| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Block on rough horizontal surface – accelerating (finding acceleration or applied force) |
| Difficulty | Moderate -0.3 This is a standard M1 friction problem requiring resolution of forces, calculation of normal reaction with an inclined applied force, and basic kinematics. The steps are routine (resolve vertically for normal reaction, horizontally for acceleration, then use v=u+at), though the inclined force adds slight complexity beyond the most basic friction questions. Slightly easier than average A-level due to straightforward application of standard methods. |
| Spec | 3.03e Resolve forces: two dimensions3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| 6 N force is at 60° to vertical, so component along vertical = \(6\cos60° = 3\) N upward | M1 | |
| Normal reaction \(N = 0.8g - 3 = 7.84 - 3 = 4.84\) N | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Friction \(F = 0.2 \times 4.84 = 0.968\) N | M1 | |
| Horizontal component of 6 N = \(6\sin60° = 5.196\) N | M1 | |
| \(0.8a = 5.196 - 0.968\) | M1 | Newton's 2nd law |
| \(a = 5.28\) m s\(^{-2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| After force removed: \(N = 0.8g = 7.84\) N | M1 | |
| Friction \(= 0.2 \times 7.84 = 1.568\) N (decelerating) | M1 | |
| \(0.8a = -1.568\), \(a = -1.96\) m s\(^{-2}\) | A1 | |
| \(0 = 4.9 - 1.96t\), \(t = 2.5\) s | A1 |
# Question 3:
**(i)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| 6 N force is at 60° to vertical, so component along vertical = $6\cos60° = 3$ N upward | M1 | |
| Normal reaction $N = 0.8g - 3 = 7.84 - 3 = 4.84$ N | A1 | |
**(ii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Friction $F = 0.2 \times 4.84 = 0.968$ N | M1 | |
| Horizontal component of 6 N = $6\sin60° = 5.196$ N | M1 | |
| $0.8a = 5.196 - 0.968$ | M1 | Newton's 2nd law |
| $a = 5.28$ m s$^{-2}$ | A1 | |
**(iii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| After force removed: $N = 0.8g = 7.84$ N | M1 | |
| Friction $= 0.2 \times 7.84 = 1.568$ N (decelerating) | M1 | |
| $0.8a = -1.568$, $a = -1.96$ m s$^{-2}$ | A1 | |
| $0 = 4.9 - 1.96t$, $t = 2.5$ s | A1 | |
---3 A block $B$ of mass 0.8 kg is pulled across a horizontal surface by a force of 6 N inclined at an angle of $60 ^ { \circ }$ to the upward vertical. The coefficient of friction between the block and the surface is 0.2 . Calculate\\
(i) the vertical component of the force exerted on $B$ by the surface,\\
(ii) the acceleration of $B$.
The 6 N force is removed when $B$ has speed $4.9 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(iii) Calculate the time taken for $B$ to decelerate from a speed of $4.9 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ to rest.
\hfill \mbox{\textit{OCR M1 2011 Q3 [10]}}