| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Motion with applied force on slope |
| Difficulty | Standard +0.3 This is a standard M1 mechanics question on forces and motion on an inclined plane. It requires resolving forces parallel and perpendicular to the slope, applying F=ma, and using work-energy principles. While it has multiple parts and includes both smooth and rough plane scenarios, all techniques are routine applications of core mechanics principles with no novel problem-solving required. The calculations are straightforward once forces are correctly resolved. |
| Spec | 3.03c Newton's second law: F=ma one dimension3.03v Motion on rough surface: including inclined planes6.02d Mechanical energy: KE and PE concepts |
| Answer | Marks | Guidance |
|---|---|---|
| \(200 - 30g\sin 20 = 30a\) | M1 | For applying Newton's second law with 3 terms parallel to the plane |
| \(a = 3.25 \text{ ms}^{-2}\) | A1 [2] | \([a = 3.2465]\) |
| Answer | Marks | Guidance |
|---|---|---|
| \([v^2 = 2 \times 3.2465 \times 12 = 77.9]\) | M1 | For using \(v^2 = u^2 + 2as\) and attempting to find KE change |
| KE change \(= 0.5 \times 30 \times 77.9 = 1170\) J | A1 [2] | \([\text{KE} = 1168.7 \text{ J}]\) |
| Answer | Marks | Guidance |
|---|---|---|
| KE change \(= 200 \times 12 - 30g \times 12\sin 20\) | M1 | Using KE gain = WD by DF \(-\) PE gain |
| KE change \(= 1170\) J | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(N = 30g\cos 20\) | B1 | \([N = 281.9]\) |
| \(F = 0.12 \times 30g\cos 20\ [= 33.8]\) | M1 | Using \(F = \mu Na\) |
| \(200 - 30g\sin 20 - 33.8 = 30a\) | M1 | For using Newton's second law with 4 terms applied to the particle |
| \(a = 2.12 \text{ ms}^{-2}\) | A1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| \(N + 200\sin 10 = 30g\cos 20\ [N = 247.2]\) | M1 | For resolving forces perpendicular to the plane. Three term equation. |
| \(F = 0.12\ N\ [= 0.12 \times 247.2 = 29.66]\) | M1 | \(N\) must be from a 3 term equation |
| \(200\cos 10 - 29.66 - 30g\sin 20 = 30a\) | M1 | For using Newton's second law with 4 terms applied to the particle |
| \(a = 2.16 \text{ ms}^{-2}\) | A1 [4] |
# Question 7:
## Part (i)(a):
$200 - 30g\sin 20 = 30a$ | M1 | For applying Newton's second law with 3 terms parallel to the plane
$a = 3.25 \text{ ms}^{-2}$ | A1 [2] | $[a = 3.2465]$
## Part (i)(b):
$[v^2 = 2 \times 3.2465 \times 12 = 77.9]$ | M1 | For using $v^2 = u^2 + 2as$ and attempting to find KE change
KE change $= 0.5 \times 30 \times 77.9 = 1170$ J | A1 [2] | $[\text{KE} = 1168.7 \text{ J}]$
**Alternative method for 7(i)(b):**
KE change $= 200 \times 12 - 30g \times 12\sin 20$ | M1 | Using KE gain = WD by DF $-$ PE gain
KE change $= 1170$ J | A1 [2] |
## Part (ii)(a):
$N = 30g\cos 20$ | B1 | $[N = 281.9]$
$F = 0.12 \times 30g\cos 20\ [= 33.8]$ | M1 | Using $F = \mu Na$
$200 - 30g\sin 20 - 33.8 = 30a$ | M1 | For using Newton's second law with 4 terms applied to the particle
$a = 2.12 \text{ ms}^{-2}$ | A1 [4] |
## Part (ii)(b):
$N + 200\sin 10 = 30g\cos 20\ [N = 247.2]$ | M1 | For resolving forces perpendicular to the plane. Three term equation.
$F = 0.12\ N\ [= 0.12 \times 247.2 = 29.66]$ | M1 | $N$ must be from a 3 term equation
$200\cos 10 - 29.66 - 30g\sin 20 = 30a$ | M1 | For using Newton's second law with 4 terms applied to the particle
$a = 2.16 \text{ ms}^{-2}$ | A1 [4] |7 A particle of mass 30 kg is on a plane inclined at an angle of $20 ^ { \circ }$ to the horizontal. Starting from rest, the particle is pulled up the plane by a force of magnitude 200 N acting parallel to a line of greatest slope.\\
(i) Given that the plane is smooth, find
\begin{enumerate}[label=(\alph*)]
\item the acceleration of the particle,
\item the change in kinetic energy after the particle has moved 12 m up the plane.\\
(ii) It is given instead that the plane is rough and the coefficient of friction between the particle and the plane is 0.12 .\\
(a) Find the acceleration of the particle.\\
(b) The direction of the force of magnitude 200 N is changed, and the force now acts at an angle of $10 ^ { \circ }$ above the line of greatest slope. Find the acceleration of the particle.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2016 Q7 [12]}}