CAIE M1 2018 June — Question 3 6 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2018
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMotion on a slope
TypeString at angle to slope
DifficultyStandard +0.3 This is a standard inclined plane mechanics problem requiring resolution of forces in two directions. Part (i) involves equilibrium conditions with straightforward trigonometry to find an angle, while part (ii) applies F=ma with given angle. The problem is slightly easier than average as it follows a routine template with clear setup and standard techniques, though it requires careful component resolution.
Spec3.03c Newton's second law: F=ma one dimension3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 0

\includegraphics{figure_3} A particle \(P\) of mass \(8 \text{ kg}\) is on a smooth plane inclined at an angle of \(30°\) to the horizontal. A force of magnitude \(100 \text{ N}\), making an angle of \(\theta°\) with a line of greatest slope and lying in the vertical plane containing the line of greatest slope, acts on \(P\) (see diagram).
  1. Given that \(P\) is in equilibrium, show that \(\theta = 66.4\), correct to \(1\) decimal place, and find the normal reaction between the plane and \(P\). [4]
  2. Given instead that \(\theta = 30\), find the acceleration of \(P\). [2]

Question 3:

AnswerMarks Guidance
3(i)M1 Attempt to resolve forces along the plane
(2 terms)
AnswerMarks Guidance
100 cos θ = 8 g sin 30 → θ = 66.4A1
[R = 8 g cos 30 + 100 sin θ]M1 Resolve forces perpendicular to the plane
(3 terms)
AnswerMarks
R = 161A1
4

AnswerMarks Guidance
3(ii)100 cos 30 – 8g sin 30 = 8a M1
plane (3 terms)
AnswerMarks
a = 5.83A1
2
AnswerMarks Guidance
QuestionAnswer Marks
Question 3:
--- 3(i) ---
3(i) | M1 | Attempt to resolve forces along the plane
(2 terms)
100 cos θ = 8 g sin 30 → θ = 66.4 | A1
[R = 8 g cos 30 + 100 sin θ] | M1 | Resolve forces perpendicular to the plane
(3 terms)
R = 161 | A1
4
--- 3(ii) ---
3(ii) | 100 cos 30 – 8g sin 30 = 8a | M1 | Apply Newton’s 2nd law parallel to the
plane (3 terms)
a = 5.83 | A1
2
Question | Answer | Marks | Guidance
\includegraphics{figure_3}

A particle $P$ of mass $8 \text{ kg}$ is on a smooth plane inclined at an angle of $30°$ to the horizontal. A force of magnitude $100 \text{ N}$, making an angle of $\theta°$ with a line of greatest slope and lying in the vertical plane containing the line of greatest slope, acts on $P$ (see diagram).

\begin{enumerate}[label=(\roman*)]
\item Given that $P$ is in equilibrium, show that $\theta = 66.4$, correct to $1$ decimal place, and find the normal reaction between the plane and $P$. [4]
\item Given instead that $\theta = 30$, find the acceleration of $P$. [2]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2018 Q3 [6]}}