| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | June |
| Marks | 1 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Find velocity from SUVAT |
| Difficulty | Easy -1.2 This is a straightforward application of a single SUVAT equation (v² = u² + 2as) with all values given directly in the question. It requires only substitution into a standard formula with no problem-solving, making it significantly easier than average A-level questions. |
| Spec | 3.02c Interpret kinematic graphs: gradient and area |
| Answer | Marks | Guidance |
|---|---|---|
| 2(i) | v= 2×2.5×5 (ms–1) | B1 |
| Total: | 1 | |
| 2(ii)(a) | M1 | Attempting PE loss or KE gain |
| Answer | Marks | Guidance |
|---|---|---|
| KE gain = 0.5 × 0.2 × (v 2 – 5 2) | A1 | Both PE and KE correct |
| Answer | Marks | Guidance |
|---|---|---|
| [6 = 0.1(v 2 – 52)] | M1 | PE loss = KE gain (3 terms) |
| v 2 = 85 → v = 9.22 ms–1 | A1 | |
| Total: | 4 | |
| Question | Answer | Marks |
| 2(ii)(b) | Max velocity at lowest point |
| Answer | Marks | Guidance |
|---|---|---|
| 0.5 × 0.2 × (v 2 – 5 2)] | M1 | PE loss = KE gain |
| v 2 = 145 → v = 12(.0) ms–1 | A1 | |
| Total: | 2 |
Question 2:
--- 2(i) ---
2(i) | v= 2×2.5×5 (ms–1) | B1 | AG Using v2 =u2 +2as
Total: | 1
2(ii)(a) | M1 | Attempting PE loss or KE gain
PE loss = 0.2 × 10 × 6 sin 30 [= 6]
and
KE gain = 0.5 × 0.2 × (v 2 – 5 2) | A1 | Both PE and KE correct
both unsimplified
[6 = 0.1(v 2 – 52)] | M1 | PE loss = KE gain (3 terms)
v 2 = 85 → v = 9.22 ms–1 | A1
Total: | 4
Question | Answer | Marks | Guidance
2(ii)(b) | Max velocity at lowest point
[0.2 × 10 × 6 =
0.5 × 0.2 × (v 2 – 5 2)] | M1 | PE loss = KE gain
v 2 = 145 → v = 12(.0) ms–1 | A1
Total: | 2\includegraphics{figure_2}
The diagram shows a wire $ABCD$ consisting of a straight part $AB$ of length $5$ m and a part $BCD$ in the shape of a semicircle of radius $6$ m and centre $O$. The diameter $BD$ of the semicircle is horizontal and $AB$ is vertical. A small ring is threaded onto the wire and slides along the wire. The ring starts from rest at $A$. The part $AB$ of the wire is rough, and the ring accelerates at a constant rate of $2.5$ m s$^{-2}$ between $A$ and $B$.
\begin{enumerate}[label=(\roman*)]
\item Show that the speed of the ring as it reaches $B$ is $5$ m s$^{-1}$.
[1]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2017 Q2 [1]}}