| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Equilibrium on slope with horizontal force |
| Difficulty | Moderate -0.3 This is a standard M1 equilibrium problem requiring resolution of forces in two perpendicular directions (parallel and perpendicular to the slope). While it involves multiple steps and careful angle work, it follows a routine textbook method with no novel insight required. Slightly easier than average due to being a straightforward application of equilibrium conditions with given numerical values. |
| Spec | 3.03d Newton's second law: 2D vectors3.03n Equilibrium in 2D: particle under forces |
| Answer | Marks | Guidance |
|---|---|---|
| resolve ↑: \(Rcos15 - 4g = 0\) \(R = \frac{4g}{cos15} = 40.6 \text{ N (3sf)}\) | M2, M1 A1 | |
| resolve // to slope: \(Fcos15 - 4sin15 = 0\) \(F = 4gtan15 = 10.5 \text{ N}\) | M1, M1 A1 | (7) |
| resolve ↑: $Rcos15 - 4g = 0$ $R = \frac{4g}{cos15} = 40.6 \text{ N (3sf)}$ | M2, M1 A1 | |
| resolve // to slope: $Fcos15 - 4sin15 = 0$ $F = 4gtan15 = 10.5 \text{ N}$ | M1, M1 A1 | (7) |1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e0de1908-cf67-460f-9473-b2dfded95b33-2_257_693_239_447}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
Figure 1 shows a particle $P$ of mass 4 kg on a smooth plane inclined at $15 ^ { \circ }$ to the horizontal. $P$ is held in equilibrium by a horizontal force, $F$.
\begin{enumerate}[label=(\alph*)]
\item Show that the normal reaction exerted by the plane on $P$ is 40.6 N correct to 3 significant figures.
\item Calculate the value of $F$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q1 [7]}}