accelerating and attempt Newton's second law for the crate only | **M1** | Must have correct number of terms (3). Allow sign errors. Allow $g$ missing. Must use $a=\pm0.4$, M0A0A0 otherwise. |
| $R-50g=50\times0.4$ or $50g-R=50\times(-0.4)$ | **A1FT** | Correct equation using their 50. |
| Force $R=520\ [\text{N}]$, **upwards** | **A1** | Must include 'upwards' OE. |
---6 An elevator is pulled vertically upwards by a cable. The elevator accelerates at $0.4 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ for 5 s , then travels at constant speed for 25 s . The elevator then decelerates at $0.2 \mathrm {~ms} ^ { - 2 }$ until it comes to rest.
\begin{enumerate}[label=(\alph*)]
\item Find the greatest speed of the elevator and hence draw a velocity-time graph for the motion of the elevator.
\item Find the total distance travelled by the elevator.\\
The mass of the elevator is 1200 kg and there is a crate of mass $m \mathrm {~kg}$ resting on the floor of the elevator.
\item Given that the tension in the cable when the elevator is decelerating is 12250 N , find the value of $m$.
\item Find the greatest magnitude of the force exerted on the crate by the floor of the elevator, and state its direction.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2023 Q6 [11]}}