| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Energy methods on slope |
| Difficulty | Standard +0.3 This is a standard mechanics problem requiring routine application of friction inequalities, energy methods, and Newton's second law on slopes. Part (i) is straightforward equilibrium, part (ii) applies work-energy theorem with given values, and part (iii) uses F=ma with resolved forces. All techniques are standard M1 content with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.03u Static equilibrium: on rough surfaces3.03v Motion on rough surface: including inclined planes6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(R = 50g\cos 10°\) and \(F = 50g\sin 10°\) | B1 | |
| \(\mu \geqslant 0.176\) | B1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| PE loss \(= 50g \times d\sin 10°\) | B1 | |
| WD against friction \(= 0.19 \times 50g\cos 10° \times d\) | B1 | |
| M1 | For using WD by 50N force \(+\) PE loss \(-\) WD against friction \(=\) KE gain | |
| \(50 \times 5 + 50g \times 10\sin 10° - 0.19 \times 50g\cos 10° \times 10 = 0.5 \times 50v^2\) | A1 | |
| Speed is \(2.70\text{ ms}^{-1}\) | A1 | [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(50g\sin 20° - 0.19 \times 50g\cos 20° = 50a\) | M1 | |
| Acceleration is \(1.63\text{ ms}^{-2}\) | A1 | [2] |
# Question 7:
## Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $R = 50g\cos 10°$ and $F = 50g\sin 10°$ | B1 | |
| $\mu \geqslant 0.176$ | B1 | [2] | $\mu \geqslant F \div R$ Allow $\mu \geqslant \tan 10°$ |
## Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| PE loss $= 50g \times d\sin 10°$ | B1 | | $d = 5$ or $d = 10$ |
| WD against friction $= 0.19 \times 50g\cos 10° \times d$ | B1 | | $d = 5$ or $d = 10$ |
| | M1 | | For using WD by 50N force $+$ PE loss $-$ WD against friction $=$ KE gain |
| $50 \times 5 + 50g \times 10\sin 10° - 0.19 \times 50g\cos 10° \times 10 = 0.5 \times 50v^2$ | A1 | | |
| Speed is $2.70\text{ ms}^{-1}$ | A1 | [5] | **SC** for candidates using Newton's Second law: max 2/5; B1 $v = 2.94\text{ ms}^{-1}$ after 5m; B1 Speed is $2.70\text{ ms}^{-1}$ |
## Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $50g\sin 20° - 0.19 \times 50g\cos 20° = 50a$ | M1 | | For using Newton's Second Law |
| Acceleration is $1.63\text{ ms}^{-2}$ | A1 | [2] | |7 A box of mass 50 kg is at rest on a plane inclined at $10 ^ { \circ }$ to the horizontal.\\
(i) Find an inequality for the coefficient of friction between the box and the plane.
In fact the coefficient of friction between the box and the plane is 0.19 .\\
(ii) A girl pushes the box with a force of 50 N , acting down a line of greatest slope of the plane, for a distance of 5 m . She then stops pushing. Use an energy method to find the speed of the box when it has travelled a further 5 m .
The box then comes to a plane inclined at $20 ^ { \circ }$ below the horizontal. The box moves down a line of greatest slope of this plane. The coefficient of friction is still 0.19 and the girl is not pushing the box.\\
(iii) Find the acceleration of the box.
\hfill \mbox{\textit{CAIE M1 2016 Q7 [9]}}