| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2021 |
| Session | March |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Applied force in addition to weights |
| Difficulty | Standard +0.3 This is a standard two-particle pulley system with inclined planes requiring resolution of forces and Newton's second law in part (a), and energy methods in part (b). While it involves multiple components (applied force, two different angles, friction in part b), the techniques are routine for M1 level with clear setup and straightforward algebra—slightly easier than average A-level mechanics. |
| Spec | 3.03k Connected particles: pulleys and equilibrium3.03l Newton's third law: extend to situations requiring force resolution6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Attempt Newton's 2nd law for either \(P\), \(Q\) or the system | M1 | Correct number of relevant terms, dimensionally correct |
| For \(P\): \(0.8 + 0.5g\sin 30 - T = 0.5a\) | A1 | For any one correct equation |
| For \(Q\): \(T - 0.3g\sin 45 = 0.3a\) | ||
| System: \(0.8 + 0.5g\sin 30 - 0.3g\sin 45 = 0.8a\) | A1 | For two correct equations |
| Attempt to solve for \(T\) | M1 | Using two equations, each with the correct number of relevant terms. \([a = 1.4733\) may be seen\(]\) |
| \(T = 2.56\) N (3sf) | A1 | Allow \(T = \dfrac{99 + 75\sqrt{2}}{80}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| KE and PE for \(m\) kg particle: \(\frac{1}{2}m \times 0.36 = 0.18m\) and \(mg\sin 45 = 5\sqrt{2}m\) | B1 | Any 2 correct PE or KE terms |
| KE and PE for 0.5 kg particle: \(\frac{1}{2} \times 0.5 \times 0.36 = 0.09\) and \(0.5g\sin 30 = 2.5\) | B1 | All 4 correct PE and KE terms |
| Apply work-energy equation to the system: PE loss + WD by \(0.8\) N = KE gain \(+ 0.5\) | M1 | Must include at least 5 relevant terms only and no extra terms. All terms dimensionally correct |
| \(0.5g \times 1 \times \sin 30 - mg \times 1 \times \sin 45 + 0.8 \times 1 = \frac{1}{2}(0.5 + m) \times 0.36 + 0.5\) | A1 | May be seen as: \(2.5 - 5\sqrt{2}m + 0.8 = 0.09 + 0.18m + 0.5\) |
| \(m = 0.374\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| KE and PE for \(m\) kg particle: \(\frac{1}{2}m \times 0.36 = 0.18m\) and \(mg\sin 45 = 5\sqrt{2}m\) | B1 | Correct KE and PE for \(m\) kg particle |
| \(a = 0.18\) and \(3.3 - T = 0.5(0.18)\) leading to \(T = 3.21\) | B1 | Evaluate tension using Newton's second law applied to 0.5 kg particle |
| For \(m\) kg particle: WD by \(T\) = KE gain + PE gain \(+ 0.5\) | M1 | At least 3 relevant terms including tension. All terms dimensionally correct |
| \(3.21 \times 1 = \frac{1}{2}m \times 0.36 + mg\sin 45 + 0.5\) | A1 | |
| \(m = 0.374\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| All four KE and PE terms as above | B1, B1 | Any 2 correct / all 4 correct PE or KE terms |
| Apply work-energy to both particles including tension: \(0.8\times1 + 0.5g\sin30 = \frac{1}{2}\times0.5\times0.36 + T\times1\) and \(T\times1 = \frac{1}{2}m\times0.36 + mg\sin45 + 0.5\) | M1 | Must include at least 5 relevant terms and tension in both. \([T=3.21]\). All terms dimensionally correct |
| \(0.8\times1 + 0.5g\sin30 - \frac{1}{2}\times0.5\times0.36 = \frac{1}{2}m\times0.36 + mg\sin45 + 0.5\) | A1 | |
| \(m = 0.374\) | A1 |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt Newton's 2nd law for either $P$, $Q$ or the system | M1 | Correct number of relevant terms, dimensionally correct |
| For $P$: $0.8 + 0.5g\sin 30 - T = 0.5a$ | A1 | For any one correct equation |
| For $Q$: $T - 0.3g\sin 45 = 0.3a$ | | |
