| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | November |
| Marks | 8 |
| 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 two-part mechanics problem requiring resolution of forces on an inclined plane and application of F=ma in both directions. While it involves multiple steps (resolving perpendicular and parallel to plane, finding friction coefficient, then reversing direction), these are routine M1 techniques with no novel insight required. Slightly above average difficulty due to the two-phase motion and coefficient calculation. |
| Spec | 3.03i Normal reaction force3.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(- F - 0.6g\sin 18° = 0.6(-4)\) | M1 | For using Newton's second law |
| Frictional component is 0.546N | A1 | |
| \([R = 0.6g\cos 18°]\) | M1 | For resolving forces normal to the plane |
| Normal component is 5.71N | A1 | 6 |
| Coefficient is 0.096 | B1ft | |
| (ii) \(0.6g\sin 18° - 0.546 = 0.6a\) or \(2(0.6g\sin 18°) = 0.6(a + 4)\) | B1ft | |
| \(a = 2.18\) | B1 | 2 |
| SR For candidates who use 'a' for the upwards acceleration, instead of as defined in the question. \(-0.6g\sin 18° + 0.546 = 0.6a \Rightarrow a = -2.18\) | B1 | |
| \(a = 2.18\) accompanied by satisfactory explanation for dropping the minus sign. | B1 |
(i) $- F - 0.6g\sin 18° = 0.6(-4)$ | M1 | For using Newton's second law
Frictional component is 0.546N | A1 |
$[R = 0.6g\cos 18°]$ | M1 | For resolving forces normal to the plane
Normal component is 5.71N | A1 | 6 |
Coefficient is 0.096 | B1ft |
(ii) $0.6g\sin 18° - 0.546 = 0.6a$ or $2(0.6g\sin 18°) = 0.6(a + 4)$ | B1ft |
$a = 2.18$ | B1 | 2 |
SR For candidates who use 'a' for the upwards acceleration, instead of as defined in the question. $-0.6g\sin 18° + 0.546 = 0.6a \Rightarrow a = -2.18$ | B1 |
$a = 2.18$ accompanied by satisfactory explanation for dropping the minus sign. | B1 |5 A particle $P$ of mass 0.6 kg moves upwards along a line of greatest slope of a plane inclined at $18 ^ { \circ }$ to the horizontal. The deceleration of $P$ is $4 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
(i) Find the frictional and normal components of the force exerted on $P$ by the plane. Hence find the coefficient of friction between $P$ and the plane, correct to 2 significant figures.
After $P$ comes to instantaneous rest it starts to move down the plane with acceleration $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
(ii) Find the value of $a$.
\hfill \mbox{\textit{CAIE M1 2009 Q5 [8]}}