| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Two particles over pulley, vertical strings |
| Difficulty | Moderate -0.3 This is a standard M1 pulley problem requiring routine resolution of forces in two directions and application of equilibrium conditions. While it involves multiple steps (resolving vertically, using 'point of lifting' condition R=0, resolving horizontally), each step follows a standard textbook approach with no novel insight required. The trigonometry is straightforward and the question guides students through the solution with clear parts. |
| Spec | 3.03n Equilibrium in 2D: particle under forces |
| Answer | Marks | Guidance |
|---|---|---|
| 3(i) | \(R + T\sin72° = 50g\) | M1 |
| 3(i) | A1 | |
| 3(i) | [2] | |
| 3(ii) | \(T = \frac{50g}{\sin72°}\) | M1 |
| 3(ii) | \(T = 515\) | A1 |
| 3(ii) | \(T = mg\) | B1 |
| 3(ii) | \(m = 52.6\) | B1 |
| 3(ii) | [4] | |
| 3(iii) | \(X = T\cos72°\) | B1 |
| 3(iii) | \(X = 159\) | B1 |
| 3(iii) | [2] |
3(i) | $R + T\sin72° = 50g$ | M1 | An equation with $R$, $T$ and 50 in linear combination.
3(i) | | A1 | $R + 0.951T = 50g$
3(i) | | [2] |
3(ii) | $T = \frac{50g}{\sin72°}$ | M1 | Using $R = 0$ (may be implied) and $T\sin72° = 50(g)$
3(ii) | $T = 515$ | A1 | Or better
3(ii) | $T = mg$ | B1 |
3(ii) | $m = 52.6$ | B1 | Accept 52.5
3(ii) | | [4] |
3(iii) | $X = T\cos72°$ | B1 | Implied by correct answer
3(iii) | $X = 159$ | B1 | Or better
3(iii) | | [2] |3\\
\includegraphics[max width=\textwidth, alt={}, center]{ae5d1e27-5853-48aa-9046-86ce1c1a154a-3_437_846_274_651}
A block of mass 50 kg is in equilibrium on smooth horizontal ground with one end of a light wire attached to its upper surface. The other end of the wire is attached to an object of mass mkg . The wire passes over a small smooth pulley, and the object hangs vertically below the pulley. The part of the wire between the block and the pulley makes an angle of $72 ^ { \circ }$ with the horizontal. A horizontal force of magnitude X N acts on the block in the vertical plane containing the wire (see diagram).
The tension in the wire is T N and the contact force exerted by the ground on the block is R N.\\
(i) By resolving forces on the block vertically, find a relationship between T and R .
It is given that the block is on the point of lifting off the ground.\\
(ii) Show that $\mathrm { T } = 515$, correct to 3 significant figures, and hence find the value of m .\\
(iii) By resolving forces on the block horizontally, write down a relationship between T and X , and hence find the value of $X$.
\hfill \mbox{\textit{OCR M1 2007 Q3 [8]}}