| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Variable mass or unknown mass |
| Difficulty | Standard +0.3 This is a standard M1 pulley problem with straightforward application of Newton's second law, SUVAT equations, and conservation of energy. While it has multiple parts and requires finding an unknown mass, each step follows directly from standard mechanics techniques without requiring novel insight or complex problem-solving. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03l Newton's third law: extend to situations requiring force resolution3.03o Advanced connected particles: and pulleys6.02e Calculate KE and PE: using formulae6.03a Linear momentum: p = mv6.03b Conservation of momentum: 1D two particles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(T - 0.4g = 0.4 \times 2.45\) | M1 | N2L on \(P\), two vertical forces, accept with \(0.4\times2.45g\) |
| \(T = 4.9 \text{ N}\) | A1 | Correct terms and signs |
| A1 | Exact, \(g=9.81\) (4.904, accept 4.9) \(g=10\) (4.98, not 5.0) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(mg - T = \pm2.45m\) | M1 | Correct terms (possible incorrect signs), and use of \(\text{cv}(T(\text{i}))\) |
| \(m = 2/3 \text{ kg}\) | A1 FT | FT \(\text{cv}(T(\text{i}))/7.35\), \(g=9.81\) (FT \(\text{cv}(T(\text{i}))/7.351 = 0.667\)) \(g=10\) (FT \(\text{cv}(T(\text{i}))/7.55 = 0.6596 = 0.66\)) |
| \(v = 2.45 \times 0.3\) \((= 0.735)\) | B1 | Must be positive |
| Momentum \(= (2/3) \times (2.45 \times 0.3)\) | M1 | Accept \(\pm\). \(\text{cv}(m) \times \text{cv}(v)\) |
| Momentum loss \(= 0.49 \text{ kgms}^{-1}\) | A1 | Exact, but accept any value which rounds to \(\pm0.490\). \(g=9.81\) (0.49) \(g=10\) (0.4848=0.485, not 0.48) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(S = 2.45 \times 0.3^2/2\) | M1 | Distance while \(Q\) descends. Watch for \(s = vt - at^2/2\). If \(v=0\), M0A0 |
| \(S = \pm0.11(025)\) | A1 | |
| OR \(S = (0 + 0.735)\times0.3/2\) | M1 Using landing speed from (ii) A1 | |
| \(0 = (2.45 \times 0.3)^2 \pm 2 \times 9.8s\) | M1 | Distance \(P\) ascends while \(Q\) at rest, must use \(g\) |
| \(s = \pm0.027(56..)\) | A1 | May be implied, \(g=9.81\) (0.02753) \(g=10\) (0.0270) |
| OR using \(t_A = 0.735/9.8 = 0.075\) | Calculating ascend time after string goes slack | |
| \(s = 0.735\times0.075 - 9.8\times0.075^2/2\) | M1 | M1 Using candidate's values of speed and \(t_A\) to find \(\pm s\) |
| \(s = \pm0.027(56..)\) | A1 May be implied | |
| Distance \(= 0.248 \text{ m}\) | A1 FT | \(2\times |
# Question 5:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $T - 0.4g = 0.4 \times 2.45$ | M1 | N2L on $P$, two vertical forces, accept with $0.4\times2.45g$ |
| $T = 4.9 \text{ N}$ | A1 | Correct terms and signs |
| | A1 | Exact, $g=9.81$ (4.904, accept 4.9) $g=10$ (4.98, not 5.0) |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $mg - T = \pm2.45m$ | M1 | Correct terms (possible incorrect signs), and use of $\text{cv}(T(\text{i}))$ |
| $m = 2/3 \text{ kg}$ | A1 FT | FT $\text{cv}(T(\text{i}))/7.35$, $g=9.81$ (FT $\text{cv}(T(\text{i}))/7.351 = 0.667$) $g=10$ (FT $\text{cv}(T(\text{i}))/7.55 = 0.6596 = 0.66$) |
| $v = 2.45 \times 0.3$ $(= 0.735)$ | B1 | Must be positive |
| Momentum $= (2/3) \times (2.45 \times 0.3)$ | M1 | Accept $\pm$. $\text{cv}(m) \times \text{cv}(v)$ |
| Momentum loss $= 0.49 \text{ kgms}^{-1}$ | A1 | Exact, but accept any value which rounds to $\pm0.490$. $g=9.81$ (0.49) $g=10$ (0.4848=0.485, not 0.48) |
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $S = 2.45 \times 0.3^2/2$ | M1 | Distance while $Q$ descends. Watch for $s = vt - at^2/2$. If $v=0$, M0A0 |
| $S = \pm0.11(025)$ | A1 | |
| OR $S = (0 + 0.735)\times0.3/2$ | | M1 Using landing speed from (ii) A1 |
| $0 = (2.45 \times 0.3)^2 \pm 2 \times 9.8s$ | M1 | Distance $P$ ascends while $Q$ at rest, must use $g$ |
| $s = \pm0.027(56..)$ | A1 | May be implied, $g=9.81$ (0.02753) $g=10$ (0.0270) |
| OR using $t_A = 0.735/9.8 = 0.075$ | | Calculating ascend time after string goes slack |
| $s = 0.735\times0.075 - 9.8\times0.075^2/2$ | M1 | M1 Using candidate's values of speed and $t_A$ to find $\pm s$ |
| $s = \pm0.027(56..)$ | | A1 May be implied |
| Distance $= 0.248 \text{ m}$ | A1 FT | $2\times|\text{cv}(S)| + |\text{cv}(s)|$. Accept 0.25. $g=9.81$ (0.248) $g=10$ (0.247511..) |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{f0813713-d677-4ed7-87e1-971a64bdb6ff-3_291_182_799_945}
Particles $P$ and $Q$, of masses 0.4 kg and $m \mathrm {~kg}$ respectively, are joined by a light inextensible string which passes over a smooth pulley. The particles are released from rest at the same height above a horizontal surface; the string is taut and the portions of the string not in contact with the pulley are vertical (see diagram). $Q$ begins to descend with acceleration $2.45 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ and reaches the surface 0.3 s after being released. Subsequently, $Q$ remains at rest and $P$ never reaches the pulley.\\
(i) Calculate the tension in the string while $Q$ is in motion.\\
(ii) Calculate the momentum lost by $Q$ when it reaches the surface.\\
(iii) Calculate the greatest height of $P$ above the surface.
\section*{[Questions 6 and 7 are printed overleaf.]}
\hfill \mbox{\textit{OCR M1 2012 Q5 [13]}}