Edexcel M1 2010 January — Question 5 15 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2010
SessionJanuary
Marks15
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMotion on a slope
TypeCoefficient of friction from motion
DifficultyStandard +0.3 This is a standard M1 mechanics question covering routine applications of Newton's laws on an inclined plane. Part (a) uses basic kinematics (suvat), part (b) applies resolving forces and friction in a familiar context, and part (c) extends to horizontal force equilibrium—all are textbook exercises requiring methodical application of standard techniques rather than problem-solving insight.
Spec3.02d Constant acceleration: SUVAT formulae3.03f Weight: W=mg3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces3.03v Motion on rough surface: including inclined planes

A particle of mass 0.8 kg is held at rest on a rough plane. The plane is inclined at 30° to the horizontal. The particle is released from rest and slides down a line of greatest slope of the plane. The particle moves 2.7 m during the first 3 seconds of its motion. Find
  1. the acceleration of the particle, [3]
  2. the coefficient of friction between the particle and the plane. [5]
The particle is now held on the same rough plane by a horizontal force of magnitude \(X\) newtons, acting in a plane containing a line of greatest slope of the plane, as shown in Figure 3. The particle is in equilibrium and on the point of moving up the plane. \includegraphics{figure_3}
  1. Find the value of \(X\). [7]

Part (a):
\(s = ut + \frac{1}{2}at^2 \Rightarrow 2.7 = \frac{1}{2}a \times 9\)
AnswerMarks Guidance
\(a = 0.6\) (m s\(^{-2}\))M1 A1 A1 (3)
Part (b):
\(R = 0.8g\cos 30°\) (\(\approx\) 6.79)
Use of \(F = \mu R\)
\(0.8g\sin 30° - \mu R = 0.8 \times a\)
\((0.8g\sin 30° - \mu 0.8g\cos 30° = 0.8 \times 0.6)\)
AnswerMarks Guidance
\(\mu \approx 0.51\) accept 0.507B1 B1 M1 A1 A1 (5)
Part (c):
\(\uparrow\): \(R\cos 30° = \mu R\cos 60° + 0.8g\)
(\(R \approx 12.8\))
\(\rightarrow\): \(X = R\sin 30° + \mu R\sin 60°\)
\(X \approx 12\) accept 12.0
AnswerMarks Guidance
Solving for X,M1 A2 (1,0) M1 A1 DM1 A1 (7) [15]
Alternative to (c):
\(R = X\sin 30° + 0.8 \times 9.8\sin 60°\)
\(\mu R + 0.8g\cos 60° = X\cos 30°\)
\(X = \frac{\mu 0.8g\sin 60° + 0.8g\cos 60°}{\cos 30° - \mu\sin 30°}\)
AnswerMarks Guidance
Solving for X, \(X \approx 12\) accept 12.0M1 A2 (1,0) M1 A1 DM1 A1 (7)
**Part (a):**
$s = ut + \frac{1}{2}at^2 \Rightarrow 2.7 = \frac{1}{2}a \times 9$
$a = 0.6$ (m s$^{-2}$) | M1 A1 A1 | (3)

**Part (b):**
$R = 0.8g\cos 30°$ ($\approx$ 6.79)
Use of $F = \mu R$
$0.8g\sin 30° - \mu R = 0.8 \times a$
$(0.8g\sin 30° - \mu 0.8g\cos 30° = 0.8 \times 0.6)$
$\mu \approx 0.51$ accept 0.507 | B1 B1 M1 A1 A1 | (5)

**Part (c):**
$\uparrow$: $R\cos 30° = \mu R\cos 60° + 0.8g$
($R \approx 12.8$)
$\rightarrow$: $X = R\sin 30° + \mu R\sin 60°$
$X \approx 12$ accept 12.0
Solving for X, | M1 A2 (1,0) M1 A1 DM1 A1 | (7) [15]

**Alternative to (c):**
$R = X\sin 30° + 0.8 \times 9.8\sin 60°$
$\mu R + 0.8g\cos 60° = X\cos 30°$

$X = \frac{\mu 0.8g\sin 60° + 0.8g\cos 60°}{\cos 30° - \mu\sin 30°}$
Solving for X, $X \approx 12$ accept 12.0 | M1 A2 (1,0) M1 A1 DM1 A1 | (7)
A particle of mass 0.8 kg is held at rest on a rough plane. The plane is inclined at 30° to the horizontal. The particle is released from rest and slides down a line of greatest slope of the plane. The particle moves 2.7 m during the first 3 seconds of its motion. Find

\begin{enumerate}[label=(\alph*)]
\item the acceleration of the particle, [3]
\item the coefficient of friction between the particle and the plane. [5]
\end{enumerate}

The particle is now held on the same rough plane by a horizontal force of magnitude $X$ newtons, acting in a plane containing a line of greatest slope of the plane, as shown in Figure 3. The particle is in equilibrium and on the point of moving up the plane.

\includegraphics{figure_3}

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the value of $X$. [7]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1 2010 Q5 [15]}}