| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Motion with applied force on slope |
| Difficulty | Standard +0.3 This is a standard M1 mechanics question requiring resolution of forces on a slope and basic kinematics. Part (i) uses equilibrium parallel to slope, part (ii) perpendicular resolution for normal force, and part (iii) applies constant acceleration equations. All techniques are routine for M1 with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03e Resolve forces: two dimensions3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(T\cos S20 = 0.25g\cos S30\) | M1 | Equates cmpt \(T\) and cmpt wt \(\parallel\) plane (doubt, see diagram and/or (ii)) |
| \(T\cos 20 = 0.25g\sin 30\) | A1 | 1.225 |
| \(T = 1.3(0)\) | A1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(R +/- T\cos S20 = +/- 0.25g\cos S30\) | M1 | Resolves perp plane, accept letter \(T\) |
| \(R + 1.3\sin 20 = 0.25g\cos 30\) | A1 ft | ft(cv(\(T\))) |
| \(R = 1.68\) N | A1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((m)\text{accn} = +/- (m)9.8\sin 30\) | M1* | N2L with single force a cmpt wt (accept cos) |
| \(a = +/-4.9\) | A1 | |
| \(u = +/-9.8\sin 30 \times 0.4\) | D*M1 | |
| \(u = 1.96\) | A1 | Must be \(+\)ve (accept loss of \(-\) sign) |
| [4] |
## Question 3(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $T\cos S20 = 0.25g\cos S30$ | M1 | Equates cmpt $T$ and cmpt wt $\parallel$ plane (doubt, see diagram and/or (ii)) |
| $T\cos 20 = 0.25g\sin 30$ | A1 | 1.225 |
| $T = 1.3(0)$ | A1 | |
| **[3]** | | |
## Question 3(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $R +/- T\cos S20 = +/- 0.25g\cos S30$ | M1 | Resolves perp plane, accept letter $T$ |
| $R + 1.3\sin 20 = 0.25g\cos 30$ | A1 ft | ft(cv($T$)) |
| $R = 1.68$ N | A1 | |
| **[3]** | | |
## Question 3(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(m)\text{accn} = +/- (m)9.8\sin 30$ | M1* | N2L with single force a cmpt wt (accept cos) |
| $a = +/-4.9$ | A1 | |
| $u = +/-9.8\sin 30 \times 0.4$ | D*M1 | |
| $u = 1.96$ | A1 | Must be $+$ve (accept loss of $-$ sign) |
| **[4]** | | |3\\
\includegraphics[max width=\textwidth, alt={}, center]{f5085265-5258-45d4-8233-6bd68f8e9034-2_300_501_799_790}
A particle $P$ of mass 0.25 kg moves upwards with constant speed $u \mathrm {~ms} ^ { - 1 }$ along a line of greatest slope on a smooth plane inclined at $30 ^ { \circ }$ to the horizontal. The pulling force acting on $P$ has magnitude $T \mathrm {~N}$ and acts at an angle of $20 ^ { \circ }$ to the line of greatest slope (see diagram). Calculate\\
(i) the value of $T$,\\
(ii) the magnitude of the contact force exerted on $P$ by the plane.
The pulling force $T \mathrm {~N}$ acting on $P$ is suddenly removed, and $P$ comes to instantaneous rest 0.4 s later.\\
(iii) Calculate $u$.
\hfill \mbox{\textit{OCR M1 2013 Q3 [10]}}