| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Connected particles with pulley |
| Difficulty | Standard +0.3 This is a standard M2 connected particles problem with straightforward energy conservation applications. Part (a) involves routine setup of energy equations for a pulley system with clear conceptual questions, while part (b) is a direct power calculation using P=Fv. The multi-part structure and 'show understanding' elements are typical, but no step requires novel insight beyond applying standard energy methods. |
| Spec | 6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.02i Conservation of energy: mechanical energy principle6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Yes. Only WD is against conservative forces. | E1 | Accept only WD is against gravity or no work done against friction. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Block has no displacement in that direction | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(0.5 \times 50 \times 1.5^2 = 20gx - 5gx\) | M1 | Use of WE with KE. Allow \(m = 25\) |
| B1 | Use of 50 | |
| M1 | At least 1 GPE term | |
| A1 | GPE terms correct signs | |
| \(x = 0.38265\ldots\) so \(0.383\) m (3 s.f.) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(0.5 \times 50 \times V^2 - 0.5 \times 50 \times 1.5^2\) | M1 | WE equation with WD term. Allow GPE terms missing |
| \(= 2 \times 20g - 2 \times 5g - 180\) | B1 | Both KE terms. Accept use of 25 |
| B1 | Either GPE term | |
| B1 | 180 with correct sign | |
| \(V = 2.6095\ldots\) so \(2.61\) m s\(^{-1}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Force down the slope is \(2000 + 450g\sin 20\) | M1 | Both terms. Allow mass not weight |
| B1 | Weight term correct | |
| Using \(P = Fv\) | M1 | |
| \(P = (2000 + 450g\sin 20) \times 2.5\) | F1 | FT their weight term |
| \(P = 8770.77\ldots\) so \(8770\) W (3 s.f.) | A1 | cao |
# Question 2:
## Part (a)(i)(A)
| Answer | Mark | Guidance |
|--------|------|----------|
| Yes. Only WD is against conservative forces. | E1 | Accept only WD is against gravity or no work done against friction. |
## Part (a)(i)(B)
| Answer | Mark | Guidance |
|--------|------|----------|
| Block has no displacement in that direction | E1 | |
## Part (a)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $0.5 \times 50 \times 1.5^2 = 20gx - 5gx$ | M1 | Use of WE with KE. Allow $m = 25$ |
| | B1 | Use of 50 |
| | M1 | At least 1 GPE term |
| | A1 | GPE terms correct signs |
| $x = 0.38265\ldots$ so $0.383$ m (3 s.f.) | A1 | cao |
## Part (a)(iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $0.5 \times 50 \times V^2 - 0.5 \times 50 \times 1.5^2$ | M1 | WE equation with WD term. Allow GPE terms missing |
| $= 2 \times 20g - 2 \times 5g - 180$ | B1 | Both KE terms. Accept use of 25 |
| | B1 | Either GPE term |
| | B1 | 180 with correct sign |
| $V = 2.6095\ldots$ so $2.61$ m s$^{-1}$ | A1 | cao |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| Force down the slope is $2000 + 450g\sin 20$ | M1 | Both terms. Allow mass not weight |
| | B1 | Weight term correct |
| Using $P = Fv$ | M1 | |
| $P = (2000 + 450g\sin 20) \times 2.5$ | F1 | FT **their** weight term |
| $P = 8770.77\ldots$ so $8770$ W (3 s.f.) | A1 | cao |
---2
\begin{enumerate}[label=(\alph*)]
\item A small block of mass 25 kg is on a long, horizontal table. Each side of the block is connected to a small sphere by means of a light inextensible string passing over a smooth pulley. Fig. 2 shows this situation. Sphere A has mass 5 kg and sphere B has mass 20 kg . Each of the spheres hangs freely.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{81efb50d-c89d-4ce1-94d7-592c946f6176-3_487_1123_466_552}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
Initially the block moves on a smooth part of the table. With the block at a point O , the system is released from rest with both strings taut.
\begin{enumerate}[label=(\roman*)]
\item (A) Is mechanical energy conserved in the subsequent motion? Give a brief reason for your answer.\\
(B) Why is no work done by the block against the reaction of the table on it?
The block reaches a speed of $1.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at point P .
\item Use an energy method to calculate the distance OP.
The block continues moving beyond P , at which point the table becomes rough. After travelling two metres beyond P , the block passes through point Q . The block does 180 J of work against resistances to its motion from P to Q .
\item Use an energy method to calculate the speed of the block at Q .
\end{enumerate}\item A tree trunk of mass 450 kg is being pulled up a slope inclined at $20 ^ { \circ }$ to the horizontal.
Calculate the power required to pull the trunk at a steady speed of $2.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ against a frictional force of 2000 N .
\end{enumerate}
\hfill \mbox{\textit{OCR MEI M2 2009 Q2 [17]}}