| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Particle on inclined plane |
| Difficulty | Standard +0.3 This is a straightforward multi-part mechanics question requiring standard application of Newton's second law (F=ma) and SUVAT equations. Parts (i)-(iv) involve routine calculations with clearly stated forces and conditions, while part (v) requires working backwards from given data—all standard M1 techniques with no novel problem-solving or geometric insight required. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(a = \frac{24}{12} = 2\) m s\(^{-2}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(a = \frac{24-15}{12} = 0.75\) m s\(^{-2}\) | M1 A1 | |
| Smooth model distance: \(s_1 = \frac{1}{2}(2)(16) = 16\) m | M1 | Using \(s=\frac{1}{2}at^2\) with \(t=4\) |
| Rough model distance: \(s_2 = \frac{1}{2}(0.75)(16) = 6\) m | A1 | |
| Difference \(= 16 - 6 = 10\) m | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Forces along slope: \(12g\sin 5° - 15 = 12a\) | M1 A1 | |
| \(12(10)(0.08716) - 15 = 12a\) | A1 | |
| \(a = \frac{10.459 - 15}{12} = -0.378\) m s\(^{-2}\) (deceleration) | A1 | Accept \(-0.38\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v = u + at\): \(0 = 1.5 - 0.378t\) | M1 | |
| \(t = 3.97\) s | A1 | |
| \(s = \frac{v^2-u^2}{2a} = \frac{0-2.25}{2(-0.378)} = 2.98\) m | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Using \(t = 3.5\) s: \(0 = 1.5 + a(3.5)\), so \(a = -\frac{3}{7}\) m s\(^{-2}\) | M1 | |
| \(12g\sin 5° - R = 12 \times \frac{3}{7}\) | M1 A1 | |
| \(R = 10.459 - 5.143 = 16.6\) N | A1 |
# Question 7:
**(i)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $a = \frac{24}{12} = 2$ m s$^{-2}$ | B1 | |
**(ii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $a = \frac{24-15}{12} = 0.75$ m s$^{-2}$ | M1 A1 | |
| Smooth model distance: $s_1 = \frac{1}{2}(2)(16) = 16$ m | M1 | Using $s=\frac{1}{2}at^2$ with $t=4$ |
| Rough model distance: $s_2 = \frac{1}{2}(0.75)(16) = 6$ m | A1 | |
| Difference $= 16 - 6 = 10$ m | A1 | |
**(iii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Forces along slope: $12g\sin 5° - 15 = 12a$ | M1 A1 | |
| $12(10)(0.08716) - 15 = 12a$ | A1 | |
| $a = \frac{10.459 - 15}{12} = -0.378$ m s$^{-2}$ (deceleration) | A1 | Accept $-0.38$ |
**(iv)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v = u + at$: $0 = 1.5 - 0.378t$ | M1 | |
| $t = 3.97$ s | A1 | |
| $s = \frac{v^2-u^2}{2a} = \frac{0-2.25}{2(-0.378)} = 2.98$ m | M1 A1 | |
**(v)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Using $t = 3.5$ s: $0 = 1.5 + a(3.5)$, so $a = -\frac{3}{7}$ m s$^{-2}$ | M1 | |
| $12g\sin 5° - R = 12 \times \frac{3}{7}$ | M1 A1 | |
| $R = 10.459 - 5.143 = 16.6$ N | A1 | |7 A horizontal force of 24 N acts on a block of mass 12 kg on a horizontal plane. The block is initially at rest.
This situation is first modelled assuming the plane is smooth.\\
(i) Write down the acceleration of the block according to this model.
The situation is now modelled assuming a constant resistance to motion of 15 N .\\
(ii) Calculate the acceleration of the block according to this new model. How much less distance does the new model predict that the block will travel in the first 4 seconds?
The 24 N force is removed and the block slides down a slope at $5 ^ { \circ }$ to the horizontal. The speed of the block at the top of the slope is $1.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, as shown in Fig. 7. The answers to parts (iii) and (iv) should be found using the assumption that the resistance to the motion of the block is still a constant 15 N .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{52d6c914-b204-4587-a82e-fbab6693fcf8-5_255_901_1128_575}
\captionsetup{labelformat=empty}
\caption{Fig. 7}
\end{center}
\end{figure}
(iii) Calculate the acceleration of the block in the direction of its motion.\\
(iv) For how much time does the block slide down the slope before coming to rest and how far does it slide in that time?
Measurements show that the block actually comes to rest in 3.5 seconds.\\
(v) Assuming that the error in the prediction is due only to the value of the resistance, calculate the true value of the resistance.
\hfill \mbox{\textit{OCR MEI M1 2007 Q7 [18]}}