| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2006 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Motion up then down slope |
| Difficulty | Standard +0.3 This is a standard three-part mechanics question on motion on a rough inclined plane. Part (i) involves resolving forces perpendicular and parallel to the plane using Newton's second law with given kinematics (straightforward application of F=ma and suvat). Part (ii) is direct calculation of μ from F=μR. Part (iii) requires recognizing that friction reverses direction on the way down and applying energy methods or kinematics again. While it requires careful bookkeeping of force directions and multiple steps, all techniques are standard M1 content with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(R = mg\cos 21° = 9.336m\) | B1, M1, A1 | For using \(\theta = 10 + at\) |
| \(a = -5\) | ||
| \([-mg\sin 21° - F = -5m]\) | M1 | For using Newton's second law; Alternative for the above 3 marks: For using WD by frictional force = KE at P – PE at highest point |
| \(WD = \frac{1}{2}m10^2 - mgs \sin 21°\) | A1 | |
| \(s = 10\) (mark available in (ii) may be given here) | ||
| For WD = Fs to find F | DM1 | |
| \(F = 1.416m\) | A1 | 5 |
| (ii) Coefficient is 0.152 | B1 | 1 |
| (iii) \(s = 10\) | B1 | From \(s = (u + v)2\) (upwards) or equivalent; may already have been scored for appropriate work in part (i) |
| \(mg\sin 21° - 1.416m = ma\) | M1 | For using Newton's second law |
| \([v^2 = 2(g\sin 21° - 1.416)10]\) | A1 | For using \(v^2 = 2as\) |
| Alternative for the above 3 marks: For using WD by frictional force (up and down) = 2Fs | M1 | |
| \(WD = 2(1.416m)10\) | A1 | |
| For using KE at P(down) = KE at P(initial) – WD by frictional force (up and down) | M1 | |
| \([\frac{1}{2}mv^2 = \frac{1}{2}m10^2 - 28.32m]\) | ||
| Second alternative for the above 3 marks | ||
| WD by frictional force (down) = 1.416m × 10 | B1 | |
| PE loss (down) = \(mg(10\sin 21°)\) | B1 | |
| For using KE at P(down) = PE loss – WD by frictional force (down) | M1 | |
| \([\frac{1}{2}mv^2 = mg(10\sin 21°) -14.16m]\) | ||
| Speed is 6.58 m\(s^{-1}\) | A1 | 5 |
(i) $R = mg\cos 21° = 9.336m$ | B1, M1, A1 | For using $\theta = 10 + at$
$a = -5$ |
$[-mg\sin 21° - F = -5m]$ | M1 | For using Newton's second law; Alternative for the above 3 marks: For using WD by frictional force = KE at P – PE at highest point | M1
| | | $WD = \frac{1}{2}m10^2 - mgs \sin 21°$ | A1
| | | $s = 10$ (mark available in (ii) may be given here) |
| | | For WD = Fs to find F | DM1
$F = 1.416m$ | A1 | 5
(ii) Coefficient is 0.152 | B1 | 1
(iii) $s = 10$ | B1 | From $s = (u + v)2$ (upwards) or equivalent; may already have been scored for appropriate work in part (i)
$mg\sin 21° - 1.416m = ma$ | M1 | For using Newton's second law
$[v^2 = 2(g\sin 21° - 1.416)10]$ | A1 | For using $v^2 = 2as$
| | | Alternative for the above 3 marks: For using WD by frictional force (up and down) = 2Fs | M1
| | | $WD = 2(1.416m)10$ | A1
| | | For using KE at P(down) = KE at P(initial) – WD by frictional force (up and down) | M1
| | | $[\frac{1}{2}mv^2 = \frac{1}{2}m10^2 - 28.32m]$ |
| | | Second alternative for the above 3 marks |
| | | WD by frictional force (down) = 1.416m × 10 | B1
| | | PE loss (down) = $mg(10\sin 21°)$ | B1
| | | For using KE at P(down) = PE loss – WD by frictional force (down) | M1
| | | $[\frac{1}{2}mv^2 = mg(10\sin 21°) -14.16m]$ |
Speed is 6.58 m$s^{-1}$ | A1 | 57 A particle of mass $m \mathrm {~kg}$ moves up a line of greatest slope of a rough plane inclined at $21 ^ { \circ }$ to the horizontal. The frictional and normal components of the contact force on the particle have magnitudes $F \mathrm {~N}$ and $R \mathrm {~N}$ respectively. The particle passes through the point $P$ with speed $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, and 2 s later it reaches its highest point on the plane.\\
(i) Show that $R = 9.336 m$ and $F = 1.416 m$, each correct to 4 significant figures.\\
(ii) Find the coefficient of friction between the particle and the plane.
After the particle reaches its highest point it starts to move down the plane.\\
(iii) Find the speed with which the particle returns to $P$.
\hfill \mbox{\textit{CAIE M1 2006 Q7 [11]}}