| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Heavier particle hits ground, lighter continues upward - inclined plane involved |
| Difficulty | Standard +0.3 This is a standard two-particle pulley system with friction on an incline. Part (i) requires equilibrium with limiting friction (straightforward force resolution). Part (ii) involves finding acceleration in two phases (before and after B hits ground) and using kinematics. While multi-step, these are routine M1 techniques with no novel insight required, making it slightly easier than average. |
| Spec | 3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys3.03u Static equilibrium: on rough surfaces3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(R = mg\cos 30\) | B1 | Resolves normally |
| \(F = 2m\cos 30\ [= m\sqrt{3}]\) | M1 | Uses \(F = \mu R\) |
| \(T = 4g\ [= 40]\) | B1 | Particle \(B\) |
| \(T = mg\sin 30 + F\) | M1 | Resolves parallel to plane for particle \(A\) |
| \(40 = 5m + m\sqrt{3}\) | A1 | Equation in \(m\) |
| \(m = \frac{40}{5+\sqrt{3}} = 5.94\) | A1 | AG All correct and no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([R = 3g\cos 30]\), \(F = 0.2 \times 3g\cos 30\ (3\sqrt{3} = 5.196)\) | (B1) | |
| \(4g - T = 4a\) or \(T - 3g\sin 30 - F = 3a\) or \(4g - 3g\sin 30 - F = 7a\) | M1 | Applies Newton's Second Law to one of the particles or forms system equation in \(a\); \((m_B g - m_A g\sin 30 - F = (m_A + m_B)a)\) |
| \(T - 3g\sin 30 - 3\sqrt{3} = 3a\) or \(40 - T = 4a\) or \(4g - 3g\sin 30 - 3\sqrt{3} = 7a \rightarrow a = \ldots\) | M1 | Applies Newton's Second Law to form second equation in \(T\) and \(a\) and solves for \(a\) or solves system equation for \(a\) |
| \(a = \frac{25 - 3\sqrt{3}}{7} = 2.83\) | A1 | |
| \(v^2 = 2 \times 2.83 \times 0.5\), \(v = 1.68\ldots\) | B1 FT | \(v\) as \(T\) becomes zero; FT on \(a\) |
| \(-3g\sin 30 - 0.2(3g\cos 30) = 3a\), \(-15 - 3\sqrt{3} = 3a \rightarrow a = \ldots(-5 - \sqrt{3} = -6.73)\) | M1 | Applies Newton's Second Law and solves for \(a\) |
| \(0 = 1.68^2 - 2 \times 6.73s\), \(s = \ldots(0.210)\) | M1 | Uses \(v^2 = u^2 + 2as\) and solves for \(s\) |
| Total distance \(= 0.710\) m | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| (implied) | M1 | For 4kg mass, uses \(PE_{loss} - WD_T = KE_{gain}\) |
| (implied) | M1 | For 3kg mass, uses \(WD_T = KE_{gain} + PE_{gain} + WD_{Fr}\) |
| \(4g(0.5) - 0.5T = \frac{1}{2}(4v^2)\) and \(0.5T = \frac{1}{2}(3v^2) + 3g(0.5\sin30) + 3\sqrt{3}(0.5)\) | A1 | |
| \(v^2 = (25 - 3\sqrt{3})/7\) or \(v = 1.68\) | B1 | |
| \(\frac{1}{2}(3)(1.68)^2 = 3g(s\sin30) + 3\sqrt{3}s\) | M1 | For 3kg mass, uses \(KE_{loss} = PE_{gain} + WD_{Fr}\) |
| \(s = \ldots(0.210)\) | M1 | Solves for \(s\) |
| Total distance \(= 0.710\) m | A1 | |
| Total: | 8 |
## Question 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $R = mg\cos 30$ | B1 | Resolves normally |
| $F = 2m\cos 30\ [= m\sqrt{3}]$ | M1 | Uses $F = \mu R$ |
