| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Find acceleration from distances/times |
| Difficulty | Moderate -0.3 This is a straightforward mechanics problem combining basic SUVAT kinematics with work-energy concepts. Part (i) requires direct application of s = ut + ½at² with given values, and part (ii) needs W = Fs cos θ where the horizontal component of tension must be found. Both parts are standard textbook exercises requiring routine application of formulas with no problem-solving insight or complex reasoning. |
| Spec | 3.02d Constant acceleration: SUVAT formulae6.02b Calculate work: constant force, resolved component |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(s = 0.3 \times 5 + \frac{1}{2} \times 0.5 \times 5^2\); complete method for finding \(s\) | M1 | For using \(s = ut + \frac{1}{2}at^2\), or using \(v = u + at\) followed by either \(v^2 = u^2 + 2as\), or \(s = \frac{(u+v)}{2}t\), or \(s = vt - \frac{1}{2}at^2\) |
| Distance \(= 7.75\) m | A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(WD = 8 \times 7.75 \times 0.5\) | M1 | For using \(WD = Td\cos 60°\) |
| Work done is \(31\) J | A1 | [2] |
## Question 1:
### Part (i)
| Working/Answer | Mark | Guidance |
|---|---|---|
| $s = 0.3 \times 5 + \frac{1}{2} \times 0.5 \times 5^2$; complete method for finding $s$ | M1 | For using $s = ut + \frac{1}{2}at^2$, or using $v = u + at$ followed by either $v^2 = u^2 + 2as$, or $s = \frac{(u+v)}{2}t$, or $s = vt - \frac{1}{2}at^2$ |
| Distance $= 7.75$ m | A1 | **[2]** |
### Part (ii)
| Working/Answer | Mark | Guidance |
|---|---|---|
| $WD = 8 \times 7.75 \times 0.5$ | M1 | For using $WD = Td\cos 60°$ |
| Work done is $31$ J | A1 | **[2]** |
---1 One end of a light inextensible string is attached to a block. The string makes an angle of $60 ^ { \circ }$ above the horizontal and is used to pull the block in a straight line on a horizontal floor with acceleration $0.5 \mathrm {~m} \mathrm {~s} ^ { - 2 }$. The tension in the string is 8 N . The block starts to move with speed $0.3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. For the first 5 s of the block's motion, find\\
(i) the distance travelled,\\
(ii) the work done by the tension in the string.
\hfill \mbox{\textit{CAIE M1 2015 Q1 [4]}}