| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2006 |
| 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 Part (i) is a straightforward SUVAT application using v² = u² + 2as with all values given. Part (ii) requires applying F = ma with the acceleration from part (i), but this is a standard textbook exercise in combining kinematics with Newton's second law. Both parts involve routine procedures with no problem-solving insight required, making this slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03c Newton's second law: F=ma one dimension |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(2s^2 = 10^2 + 1050a\) | M1 | For using \(v^2 = u^2 + 2as\) |
| \(a = 0.5\) | A1 | 2 |
| \(900 - R = 1200 \times 0.5\) | M1 | For using Newton's second law |
| \(R = 300\) | A1 ft | 2 |
| (ii) |
**(i)** $2s^2 = 10^2 + 1050a$ | M1 | For using $v^2 = u^2 + 2as$ |
$a = 0.5$ | A1 | 2 |
| | | |
$900 - R = 1200 \times 0.5$ | M1 | For using Newton's second law |
$R = 300$ | A1 ft | 2 |
**(ii)** | | |
---1 A car of mass 1200 kg travels on a horizontal straight road with constant acceleration $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
(i) Given that the car's speed increases from $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ to $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ while travelling a distance of 525 m , find the value of $a$.
The car's engine exerts a constant driving force of 900 N . The resistance to motion of the car is constant and equal to $R \mathrm {~N}$.\\
(ii) Find $R$.
\hfill \mbox{\textit{CAIE M1 2006 Q1 [4]}}