| $T = 4g\ [= 40]$ | B1 | Particle $B$ |
| $T = mg\sin 30 + F$ | M1 | Resolves parallel to plane for particle $A$ |
| $40 = 5m + m\sqrt{3}$ | A1 | Equation in $m$ |
| $m = \frac{40}{5+\sqrt{3}} = 5.94$ | A1 | AG All correct and no errors seen |
## Question 7(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[R = 3g\cos 30]$, $F = 0.2 \times 3g\cos 30\ (3\sqrt{3} = 5.196)$ | (B1) | |
| $4g - T = 4a$ or $T - 3g\sin 30 - F = 3a$ or $4g - 3g\sin 30 - F = 7a$ | M1 | Applies Newton's Second Law to one of the particles or forms system equation in $a$; $(m_B g - m_A g\sin 30 - F = (m_A + m_B)a)$ |
| $T - 3g\sin 30 - 3\sqrt{3} = 3a$ or $40 - T = 4a$ or $4g - 3g\sin 30 - 3\sqrt{3} = 7a \rightarrow a = \ldots$ | M1 | Applies Newton's Second Law to form second equation in $T$ and $a$ and solves for $a$ or solves system equation for $a$ |
| $a = \frac{25 - 3\sqrt{3}}{7} = 2.83$ | A1 | |
| $v^2 = 2 \times 2.83 \times 0.5$, $v = 1.68\ldots$ | B1 FT | $v$ as $T$ becomes zero; FT on $a$ |
| $-3g\sin 30 - 0.2(3g\cos 30) = 3a$, $-15 - 3\sqrt{3} = 3a \rightarrow a = \ldots(-5 - \sqrt{3} = -6.73)$ | M1 | Applies Newton's Second Law and solves for $a$ |
| $0 = 1.68^2 - 2 \times 6.73s$, $s = \ldots(0.210)$ | M1 | Uses $v^2 = u^2 + 2as$ and solves for $s$ |
| Total distance $= 0.710$ m | A1 | |
## Mark Scheme Content
**Question (continued):**
| Answer | Mark | Guidance |
|--------|------|----------|
| (implied) | **M1** | For 4kg mass, uses $PE_{loss} - WD_T = KE_{gain}$ |
| (implied) | **M1** | For 3kg mass, uses $WD_T = KE_{gain} + PE_{gain} + WD_{Fr}$ |
| $4g(0.5) - 0.5T = \frac{1}{2}(4v^2)$ and $0.5T = \frac{1}{2}(3v^2) + 3g(0.5\sin30) + 3\sqrt{3}(0.5)$ | **A1** | |
| $v^2 = (25 - 3\sqrt{3})/7$ or $v = 1.68$ | **B1** | |
| $\frac{1}{2}(3)(1.68)^2 = 3g(s\sin30) + 3\sqrt{3}s$ | **M1** | For 3kg mass, uses $KE_{loss} = PE_{gain} + WD_{Fr}$ |
| $s = \ldots(0.210)$ | **M1** | Solves for $s$ |
| Total distance $= 0.710$ m | **A1** | |
| **Total:** | **8** | |7\\
\includegraphics[max width=\textwidth, alt={}, center]{3d7f53af-dbf2-499b-9966-ae85514cef02-10_336_803_258_671}
Two particles $A$ and $B$ of masses $m \mathrm {~kg}$ and 4 kg respectively are connected by a light inextensible string that passes over a fixed smooth pulley. Particle $A$ is on a rough fixed slope which is at an angle of $30 ^ { \circ }$ to the horizontal ground. Particle $B$ hangs vertically below the pulley and is 0.5 m above the ground (see diagram). The coefficient of friction between the slope and particle $A$ is 0.2 .\\
(i) In the case where the system is in equilibrium with particle $A$ on the point of moving directly up the slope, show that $m = 5.94$, correct to 3 significant figures.\\
(ii) In the case where $m = 3$, the system is released from rest with the string taut. Find the total distance travelled by $A$ before coming to instantaneous rest. You may assume that $A$ does not reach the pulley.\\
\hfill \mbox{\textit{CAIE M1 2017 Q7 [14]}}