Edexcel M1 — Question 3 9 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMotion on a slope
TypeEquilibrium on slope with force at angle to slope
DifficultyStandard +0.3 This is a standard M1 equilibrium problem on a slope with a force at an angle. It requires resolving forces in two directions (parallel and perpendicular to the slope) and applying equilibrium conditions. The modeling part (a) is trivial (particle model), and parts (b) and (c) involve straightforward resolution and simultaneous equations. The angle being given as sin α = 3/5 simplifies calculations. This is slightly easier than average as it's a textbook-style mechanics question with no novel problem-solving required.
Spec1.05g Exact trigonometric values: for standard angles3.03e Resolve forces: two dimensions3.03i Normal reaction force3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces

3. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{c762bd90-5b57-428a-a7a8-291a1a643a14-3_309_590_196_518} \captionsetup{labelformat=empty} \caption{Fig. 2}
\end{figure} Figure 2 shows a ball of mass 3 kg lying on a smooth plane inclined at an angle \(\alpha\) to the horizontal where \(\sin \alpha = \frac { 3 } { 5 }\). The ball is held in equilibrium by a force of magnitude \(P\) newtons, which acts at an angle of \(10 ^ { \circ }\) to the line of greatest slope of the plane.
  1. Suggest a suitable model for the ball. Giving your answers correct to 1 decimal place,
  2. find the value of \(P\),
  3. find the magnitude of the reaction between the ball and the plane.

AnswerMarks Guidance
(a) particleB1
(b) resolve // to plane: \(P\cos 10° - 3g\sin\alpha = 0\)M1 A1
\(P\cos 10° = 3g(\frac{3}{5})\) \(\therefore P = 17.9\)M1 A1 (1dp)
(c) resolve perp. to plane: \(R + P\sin 10° - 3g\cos\alpha = 0\)M1 A1
\(R = 3g(\frac{4}{5}) - P\sin 10° = 20.4\)M1 A1 (1dp)
**(a)** particle | B1 |

**(b)** resolve // to plane: $P\cos 10° - 3g\sin\alpha = 0$ | M1 A1 |
$P\cos 10° = 3g(\frac{3}{5})$ $\therefore P = 17.9$ | M1 A1 | (1dp)

**(c)** resolve perp. to plane: $R + P\sin 10° - 3g\cos\alpha = 0$ | M1 A1 |
$R = 3g(\frac{4}{5}) - P\sin 10° = 20.4$ | M1 A1 | (1dp) | (9)
3.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{c762bd90-5b57-428a-a7a8-291a1a643a14-3_309_590_196_518}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}

Figure 2 shows a ball of mass 3 kg lying on a smooth plane inclined at an angle $\alpha$ to the horizontal where $\sin \alpha = \frac { 3 } { 5 }$. The ball is held in equilibrium by a force of magnitude $P$ newtons, which acts at an angle of $10 ^ { \circ }$ to the line of greatest slope of the plane.
\begin{enumerate}[label=(\alph*)]
\item Suggest a suitable model for the ball.

Giving your answers correct to 1 decimal place,
\item find the value of $P$,
\item find the magnitude of the reaction between the ball and the plane.
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1  Q3 [9]}}