| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Motion up rough slope |
| Difficulty | Moderate -0.3 This is a straightforward kinematics problem on a smooth slope requiring application of standard SUVAT equations and resolution of forces. Students need to find deceleration using v² = u² + 2as, then equate this to g sin α. It's slightly easier than average because it's a direct two-step application of standard M1 techniques with no conceptual complications or problem-solving insight required. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03g Gravitational acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \([1.5^2 = 2.5^2 + 2a \times 4]\) | M1 | For using \(v^2 = u^2 + 2as\) |
| Deceleration is \(0.5\ \text{ms}^{-2}\) | A1 | Accept \(a = -0.5\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| M1 | For using Newton's second law or \(a = (-)g\sin\alpha\) or \(\frac{1}{2}m(v_B^2 - v_A^2) = mg(AB)\sin\alpha\) | |
| \(\alpha = 2.9\) | A1ft | ft \(\alpha = \sin^{-1}(-0.1a)\) |
## Question 1:
### Part (i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $[1.5^2 = 2.5^2 + 2a \times 4]$ | M1 | For using $v^2 = u^2 + 2as$ |
| Deceleration is $0.5\ \text{ms}^{-2}$ | A1 | Accept $a = -0.5$ |
### Part (ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| | M1 | For using Newton's second law or $a = (-)g\sin\alpha$ or $\frac{1}{2}m(v_B^2 - v_A^2) = mg(AB)\sin\alpha$ |
| $\alpha = 2.9$ | A1ft | ft $\alpha = \sin^{-1}(-0.1a)$ |
---1\\
\includegraphics[max width=\textwidth, alt={}, center]{f7a22c07-44e3-4891-be60-cbab772f45df-2_203_1200_264_475}
A particle slides up a line of greatest slope of a smooth plane inclined at an angle $\alpha ^ { \circ }$ to the horizontal. The particle passes through the points $A$ and $B$ with speeds $2.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and $1.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ respectively. The distance $A B$ is 4 m (see diagram). Find\\
(i) the deceleration of the particle,\\
(ii) the value of $\alpha$.
\hfill \mbox{\textit{CAIE M1 2007 Q1 [4]}}