| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Variable mass or unknown mass |
| Difficulty | Standard +0.8 This is a multi-part pulley problem requiring kinematics (finding acceleration from v²=u²+2as), Newton's second law for connected particles (finding tension and unknown mass m), energy considerations (maximum height after P hits ground), and analysis of forces on the pulley system itself. While each individual step uses standard M1 techniques, the question requires careful sequencing through multiple phases of motion and the final part about chain tension requires insight about the pulley as a separate body. This is more demanding than typical M1 pulley questions but doesn't require novel mathematical insight. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03k Connected particles: pulleys and equilibrium6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1.4^2 = 2 \times a \times 0.2\) OR \(0.2=(0+1.4)t/2\) and \(1.4=0+at\) | M1 | Any use of \(a = g\) is M0 |
| \(a = 4.9\) m s\(^{-2}\) | A1 | \(t = 2/7\) hence \(1.4 = a \times 2/7\) |
| \(0.3g - T = +/- 0.3 \times 4.9\) | M1 | N2L diff of weight and tension. Any use of \(a = g\) is M0 |
| \(T = 1.47\) N | A1 | |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(+/-4.9m = 1.47 - mg\) | M1 | N2L for \(Q\) using values from (i), \(a\) not \(g\); accept \(a = g\Delta M/\Sigma M\) |
| \(4.9m = 1.47 - mg\) | A1ft | Diff cv(\(T\)) and \(mg\) correct way round; ft cv(\(T,a\)); \(4.9 = g(0.3-m)/(0.3+m)\) M1A1; ftcv(\(a\)) |
| \(m = 0.1\) | A1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1.4^2 = 2gs\) | M1 | Accn \(= g\) |
| \(s = 0.1\) | A1 | may be implied (eg \(H = 0.3\)) BoD sign uncertainty |
| \(H = 0.2 + 0.2 + 0.1\) | M1 | Needs 0.2 twice |
| \(H = 0.5\) m | A1 | |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Tension \(= 0.5g + 2 \times 1.47\) | M1 | |
| Tension \(= 7.84\) N | A1 | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Tension \((= 0.5g) = 4.9\) N | B1 | |
| [1] |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1.4^2 = 2 \times a \times 0.2$ OR $0.2=(0+1.4)t/2$ and $1.4=0+at$ | M1 | Any use of $a = g$ is M0 |
| $a = 4.9$ m s$^{-2}$ | A1 | $t = 2/7$ hence $1.4 = a \times 2/7$ |
| $0.3g - T = +/- 0.3 \times 4.9$ | M1 | N2L diff of weight and tension. Any use of $a = g$ is M0 |
| $T = 1.47$ N | A1 | |
| **[4]** | | |
## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $+/-4.9m = 1.47 - mg$ | M1 | N2L for $Q$ using values from (i), $a$ not $g$; accept $a = g\Delta M/\Sigma M$ |
| $4.9m = 1.47 - mg$ | A1ft | Diff cv($T$) and $mg$ correct way round; ft cv($T,a$); $4.9 = g(0.3-m)/(0.3+m)$ M1A1; ftcv($a$) |
| $m = 0.1$ | A1 | |
| **[3]** | | |
## Question 5(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1.4^2 = 2gs$ | M1 | Accn $= g$ |
| $s = 0.1$ | A1 | may be implied (eg $H = 0.3$) BoD sign uncertainty |
| $H = 0.2 + 0.2 + 0.1$ | M1 | Needs 0.2 twice |
| $H = 0.5$ m | A1 | |
| **[4]** | | |
## Question 5(iv)(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Tension $= 0.5g + 2 \times 1.47$ | M1 | |
| Tension $= 7.84$ N | A1 | |
| **[2]** | | |
## Question 5(iv)(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Tension $(= 0.5g) = 4.9$ N | B1 | |
| **[1]** | | |5\\
\includegraphics[max width=\textwidth, alt={}, center]{f5085265-5258-45d4-8233-6bd68f8e9034-3_462_405_258_845}
A small smooth pulley is suspended from a fixed point by a light chain. A light inextensible string passes over the pulley. Particles $P$ and $Q$, of masses 0.3 kg and $m \mathrm {~kg}$ respectively, are attached to the opposite ends of the string. The particles are released from rest at a height of 0.2 m above horizontal ground with the string taut; the portions of the string not in contact with the pulley are vertical (see diagram). $P$ strikes the ground with speed $1.4 \mathrm {~ms} ^ { - 1 }$. Subsequently $P$ remains on the ground, and $Q$ does not reach the pulley.\\
(i) Calculate the acceleration of $P$ while it is in motion and the corresponding tension in the string.\\
(ii) Find the value of $m$.\\
(iii) Calculate the greatest height of $Q$ above the ground.\\
(iv) It is given that the mass of the pulley is 0.5 kg . State the magnitude of the tension in the chain which supports the pulley
\begin{enumerate}[label=(\alph*)]
\item when $P$ is in motion,
\item when $P$ is at rest on the ground and $Q$ is moving upwards.
\end{enumerate}
\hfill \mbox{\textit{OCR M1 2013 Q5 [14]}}