| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2008 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Particle on inclined plane |
| Difficulty | Moderate -0.8 This is a straightforward two-part SUVAT question requiring only direct application of v = u + at to find acceleration, then resolving forces (mg sin α = ma) to find the angle. Both parts are standard textbook exercises with no problem-solving insight needed, making it easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03g Gravitational acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([4.5 = 1.5 + 1.2a]\) | M1 | For using \(v = u + at\) |
| Acceleration is \(2.5 \text{ ms}^{-2}\) | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\alpha = 14.5\) | M1, A1 [2] | For using \((m)g\sin\alpha° = (m)a\) |
## Question 1:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[4.5 = 1.5 + 1.2a]$ | M1 | For using $v = u + at$ |
| Acceleration is $2.5 \text{ ms}^{-2}$ | A1 [2] | |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\alpha = 14.5$ | M1, A1 [2] | For using $(m)g\sin\alpha° = (m)a$ |
---1 A particle slides down a smooth plane inclined at an angle of $\alpha ^ { \circ }$ to the horizontal. The particle passes through the point $A$ with speed $1.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, and 1.2 s later it passes through the point $B$ with speed $4.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Find\\
(i) the acceleration of the particle,\\
(ii) the value of $\alpha$.
\hfill \mbox{\textit{CAIE M1 2008 Q1 [4]}}