| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Single particle, Newton's second law – scalar (1D, horizontal or inclined) |
| Difficulty | Moderate -0.3 This is a straightforward M1 mechanics question requiring standard techniques: constant acceleration formula (s=ut+½at²), resolving forces in two perpendicular directions, and Newton's second law. Part (iii) requires basic conceptual understanding that friction provides the perpendicular component. All steps are routine applications of standard methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03c Newton's second law: F=ma one dimension3.03e Resolve forces: two dimensions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(s = ut + \frac{1}{2}at^2\) | M1 | Substitution required |
| \(7.2 = \frac{1}{2} \times a \times 6^2\) | A1 | |
| \(a = 0.4 \text{ ms}^{-2}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(F = ma\) | M1 | Attempt at Newton's second law |
| M1 | Attempt at resolving both \(S\) and \(T\) | |
| \(300\cos30° + 175\cos15° - R = 1000 \times 0.4\) | A1 | Correct elements present and no extras; follow through for \(a\) |
| \(R = 28.8 \text{ N}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| The resistance perpendicular to the line of motion has been ignored. | B1 | There is also a sideways resistance force |
## Question 2:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $s = ut + \frac{1}{2}at^2$ | M1 | Substitution required |
| $7.2 = \frac{1}{2} \times a \times 6^2$ | A1 | |
| $a = 0.4 \text{ ms}^{-2}$ | A1 | cao |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $F = ma$ | M1 | Attempt at Newton's second law |
| | M1 | Attempt at resolving both $S$ and $T$ |
| $300\cos30° + 175\cos15° - R = 1000 \times 0.4$ | A1 | Correct elements present and no extras; follow through for $a$ |
| $R = 28.8 \text{ N}$ | A1 | cao |
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| The resistance perpendicular to the line of motion has been ignored. | B1 | There is also a sideways resistance force |
---2 Fig. 3 shows two people, Sam and Tom, pushing a car of mass 1000 kg along a straight line $l$ on level ground.
Sam pushes with a constant horizontal force of 300 N at an angle of $30 ^ { \circ }$ to the line $l$.\\
Tom pushes with a constant horizontal force of 175 N at an angle of $15 ^ { \circ }$ to the line $l$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d5a09ed4-a32f-4ff7-aa08-6e54c2ab26a0-2_289_1132_571_507}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}
(i) The car starts at rest and moves with constant acceleration. After 6 seconds it has travelled 7.2 m .
Find its acceleration.\\
(ii) Find the resistance force acting on the car along the line $l$.\\
(iii) The resultant of the forces exerted by Sam and Tom is not in the direction of the car's acceleration. Explain briefly why.
\hfill \mbox{\textit{OCR MEI M1 Q2 [8]}}