| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Motion on rough inclined plane |
| Difficulty | Standard +0.3 This is a straightforward mechanics problem requiring standard kinematics equations (v² = u² + 2as), work-energy principle, and resolving forces on an inclined plane. Part (iii) adds a minor twist with average speed, but all steps follow routine procedures with no novel insight required. Slightly easier than average due to clear structure and standard techniques. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02a Work done: concept and definition6.02d Mechanical energy: KE and PE concepts |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \([v^2 = 3^2 + 2 \times 2.5 \times 8]\) | M1 | For using \(v^2 = u^2 + 2as\) |
| Speed is 7 ms⁻¹ | A1 | [2] |
| (ii) \(KE \text{ gain} = \frac{1}{2}(0.8)(7^2 - 5^2)(= 16)\) | B1ft | (if incorrect speed) |
| \(PE \text{ loss} = 16 + 7\) | B1ft | (if incorrect expression for KE) |
| \([0.8 \times 10 \times 8\sin\alpha = 23]\) | M1 | For using \(PE \text{ loss} = mg L\sin\alpha\) |
| Angle is 21.1° or 0.368° | A1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| \(F = 7/8\) | B1 | |
| \([0.8 \times 10\sin\alpha - F = 0.8 \times 2.5]\) | M1 | For using Newton's second law |
| \(0.8 \times 10\sin\alpha - 0.875 = 0.8 \times 2.5\) | A1 | |
| Angle is 21.1° or 0.368° | A1 | |
| (iii) \(s^2 = 3^2 + 2 \times 2.5s(s - 3.2)\) | B1 | |
| \([WD/7 = 3.2/8\) or \(WD = 0.875 \times 3.2\) or \(WD = 8 \times 3.2 \times (23/64) - \frac{1}{2}0.8(5 - 3^2)]\) | M1 | For using WD [proprt'l] to dist. or \(WD = F(AX)\) or \(WD = PE \text{ loss} - KE \text{ gain}\) |
| Work done is 2.8 J | A1 | [3] |
**(i)** $[v^2 = 3^2 + 2 \times 2.5 \times 8]$ | M1 | For using $v^2 = u^2 + 2as$
Speed is 7 ms⁻¹ | A1 | [2]
**(ii)** $KE \text{ gain} = \frac{1}{2}(0.8)(7^2 - 5^2)(= 16)$ | B1ft | (if incorrect speed)
$PE \text{ loss} = 16 + 7$ | B1ft | (if incorrect expression for KE)
$[0.8 \times 10 \times 8\sin\alpha = 23]$ | M1 | For using $PE \text{ loss} = mg L\sin\alpha$
Angle is 21.1° or 0.368° | A1 | [4]
**(iii) ALTERNATIVELY**
$F = 7/8$ | B1 |
$[0.8 \times 10\sin\alpha - F = 0.8 \times 2.5]$ | M1 | For using Newton's second law
$0.8 \times 10\sin\alpha - 0.875 = 0.8 \times 2.5$ | A1 |
Angle is 21.1° or 0.368° | A1 |
**(iii)** $s^2 = 3^2 + 2 \times 2.5s(s - 3.2)$ | B1 |
$[WD/7 = 3.2/8$ or $WD = 0.875 \times 3.2$ or $WD = 8 \times 3.2 \times (23/64) - \frac{1}{2}0.8(5 - 3^2)]$ | M1 | For using WD [proprt'l] to dist. or $WD = F(AX)$ or $WD = PE \text{ loss} - KE \text{ gain}$
Work done is 2.8 J | A1 | [3]5 A particle of mass 0.8 kg slides down a rough inclined plane along a line of greatest slope $A B$. The distance $A B$ is 8 m . The particle starts at $A$ with speed $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and moves with constant acceleration $2.5 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
(i) Find the speed of the particle at the instant it reaches $B$.\\
(ii) Given that the work done against the frictional force as the particle moves from $A$ to $B$ is 7 J , find the angle of inclination of the plane.
When the particle is at the point $X$ its speed is the same as the average speed for the motion from $A$ to $B$.\\
(iii) Find the work done by the frictional force for the particle's motion from $A$ to $X$.
\hfill \mbox{\textit{CAIE M1 2010 Q5 [9]}}