| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics B AS (Further Mechanics B AS) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Radial and transverse acceleration |
| Difficulty | Standard +0.3 This is a straightforward circular motion question requiring standard techniques: converting linear to angular speed (v=rΟ), integrating to find distance, and applying F=mvΒ²/r for the normal force. Part (c) involves solving a quadratic equation but all steps are routine applications of formulas with no novel problem-solving required. Slightly above average due to the multi-part nature and algebraic manipulation needed. |
| Spec | 3.03b Newton's first law: equilibrium3.03c Newton's second law: F=ma one dimension6.05a Angular velocity: definitions6.05b Circular motion: v=r*omega and a=v^2/r |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (a) | 0.2 0.4 0.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 0.8 0.8 0.8 | M1 | 1.2 |
| Answer | Marks | Guidance |
|---|---|---|
| π(5) = 8.875 (rad sβ1) | A1 | 1.1 |
| Answer | Marks |
|---|---|
| (b) | 0.2 0.4 |
| Answer | Marks | Guidance |
|---|---|---|
| 3 2 | M1 | 1.1a |
| Answer | Marks | Guidance |
|---|---|---|
| Dist = 0.367 (m) | A1 | 1.1 |
| Answer | Marks |
|---|---|
| (i) | 0.5π£2 |
| Answer | Marks | Guidance |
|---|---|---|
| 0.8 | M1 | 3.1b |
| v = 3.2 (m sβ1) | Or Ο = 4 | |
| 3.2 = 0.2π‘2+0.4π‘+0.1 and attempt to solve | M1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | A1 | 1.1 |
| Accept 3.1 | BC β donβt need to see the |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) | π = 0.4π‘+0.4 | M1 |
| Tang acc = 1.62 (m sβ2) | A1 | 1.1 |
| [1.6, 1.65] | Allow 2.0(31) rad sβ2 |
Question 1:
1 | (a) | 0.2 0.4 0.1
π = π‘2 + π‘+
0.8 0.8 0.8 | M1 | 1.2 | Or π£(5) = 0.2Γ52 +0.4Γ5+
0.1(=7.1)
π(5) = 8.875 (rad sβ1) | A1 | 1.1 | Accept 8.9
[2]
(b) | 0.2 0.4
π = π‘3 + π‘2 +0.1π‘ (+π)
3 2 | M1 | 1.1a | All powers and at least one coeff
correct
Dist = 0.367 (m) | A1 | 1.1 | 0.366666β¦ Accept 0.37 | BC allow
1
β« 0.2π‘2 +0.4π‘+0.1dπ‘
0
= 0.367
[2]
(c)
(i) | 0.5π£2
6.4 =
0.8 | M1 | 3.1b | OR 6.4 = 0.5Γ0.8Γπ2
v = 3.2 (m sβ1) | Or Ο = 4
3.2 = 0.2π‘2+0.4π‘+0.1 and attempt to solve | M1 | 1.1 | Or 4= 0.25π‘2+0.5π‘+0.125 | Attempt to solve can be
implied by correct answer
1
T = 3.06 orβ1+ β66 (s)
2 | A1 | 1.1 | 3.062019β¦ Not β5.06
Accept 3.1 | BC β donβt need to see the
method used to find root(s).
[3]
(ii) | π = 0.4π‘+0.4 | M1 | 3.1b | Or πΜ = 0.5π‘+0.5
Tang acc = 1.62 (m sβ2) | A1 | 1.1 | 1.6248β¦ Accept answers in range
[1.6, 1.65] | Allow 2.0(31) rad sβ2
[2]1 A small smooth ring of mass 0.5 kg is travelling round a smooth circular wire, with centre O and radius 0.8 m . The circle of wire is in a horizontal plane.
The speed of the ring, $v \mathrm {~ms} ^ { - 1 }$, at time $t \mathrm {~s}$ after passing through a point A on the wire is given by $\mathrm { v } = 0.2 \mathrm { t } ^ { 2 } + 0.4 \mathrm { t } + 0.1$.
\begin{enumerate}[label=(\alph*)]
\item Find the angular speed of the ring 5 seconds after it passes through A .
\item Find the distance the ring travels along the wire in the first second after passing through A .
At time $T$ s after the ring passes through A the magnitude of the force exerted on the ring by the wire is 6.4 N . You may assume that any forces acting on the ring other than the force exerted on the ring by the wire and gravity can be ignored.
\item \begin{enumerate}[label=(\roman*)]
\item Determine the value of $T$.
\item Hence find the tangential acceleration of the ring at this time.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics B AS 2022 Q1 [9]}}