| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics A AS (Further Mechanics A AS) |
| Year | 2020 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Energy methods for pulley systems |
| Difficulty | Standard +0.3 This is a standard Further Mechanics energy method problem with connected particles. Part (a) requires straightforward application of energy conservation with one friction force, while part (b) involves friction on an inclined plane. The calculations are multi-step but follow routine procedures for A-level Further Maths mechanics, making it slightly easier than average. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (a) | 5.2g×1.6−1.6F = 1×4.4×3.52 + 1×5.2×3.52 |
| 2 2 | B1 |
| Answer | Marks |
|---|---|
| M1 | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | Either kinetic energy term correct |
| Answer | Marks | Guidance |
|---|---|---|
| ⇒F =14.21 | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | Block A gains 4.4g×1.6sinθ(=68.992sinθ) of GPE | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 11 | B1 | 3.1b |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | B1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | M1 | 3.3 |
| Answer | Marks | Guidance |
|---|---|---|
| ⇒3cosθ+11sinθ=9 | A1 | 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| Acceleration a = 0.153125 (m s-2) | B1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 11 | B1 | 3.1b |
| 5.2g−T −14.21=5.2a | M1 | 3.3 |
| Answer | Marks | Guidance |
|---|---|---|
| 11 | M1 | 3.3 |
| ⇒3cosθ+11sinθ=9 | A1 | 2.2a |
Question 7:
7 | (a) | 5.2g×1.6−1.6F = 1×4.4×3.52 + 1×5.2×3.52
2 2 | B1
B1
M1 | 1.1
1.1
1.1 | Either kinetic energy term correct
Correct GPE term
Attempt at WEP: all terms present; signs may not all
be correct.
No credit for non-energy methods
⇒F =14.21 | A1 | 1.1 | AG
[4]
(b) | Block A gains 4.4g×1.6sinθ(=68.992sinθ) of GPE | B1 | 1.1
Friction on A = 3 ×4.4gcosθ(=11.76cosθ)
11 | B1 | 3.1b | Need not be seen separately
1×(4.4+5.2)×0.72
2 | B1 | 1.1 | Gain in KE of both blocks
14.21×1.6+1.6× 3 ×4.4gcosθ
11
=−1×(4.4+5.2)×0.72 −68.992sinθ+5.2g×1.6
2 | M1 | 3.3 | Attempt at WEP: all terms present; signs may not all
be correct
⇒3cosθ+11sinθ=9 | A1 | 2.2a | ‘k =9’ need not be explicitly stated
[5]
Non-energy method
Acceleration a = 0.153125 (m s-2) | B1 | 1.1
Friction on A = 3 ×4.4gcosθ(=11.76cosθ)
11 | B1 | 3.1b | Need not be seen separately
5.2g−T −14.21=5.2a | M1 | 3.3 | N2L for B: all terms present (T = 35.95375)
T −4.4gsinθ− 3 ×4.4gcosθ=4.4a
11 | M1 | 3.3 | N2L for A: all terms present
⇒3cosθ+11sinθ=9 | A1 | 2.2a | ‘k =9’ need not be explicitly stated
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recorded or monitored7 Fig. 7.1 shows one end of a light inextensible string attached to a block A of mass 4.4 kg . The other end of the string is attached to a block B of mass 5.2 kg .
Block A is in contact with a smooth horizontal plane. The string is taut and passes over a small smooth pulley at the end of the plane. Block B is inside a hollow vertical tube and the vertical sides of B are in contact with the tube. Initially B is 1.6 m above the horizontal base of the tube.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b20e2254-955e-466c-8161-9614d8ccdba0-7_641_771_559_264}
\captionsetup{labelformat=empty}
\caption{Fig. 7.1}
\end{center}
\end{figure}
The blocks are released from rest. It may be assumed that in the subsequent motion A does not reach the pulley and the string remains taut.
Block B reaches the base of the tube with speed $3.5 \mathrm {~ms} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Given that the frictional force exerted by the tube on B is constant, use an energy method to show that the magnitude of this force is 14.21 N .
Blocks A and B remain attached to the opposite ends of a light inextensible string, but A is now in contact with a rough plane inclined at $\theta ^ { \circ }$ to the horizontal, as shown in Fig. 7.2.
The string connecting A and B is taut and passes over a small smooth pulley at the top of the plane. Block B is inside the same hollow vertical tube as before with the vertical sides of B in contact with the tube. It may be assumed that the frictional force exerted by the tube on B remains unchanged.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b20e2254-955e-466c-8161-9614d8ccdba0-8_623_723_552_260}
\captionsetup{labelformat=empty}
\caption{Fig. 7.2}
\end{center}
\end{figure}
The coefficient of friction between block A and the plane is $\frac { 3 } { 11 }$.\\
The blocks are released from rest, with block B 1.6 m above the base of the tube. It may be assumed that in the subsequent motion A does not reach the pulley and the string remains taut.
\item Given that block B reaches the base of the tube with speed $0.7 \mathrm {~ms} ^ { - 1 }$, show that $\theta$ satisfies the equation\\
$3 \cos \theta + 11 \sin \theta = k$,\\
where $k$ is a constant to be determined.
\section*{END OF QUESTION PAPER}
\section*{}
\begin{itemize}
\item \item \item \end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics A AS 2020 Q7 [9]}}