| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Two connected particles, horizontal surface |
| Difficulty | Standard +0.3 This is a standard M1 connected particles problem requiring Newton's second law applied to two bodies with given acceleration. Part (i) involves straightforward resolution of forces on the block, and part (ii) requires resolving forces on the particle including friction and the inclined string tension. While it requires careful component resolution (the 10° angle adds mild complexity), it follows a well-practiced procedure with no novel insight needed—slightly above average due to the two-part structure and angled string. |
| Spec | 3.03d Newton's second law: 2D vectors3.03t Coefficient of friction: F <= mu*R model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2 - F = 0.8 \times 0.2\) \(F = T\cos10\) \(T = 1.87 \text{ N}\) OR \(2 - T\cos10 = 0.8 \times 0.2\) \(T = 1.87 \text{ N}\) | M1 M1 A1 [3] | N2L 2 force terms and ma (\(F = 1.84\) N) \(F = T\Cor S10\) 1.8683.. N2L 2 force terms and ma T\(\Cor S10\) |
| \(R - 0.3 \times 9.8 = 0\) \(R = 0.3 \times 9.8 - 1.87\sin10\) \(R = 2.62\) T\(\cos10 - Fr = 0.3 \times 0.2\) Fr = 1.78 \(\mu = 1.78 / 2.62\) or 1.78 = 2.62 μ \(\mu = 0.68\) | M1 A1ft A1ft A1 [7] | 3 term equation, vertically cv(T(i)) 2.61(5..) seen or implied N2L 2 forces for F; component of T cv(T(i)) seen or implied both terms same sign |
| **Answer/Working** | **Marks** | **Guidance** |
|---|---|---|
| $2 - F = 0.8 \times 0.2$ $F = T\cos10$ $T = 1.87 \text{ N}$ OR $2 - T\cos10 = 0.8 \times 0.2$ $T = 1.87 \text{ N}$ | M1 M1 A1 [3] | N2L 2 force terms and ma ($F = 1.84$ N) $F = T\Cor S10$ 1.8683.. N2L 2 force terms and ma T$\Cor S10$ | **m is the block mass, award if T not F** |
| $R - 0.3 \times 9.8 = 0$ $R = 0.3 \times 9.8 - 1.87\sin10$ $R = 2.62$ T$\cos10 - Fr = 0.3 \times 0.2$ Fr = 1.78 $\mu = 1.78 / 2.62$ or 1.78 = 2.62 μ $\mu = 0.68$ | M1 A1ft A1ft A1 [7] | 3 term equation, vertically cv(T(i)) 2.61(5..) seen or implied N2L 2 forces for F; component of T cv(T(i)) seen or implied both terms same sign | **Treat as a mis-read R-0.8x9.8 = -1T$\Cor S10 = 0$ leading to R=8.16 (i.e.works on block[2/3])** OR **N2L 2 forces for P+Q:** $2 - Fr = (0.8+0.3) \times 0.2$ Fr unequal to T From correct value of T = 1.87 only |4\\
\includegraphics[max width=\textwidth, alt={}, center]{4c6c9323-8238-4ec2-94a1-6e8188a34521-03_156_1141_258_502}
A block $B$ of mass 0.8 kg and a particle $P$ of mass 0.3 kg are connected by a light inextensible string inclined at $10 ^ { \circ }$ to the horizontal. They are pulled across a horizontal surface with acceleration $0.2 \mathrm {~m} \mathrm {~s} ^ { - 2 }$, by a horizontal force of 2 N applied to $B$ (see diagram).\\
(i) Given that contact between $B$ and the surface is smooth, calculate the tension in the string.\\
(ii) Calculate the coefficient of friction between $P$ and the surface.
\hfill \mbox{\textit{OCR M1 2011 Q4 [10]}}