| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2006 |
| Session | January |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Connected particles pulley energy method |
| Difficulty | Standard +0.3 This is a standard M2 connected particles problem requiring energy methods. Part (i) is straightforward work-energy calculation. Parts (ii) and (iii) involve routine application of energy conservation with gravitational PE changes, friction work, and kinetic energy - all standard textbook techniques with clear scaffolding. The multi-part structure and numerical complexity place it slightly above average difficulty, but no novel insight is required. |
| Spec | 6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\dfrac{0.5 \times 20 \times 8^2 - 0.5 \times 20 \times 5^2 + 510}{6}\) | M1 | Use of \(P = \text{WD}/t\) |
| B1 | \(\Delta\)KE. Accept \(\pm 390\) soi | |
| A1 | All correct including signs | |
| \(= 150\) W | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(20g \times \dfrac{3}{5}x - 5gx\) | M1 | Use of \(mgh\) on both terms |
| B1 | Either term (neglecting signs) | |
| A1 | \(\pm 7gx\) in any form | |
| \(7gx\) (68.6\(x\)) gain | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(11gx\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(0.5 \times 25 \times 4^2 = 7gx + 11gx = 18gx\) | M1 | Use of work-energy equation. Allow 1 RHS term omitted. |
| B1 | KE term correct | |
| \(x = 1.13378...\) so 1.13 m (3 s.f.) | A1 | cao. Except follow wrong sign for \(7gx\) only. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| either | ||
| \(0.5 \times 35 \times v^2 - 0.5 \times 35 \times 16 = 15g \times 0.5 - 11g \times 0.5 - 12g \times 0.5\) | M1 | Use of work-energy. KE, GPE and WD against friction terms present. |
| B1 | \(\Delta\)GPE correct inc sign (1.5\(g\) J loss) | |
| A1 | All correct | |
| \(v^2 = 13.76\) so \(v = 3.70944...\) so \(3.71\ \text{m s}^{-1}\) (3 s.f.) | A1 | cao |
| or | ||
| \(15g - T = 15a \quad T - 12g - 11g = 20a\) | M1 | N2L in 1 or 2 equations. All terms present |
| so \(a = -2.24\) | A1 | cao |
| \(v^2 = 4^2 + 2 \times (-2.24) \times 0.5\) | M1 | Use of appropriate (sequence of) \(uvast\) |
| so \(3.71\ \text{m s}^{-1}\) (3 s.f.) | A1 | cao |
# Question 4:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\dfrac{0.5 \times 20 \times 8^2 - 0.5 \times 20 \times 5^2 + 510}{6}$ | M1 | Use of $P = \text{WD}/t$ |
| | B1 | $\Delta$KE. Accept $\pm 390$ soi |
| | A1 | All correct including signs |
| $= 150$ W | A1 | |
## Part (ii)(A)
| Answer | Mark | Guidance |
|--------|------|----------|
| $20g \times \dfrac{3}{5}x - 5gx$ | M1 | Use of $mgh$ on both terms |
| | B1 | Either term (neglecting signs) |
| | A1 | $\pm 7gx$ in any form |
| $7gx$ (68.6$x$) gain | A1 | cao |
## Part (ii)(B)
| Answer | Mark | Guidance |
|--------|------|----------|
| $11gx$ | B1 | |
## Part (ii)(C)
| Answer | Mark | Guidance |
|--------|------|----------|
| $0.5 \times 25 \times 4^2 = 7gx + 11gx = 18gx$ | M1 | Use of work-energy equation. Allow 1 RHS term omitted. |
| | B1 | KE term correct |
| $x = 1.13378...$ so 1.13 m (3 s.f.) | A1 | cao. Except follow wrong sign for $7gx$ only. |
## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| **either** | | |
| $0.5 \times 35 \times v^2 - 0.5 \times 35 \times 16 = 15g \times 0.5 - 11g \times 0.5 - 12g \times 0.5$ | M1 | Use of work-energy. KE, GPE and WD against friction terms present. |
| | B1 | $\Delta$GPE correct inc sign (1.5$g$ J loss) |
| | A1 | All correct |
| $v^2 = 13.76$ so $v = 3.70944...$ so $3.71\ \text{m s}^{-1}$ (3 s.f.) | A1 | cao |
| **or** | | |
| $15g - T = 15a \quad T - 12g - 11g = 20a$ | M1 | N2L in 1 or 2 equations. All terms present |
| so $a = -2.24$ | A1 | cao |
| $v^2 = 4^2 + 2 \times (-2.24) \times 0.5$ | M1 | Use of appropriate (sequence of) $uvast$ |
| so $3.71\ \text{m s}^{-1}$ (3 s.f.) | A1 | cao |4 A block of mass 20 kg is pulled by a light, horizontal string over a rough, horizontal plane. During 6 seconds, the work done against resistances is 510 J and the speed of the block increases from $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ to $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\roman*)]
\item Calculate the power of the pulling force.
The block is now put on a rough plane that is at an angle $\alpha$ to the horizontal, where $\sin \alpha = \frac { 3 } { 5 }$. The frictional resistance to sliding is $11 g \mathrm {~N}$. A light string parallel to the plane is connected to the block. The string passes over a smooth pulley and is connected to a freely hanging sphere of mass $m \mathrm {~kg}$, as shown in Fig. 4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c1785fde-a6ce-4f8b-9948-4b4dd973ce84-6_348_855_847_605}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}
In parts (ii) and (iii), the sphere is pulled downwards and then released when travelling at a speed of $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ vertically downwards. The block never reaches the pulley.
\item Suppose that $m = 5$ and that after the sphere is released the block moves $x \mathrm {~m}$ up the plane before coming to rest.\\
(A) Find an expression in terms of $x$ for the change in gravitational potential energy of the system, stating whether this is a gain or a loss.\\
(B) Find an expression in terms of $x$ for the work done against friction.\\
(C) Making use of your answers to parts (A) and (B), find the value of $x$.
\item Suppose instead that $m = 15$. Calculate the speed of the sphere when it has fallen a distance 0.5 m from its point of release.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI M2 2006 Q4 [16]}}