| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2021 |
| Session | March |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Multi-stage motion with velocity-time graph given |
| Difficulty | Moderate -0.3 This is a standard M1 kinematics question requiring straightforward application of SUVAT equations and area-under-graph interpretation. Part (a) is direct calculation from given values, part (b) uses the principle that total displacement equals zero (areas above and below axis must balance), and part (c) applies Newton's second law with weight and tension. All techniques are routine for M1 students with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae6.01a Dimensions: M, L, T notation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Acceleration \(= \frac{4}{3} \text{ ms}^{-2}\) | B1 | Allow \(= 1.33 \text{ ms}^{-2}\) |
| 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{2}(7+4.5)\times 2 = \frac{1}{2}(8.5+5)\times V\) | M1 | Equate expressions for the two areas (distances) leading to an equation in \(V\) |
| \(V = 1.7[0]\) (3sf) | A1 | Allow \(V = \frac{46}{27}\) |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Acceleration \(= -2 \text{ ms}^{-2}\) | B1 | Or Deceleration \(= 2\) |
| \(T - 1500g = 1500\times(-2)\) | M1 | Apply Newton's second law to the lift, using an acceleration \(\neq \frac{4}{3}\) or *their* 4(a). Correct dimensions and number of relevant terms |
| \(T = 12\,000\) N | A1 | |
| 3 |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Acceleration $= \frac{4}{3} \text{ ms}^{-2}$ | B1 | Allow $= 1.33 \text{ ms}^{-2}$ |
| | **1** | |
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2}(7+4.5)\times 2 = \frac{1}{2}(8.5+5)\times V$ | M1 | Equate expressions for the two areas (distances) leading to an equation in $V$ |
| $V = 1.7[0]$ (3sf) | A1 | Allow $V = \frac{46}{27}$ |
| | **2** | |
## Question 4(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Acceleration $= -2 \text{ ms}^{-2}$ | B1 | Or Deceleration $= 2$ |
| $T - 1500g = 1500\times(-2)$ | M1 | Apply Newton's second law to the lift, using an acceleration $\neq \frac{4}{3}$ or *their* **4(a)**. Correct dimensions and number of relevant terms |
| $T = 12\,000$ N | A1 | |
| | **3** | |4\\
\includegraphics[max width=\textwidth, alt={}, center]{a96ca3b4-6d35-4512-a0a1-3f28443fd051-06_661_1529_260_306}
An elevator moves vertically, supported by a cable. The diagram shows a velocity-time graph which models the motion of the elevator. The graph consists of 7 straight line segments.
The elevator accelerates upwards from rest to a speed of $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ over a period of 1.5 s and then travels at this speed for 4.5 s , before decelerating to rest over a period of 1 s .
The elevator then remains at rest for 6 s , before accelerating to a speed of $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$ downwards over a period of 2 s . The elevator travels at this speed for a period of 5 s , before decelerating to rest over a period of 1.5 s .
\begin{enumerate}[label=(\alph*)]
\item Find the acceleration of the elevator during the first 1.5 s .
\item Given that the elevator starts and finishes its journey on the ground floor, find $V$.
\item The combined weight of the elevator and passengers on its upward journey is 1500 kg . Assuming that there is no resistance to motion, find the tension in the elevator cable on its upward journey when the elevator is decelerating.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2021 Q4 [6]}}