| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Sketch velocity-time graph |
| Difficulty | Moderate -0.3 This is a straightforward two-stage SUVAT problem with all necessary information provided. Part (i) uses standard equations with given u, v, t to find s and a. Part (ii) applies the same methods to the return journey. Part (iii) requires sketching a simple piecewise-linear velocity-time graph. While multi-part, each stage follows routine procedures without requiring problem-solving insight or dealing with ambiguous setups—slightly easier than average due to its mechanical nature. |
| Spec | 3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(s = \frac{1}{2}(1.4+1.1)\times1.2\); \(1.1 = 1.4 + (-d)\times1.2\) | M1 | For using \(s = \frac{1}{2}(u+v)t\) to find AB or \(v = u + at\) to find \(d\) |
| \(AB = 1.5\text{ m}\) or \(d = 0.25\) | A1 | |
| \(d = 0.25\) or \(AB = 1.5\text{ m}\) | B1ft | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(0 = u^2 + 2(-0.25)\times2\); \(2 = 0 - \frac{1}{2}(-0.25)t^2\) | M1 | For using \(0 = u^2 + 2(-d)s\) to find \(u\) or \(s = 0 - \frac{1}{2}(-d)t^2\) to find \(t\) |
| Speed is \(1\text{ ms}^{-1}\) or time is 4 s | A1 | |
| Time is 4 s or speed is \(1\text{ ms}^{-1}\) | B1ft | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| For line joining \((0, 1.4)\) and \((1.2, 1.1)\) | B1 | |
| For line joining \((1.2, -1)\) and \((5.2, 0)\) | B1ft | ft wrong answer(s) in (ii) |
| [2] | ||
| SR (max 1/2): For two correct lines with values missing | B1ft |
## Question 5:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $s = \frac{1}{2}(1.4+1.1)\times1.2$; $1.1 = 1.4 + (-d)\times1.2$ | M1 | For using $s = \frac{1}{2}(u+v)t$ to find AB **or** $v = u + at$ to find $d$ |
| $AB = 1.5\text{ m}$ **or** $d = 0.25$ | A1 | |
| $d = 0.25$ **or** $AB = 1.5\text{ m}$ | B1ft | |
| **[3]** | | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0 = u^2 + 2(-0.25)\times2$; $2 = 0 - \frac{1}{2}(-0.25)t^2$ | M1 | For using $0 = u^2 + 2(-d)s$ to find $u$ **or** $s = 0 - \frac{1}{2}(-d)t^2$ to find $t$ |
| Speed is $1\text{ ms}^{-1}$ **or** time is 4 s | A1 | |
| Time is 4 s **or** speed is $1\text{ ms}^{-1}$ | B1ft | |
| **[3]** | | |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| For line joining $(0, 1.4)$ and $(1.2, 1.1)$ | B1 | |
| For line joining $(1.2, -1)$ and $(5.2, 0)$ | B1ft | ft wrong answer(s) in (ii) |
| **[2]** | | |
| SR (max 1/2): For two correct lines with values missing | B1ft | |
---5 A ball moves on the horizontal surface of a billiards table with deceleration of constant magnitude $d \mathrm {~m} \mathrm {~s} ^ { - 2 }$. The ball starts at $A$ with speed $1.4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and reaches the edge of the table at $B , 1.2 \mathrm {~s}$ later, with speed $1.1 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) Find the distance $A B$ and the value of $d$.\\
$A B$ is at right angles to the edge of the table containing $B$. The table has a low wall along each of its edges and the ball rebounds from the wall at $B$ and moves directly towards $A$. The ball comes to rest at $C$ where the distance $B C$ is 2 m .\\
(ii) Find the speed with which the ball starts to move towards $A$ and the time taken for the ball to travel from $B$ to $C$.\\
(iii) Sketch a velocity-time graph for the motion of the ball, from the time the ball leaves $A$ until it comes to rest at $C$, showing on the axes the values of the velocity and the time when the ball is at $A$, at $B$ and at $C$.
\hfill \mbox{\textit{CAIE M1 2010 Q5 [8]}}