| System: $0.8 + 0.5g\sin 30 - 0.3g\sin 45 = 0.8a$ | A1 | For two correct equations |
| Attempt to solve for $T$ | M1 | Using two equations, each with the correct number of relevant terms. $[a = 1.4733$ may be seen$]$ |
| $T = 2.56$ N (3sf) | A1 | Allow $T = \dfrac{99 + 75\sqrt{2}}{80}$ |
---
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| KE and PE for $m$ kg particle: $\frac{1}{2}m \times 0.36 = 0.18m$ and $mg\sin 45 = 5\sqrt{2}m$ | B1 | Any 2 correct PE or KE terms |
| KE and PE for 0.5 kg particle: $\frac{1}{2} \times 0.5 \times 0.36 = 0.09$ and $0.5g\sin 30 = 2.5$ | B1 | All 4 correct PE and KE terms |
| Apply work-energy equation to the system: PE loss + WD by $0.8$ N = KE gain $+ 0.5$ | M1 | Must include at least 5 relevant terms only and no extra terms. All terms dimensionally correct |
| $0.5g \times 1 \times \sin 30 - mg \times 1 \times \sin 45 + 0.8 \times 1 = \frac{1}{2}(0.5 + m) \times 0.36 + 0.5$ | A1 | May be seen as: $2.5 - 5\sqrt{2}m + 0.8 = 0.09 + 0.18m + 0.5$ |
| $m = 0.374$ | A1 | |
---
## Question 7(b) Alternative Method 1:
| Answer | Marks | Guidance |
|--------|-------|----------|
| KE and PE for $m$ kg particle: $\frac{1}{2}m \times 0.36 = 0.18m$ and $mg\sin 45 = 5\sqrt{2}m$ | B1 | Correct KE and PE for $m$ kg particle |
| $a = 0.18$ and $3.3 - T = 0.5(0.18)$ leading to $T = 3.21$ | B1 | Evaluate tension using Newton's second law applied to 0.5 kg particle |
| For $m$ kg particle: WD by $T$ = KE gain + PE gain $+ 0.5$ | M1 | At least 3 relevant terms including tension. All terms dimensionally correct |
| $3.21 \times 1 = \frac{1}{2}m \times 0.36 + mg\sin 45 + 0.5$ | A1 | |
| $m = 0.374$ | A1 | |
---
## Question 7(b) Alternative Method 2:
| Answer | Marks | Guidance |
|--------|-------|----------|
| All four KE and PE terms as above | B1, B1 | Any 2 correct / all 4 correct PE or KE terms |
| Apply work-energy to both particles including tension: $0.8\times1 + 0.5g\sin30 = \frac{1}{2}\times0.5\times0.36 + T\times1$ and $T\times1 = \frac{1}{2}m\times0.36 + mg\sin45 + 0.5$ | M1 | Must include at least 5 relevant terms and tension in both. $[T=3.21]$. All terms dimensionally correct |
| $0.8\times1 + 0.5g\sin30 - \frac{1}{2}\times0.5\times0.36 = \frac{1}{2}m\times0.36 + mg\sin45 + 0.5$ | A1 | |
| $m = 0.374$ | A1 | |7\\
\includegraphics[max width=\textwidth, alt={}, center]{a96ca3b4-6d35-4512-a0a1-3f28443fd051-12_439_1095_258_525}
Two particles $P$ and $Q$ of masses 0.5 kg and $m \mathrm {~kg}$ respectively are attached to the ends of a light inextensible string. The string passes over a fixed smooth pulley which is attached to the top of two inclined planes. The particles are initially at rest with $P$ on a smooth plane inclined at $30 ^ { \circ }$ to the horizontal and $Q$ on a plane inclined at $45 ^ { \circ }$ to the horizontal. The string is taut and the particles can move on lines of greatest slope of the two planes. A force of magnitude 0.8 N is applied to $P$ acting down the plane, causing $P$ to move down the plane (see diagram).
\begin{enumerate}[label=(\alph*)]
\item It is given that $m = 0.3$, and that the plane on which $Q$ rests is smooth.
Find the tension in the string.
\item It is given instead that the plane on which $Q$ rests is rough, and that after each particle has moved a distance of 1 m , their speed is $0.6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The work done against friction in this part of the motion is 0.5 J .
Use an energy method to find the value of $m$.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2021 Q7 [10]}}