| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Particle on inclined plane |
| Difficulty | Moderate -0.3 This is a straightforward M1 mechanics question requiring standard application of Newton's second law in two scenarios. Part (a) uses equilibrium (constant speed means zero acceleration) to find resistance forces, and part (b) applies F=ma on horizontal ground. The calculations are routine with clearly given values and no conceptual subtleties, making it slightly easier than average but still requiring proper force resolution and understanding of equilibrium vs acceleration. |
| Spec | 3.03c Newton's second law: F=ma one dimension3.03f Weight: W=mg |
| Answer | Marks |
|---|---|
| resolve // to plane: \(1800 - 1250g\sin\alpha - R = 0\) | M1 A1 |
| \(R = 1800 - 1250(9.8)\frac{3}{10} = 575\) N | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| at instant car reaches top of hill, \(F = 1800\), \(R = 575\) | M1 | |
| \(1800 - 575 = 1250a \Rightarrow a = 0.98\) ms\(^{-2}\) | M1 A1 | (7) |
**Part (a):**
resolve // to plane: $1800 - 1250g\sin\alpha - R = 0$ | M1 A1 |
$R = 1800 - 1250(9.8)\frac{3}{10} = 575$ N | M1 A1 |
**Part (b):**
at instant car reaches top of hill, $F = 1800$, $R = 575$ | M1 |
$1800 - 575 = 1250a \Rightarrow a = 0.98$ ms$^{-2}$ | M1 A1 | (7)
---A car of mass 1250 kg is moving at constant speed up a hill, inclined at an angle $\alpha$ to the horizontal, where $\sin \alpha = \frac{1}{10}$. The driving force produced by the engine is 1800 N.
\begin{enumerate}[label=(\alph*)]
\item Calculate the resistance to motion which the car experiences. [4 marks]
\end{enumerate}
At the top of the hill, the road becomes horizontal.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the initial acceleration of the car. [3 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q3 [7]}}