| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Motion up then down slope |
| Difficulty | Standard +0.3 This is a standard M1 mechanics problem requiring resolution of forces on a slope, application of equations of motion in both directions (friction opposes motion each way), and momentum calculation. While it has multiple parts and requires careful sign management, it follows a well-established template with no novel problem-solving required beyond routine application of SUVAT and force resolution. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03v Motion on rough surface: including inclined planes6.03a Linear momentum: p = mv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(R = 0.3g\cos 30\) | B1 | \(R = 2.546\) N. May be shown on diagram |
| \(Fr = 0.15 \times 0.3g\cos 30\) | M1 | \(0.15 \times \text{cv}(R)\), \(Fr = 0.382\) |
| \(0.3a = -0.3g\sin 30 - 0.15 \times 0.3g\cos 30\) | M1 | N2L, two forces inc. \(0.3g\text{Cor}S30\) and friction |
| \(a = -6.17\) | A1 | Accept positive value |
| \(0 = 4^2 - 2 \times 6.17s\) | M1 | Using \(a\) from above |
| \(s = 1.3(0) \text{ m}\) | A1 ft [6] | ft\((8/ |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.3a = 0.3g\sin 30 - 0.382\) | M1 | N2L, diff. of two forces inc. \(0.3g\text{Cor}S30\) and friction |
| \(a = 3.63\) | A1 | |
| \(1.3 = 3.63t^2/2\) | M1 | Using cv\((s(\mathbf{i}))\), and \(a\) not \(a(\mathbf{i})\) nor 9.8 |
| \(t = 0.845 \text{ s}\) | A1 [4] | Rounds to 0.85 if 2 sig fig. CorS30 means cos30 or sin30 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(V = 3.63 \times 0.845\) OR \(V = \sqrt{2 \times 3.63 \times 1.3}\) OR \(V = 2 \times 1.3/0.845\) | M1 | cv\((a(\mathbf{ii})) \times t(\mathbf{ii})\) OR cv\((\sqrt{2 \times a(\mathbf{ii})} \times s(\mathbf{i}))\) OR cv\((2 \times s(\mathbf{i})/t(\mathbf{ii}))\), \(a(\mathbf{ii})\) not \(a(\mathbf{i})\) nor 9.8 |
| \((V = 3.07)\) | ||
| Mom change \(= +/-(0.3 \times 4 + 0.3 \times 3.07)\) | M1 | \(+/-(0.3 \times 4 +/- 0.3 \times \text{speed(return)})\), \(0 < \text{speed(return)} < 4\), \(g\) omitted |
| Mom change \(= +/-2.12 \text{ kgms}^{-1}\) | A1 [3] |
# Question 6:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $R = 0.3g\cos 30$ | B1 | $R = 2.546$ N. May be shown on diagram |
| $Fr = 0.15 \times 0.3g\cos 30$ | M1 | $0.15 \times \text{cv}(R)$, $Fr = 0.382$ |
| $0.3a = -0.3g\sin 30 - 0.15 \times 0.3g\cos 30$ | M1 | N2L, two forces inc. $0.3g\text{Cor}S30$ and friction |
| $a = -6.17$ | A1 | Accept positive value |
| $0 = 4^2 - 2 \times 6.17s$ | M1 | Using $a$ from above |
| $s = 1.3(0) \text{ m}$ | A1 ft **[6]** | ft$(8/|\text{cv}(a)|)$. CorS30 means cos30 or sin30 |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.3a = 0.3g\sin 30 - 0.382$ | M1 | N2L, diff. of two forces inc. $0.3g\text{Cor}S30$ and friction |
| $a = 3.63$ | A1 | |
| $1.3 = 3.63t^2/2$ | M1 | Using cv$(s(\mathbf{i}))$, and $a$ not $a(\mathbf{i})$ nor 9.8 |
| $t = 0.845 \text{ s}$ | A1 **[4]** | Rounds to 0.85 if 2 sig fig. CorS30 means cos30 or sin30 |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $V = 3.63 \times 0.845$ OR $V = \sqrt{2 \times 3.63 \times 1.3}$ OR $V = 2 \times 1.3/0.845$ | M1 | cv$(a(\mathbf{ii})) \times t(\mathbf{ii})$ OR cv$(\sqrt{2 \times a(\mathbf{ii})} \times s(\mathbf{i}))$ OR cv$(2 \times s(\mathbf{i})/t(\mathbf{ii}))$, $a(\mathbf{ii})$ not $a(\mathbf{i})$ nor 9.8 |
| $(V = 3.07)$ | | |
| Mom change $= +/-(0.3 \times 4 + 0.3 \times 3.07)$ | M1 | $+/-(0.3 \times 4 +/- 0.3 \times \text{speed(return)})$, $0 < \text{speed(return)} < 4$, $g$ omitted |
| Mom change $= +/-2.12 \text{ kgms}^{-1}$ | A1 **[3]** | |
---6 A particle $P$ of mass 0.3 kg is projected upwards along a line of greatest slope from the foot of a plane inclined at $30 ^ { \circ }$ to the horizontal. The initial speed of $P$ is $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and the coefficient of friction is 0.15 . The particle $P$ comes to instantaneous rest before it reaches the top of the plane.\\
(i) Calculate the distance $P$ moves up the plane.\\
(ii) Find the time taken by $P$ to return from its highest position on the plane to the foot of the plane.\\
(iii) Calculate the change in the momentum of $P$ between the instant that $P$ leaves the foot of the plane and the instant that $P$ returns to the foot of the plane.
\hfill \mbox{\textit{OCR M1 2012 Q6 [13]}}