| Exam Board | OCR |
|---|---|
| Module | Further Mechanics AS (Further Mechanics AS) |
| Session | Specimen |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Particle on circular wire/arc |
| Difficulty | Standard +0.3 This is a standard circular motion energy conservation problem with straightforward application of formulas. Part (i) uses basic energy conservation (KE + PE = constant), part (ii) is immediate from the result, and part (iii) requires simple calculus or recognition that v is maximum when cos θ = -1. All steps are routine for Further Maths students with no novel problem-solving required. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05d Variable speed circles: energy methods |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (i) | 1 |
| Answer | Marks |
|---|---|
| v2 (cid:32)33.32(cid:16)15.68cos(cid:84) | B1 |
| Answer | Marks |
|---|---|
| [4] | 1.1 |
| Answer | Marks |
|---|---|
| 2.1 | m |
| Answer | Marks |
|---|---|
| must be shown | Allow m used instead of 0.03 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (ii) | S |
| e.g. v(cid:32)0(cid:159)cos(cid:84)(cid:32)2.125(cid:33)1 so never at rest | E1 | |
| [1] | 2.2a | e.g. The speed of the bead at any point |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (iii) | v2 (cid:32)33.32(cid:16)15.68(cid:117)(cid:11)(cid:16)1(cid:12) |
| v(cid:32)7 | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 1.1 | Setting cos(cid:84)(cid:32)(cid:16)1 and solving for v |
Question 2:
2 | (i) | 1
(0.03)(cid:11)4.2(cid:12)2
Initial kinetic energy (cid:32)
2
At angle(cid:84),potential energy (cid:32)
(cid:16)0.03g(0.8(cid:16)0.8cos(cid:84))
1(0.03)(cid:11)4.2(cid:12)2 (cid:32)(cid:16)0.03g(0.8(cid:16)0.8cos(cid:84))(cid:14)1(0.03)v2
2 2
p
v2 (cid:32)33.32(cid:16)15.68cos(cid:84) | B1
B1
eM1
E1
[4] | 1.1
i
3.3
c
3.4
2.1 | m
Attempt at conservation of mechanical
energy (correct number of terms) and
rearranging to make v2 the subject
www; AG at least one intermediate step
must be shown | Allow m used instead of 0.03
Allow m used instead of 0.03
2 | (ii) | S
e.g. v(cid:32)0(cid:159)cos(cid:84)(cid:32)2.125(cid:33)1 so never at rest | E1
[1] | 2.2a | e.g. The speed of the bead at any point
below the top of the circle must be higher
than its initial speed because the bead has
lost PE and thus gained KE
2 | (iii) | v2 (cid:32)33.32(cid:16)15.68(cid:117)(cid:11)(cid:16)1(cid:12)
v(cid:32)7 | M1
A1
[2] | 1.1
1.1 | Setting cos(cid:84)(cid:32)(cid:16)1 and solving for v
--- 2(i) ---
2(i)
t
f
a
r
--- 2(ii) ---
2(ii)
D
--- 2(iii) ---
2(iii)\includegraphics{figure_2}
A smooth wire is shaped into a circle of centre $O$ and radius 0.8 m. The wire is fixed in a vertical plane. A small bead $P$ of mass 0.03 kg is threaded on the wire and is projected along the wire from the highest point with a speed of $4.2 \, \text{m s}^{-1}$. When $OP$ makes an angle $\theta$ with the upward vertical the speed of $P$ is $v \, \text{m s}^{-1}$ (see diagram).
\begin{enumerate}[label=(\roman*)]
\item Show that $v^2 = 33.32 - 15.68\cos\theta$. [4]
\item Prove that the bead is never at rest. [1]
\item Find the maximum value of $v$. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics AS Q2 [7]}}