| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Friction |
| Type | Coefficient from constant speed |
| Difficulty | Moderate -0.8 This is a straightforward mechanics problem requiring standard application of Newton's laws and friction formulas. Part (i) involves resolving forces at equilibrium (constant speed) and calculating μ from F = μR, while part (ii) uses basic kinematics with constant deceleration. Both parts follow textbook procedures with no problem-solving insight required, making it easier than average but not trivial due to the two-part structure and need for careful force resolution. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(Fr = 14\cos30\) | B1 | 12.1(24..) |
| \(R = 28 - 14\sin30\) | B1 | 21 |
| \((14\cos30) = \mu(28 - 14\sin30)\) | M1 | 12.1(24..)/21. Allow \ |
| \(\mu = 0.577\) AG | A1 | 0.577(35..) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{Mass} = 28/g\) | B1 | 2.857.. Award here if seen in (i) and used in (ii) |
| \(Fr = 0.577 \times 28\) | B1 | 16.156 or \(0.57735.. \times 28 = 16.1658..\) |
| \((28/9.8)a = \pm0.577 \times 28\) | M1 | Award also for \(\text{cv}(m)\), \(m=28\). Must be only one force (friction), allow \(Fr(\text{i})\) |
| \(a = \pm5.66\) from exact \(\mu\), \(a = \pm5.65\) from \(\mu = 0.577\) | A1 | \(g=10\) \((\pm5.77)\) |
| \(0 = u^2 - 2 \times 5.66 \times 3.2\) | M1 | Valid signs with cv(5.66) |
| \(u = 6.02 \text{ m s}^{-1}\) | A1 | Accept any answer rounding to 6.0 (inc 6.0, not 6) or 6.1 from \(g=10\) |
# Question 4:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $Fr = 14\cos30$ | B1 | 12.1(24..) |
| $R = 28 - 14\sin30$ | B1 | 21 |
| $(14\cos30) = \mu(28 - 14\sin30)$ | M1 | 12.1(24..)/21. Allow \|component of $14$\|/\|cv$(R)$\| for M1 |
| $\mu = 0.577$ AG | A1 | 0.577(35..) |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{Mass} = 28/g$ | B1 | 2.857.. Award here if seen in (i) and used in (ii) |
| $Fr = 0.577 \times 28$ | B1 | 16.156 or $0.57735.. \times 28 = 16.1658..$ |
| $(28/9.8)a = \pm0.577 \times 28$ | M1 | Award also for $\text{cv}(m)$, $m=28$. Must be only one force (friction), allow $Fr(\text{i})$ |
| $a = \pm5.66$ from exact $\mu$, $a = \pm5.65$ from $\mu = 0.577$ | A1 | $g=10$ $(\pm5.77)$ |
| $0 = u^2 - 2 \times 5.66 \times 3.2$ | M1 | Valid signs with cv(5.66) |
| $u = 6.02 \text{ m s}^{-1}$ | A1 | Accept any answer rounding to 6.0 (inc 6.0, not 6) or 6.1 from $g=10$ |
---4 A block $B$ of weight 28 N is pulled at constant speed across a rough horizontal surface by a force of magnitude 14 N inclined at $30 ^ { \circ }$ above the horizontal.\\
(i) Show that the coefficient of friction between the block and the surface is 0.577 , correct to 3 significant figures.
The 14 N force is suddenly removed, and the block decelerates, coming to rest after travelling a further 3.2 m .\\
(ii) Calculate the speed of the block at the instant the 14 N force was removed.
\hfill \mbox{\textit{OCR M1 2012 Q4 [10]}}