CAIE M1 2002 November — Question 5 8 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2002
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMotion on a slope
TypeLimiting equilibrium both directions
DifficultyStandard +0.3 This is a standard M1 mechanics question covering limiting equilibrium on an inclined plane with friction in both directions. Part (i) is routine resolution perpendicular to the plane. Part (ii) requires setting up two friction equations (F = μR) for limiting equilibrium in opposite directions and solving simultaneously - a textbook exercise. The particle motion in part (iii) involves standard applications of F=ma, friction, and kinematics/energy. All techniques are straightforward applications of core M1 content with no novel problem-solving required, making it slightly easier than average.
Spec3.03u Static equilibrium: on rough surfaces3.03v Motion on rough surface: including inclined planes

5 \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{fcd2b219-d9b4-4972-b8fe-25cf543b9054-3_245_335_580_906} \captionsetup{labelformat=empty} \caption{Fig. 1}
\end{figure} A force, whose direction is upwards parallel to a line of greatest slope of a plane inclined at \(35 ^ { \circ }\) to the horizontal, acts on a box of mass 15 kg which is at rest on the plane. The normal component of the contact force on the box has magnitude \(R\) newtons (see Fig. 1).
  1. Show that \(R = 123\), correct to 3 significant figures. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{fcd2b219-d9b4-4972-b8fe-25cf543b9054-3_369_1045_1247_555} \captionsetup{labelformat=empty} \caption{Fig. 2}
    \end{figure} When the force parallel to the plane acting on the box has magnitude \(X\) newtons the box is about to move down the plane, and when this force has magnitude \(5 X\) newtons the box is about to move up the plane (see Fig. 2).
  2. Find the value of \(X\) and the coefficient of friction between the box and the plane.
  3. A particle \(P\) of mass 1.2 kg is released from rest at the top of a slope and starts to move. The slope has length 4 m and is inclined at \(25 ^ { \circ }\) to the horizontal. The coefficient of friction between \(P\) and the slope is \(\frac { 1 } { 4 }\). Find
    1. the frictional component of the contact force on \(P\),
    2. the acceleration of \(P\),
    3. the speed with which \(P\) reaches the bottom of the slope.
    4. After reaching the bottom of the slope, \(P\) moves freely under gravity and subsequently hits a horizontal floor which is 3 m below the bottom of the slope.
      (a) Find the loss in gravitational potential energy of \(P\) during its motion from the bottom of the slope until it hits the floor.
      (b) Find the speed with which \(P\) hits the floor.
      [0pt] [1]
      [0pt] [3] \(7 \quad\) A particle \(P\) starts to move from a point \(O\) and travels in a straight line. At time \(t\) s after \(P\) starts to move its velocity is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), where \(v = 0.12 t - 0.0006 t ^ { 2 }\).
      1. Verify that \(P\) comes to instantaneous rest when \(t = 200\), and find the acceleration with which it starts to return towards \(O\).
      2. Find the maximum speed of \(P\) for \(0 \leqslant t \leqslant 200\).
      3. Find the displacement of \(P\) from \(O\) when \(t = 200\).
      4. Find the value of \(t\) when \(P\) reaches \(O\) again.

Question 5:
Part (i):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(R = 15 \times 10 \times \cos 35° = 123\) (AG)B1 Do not allow answers from \(g = 9.81\) or \(g = 9.8\); accept sin instead of cos for first M1 in (ii)
Part (ii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
For resolving forces along the plane (either case)M1
\(150\sin 35° = X + F\) and \(150\sin 35° = 5X - F\)A1
For eliminating \(F\) or \(X\)M1
\(X = 28.7\) (ft from wrong \(F\) or wrong positive \(\mu\)); (\(28.1\) from \(g = 9.81\) or \(g = 9.8\))A1ft
\(F\) or \(\mu R = 10g\sin 35°\) or equivalent (may be implied) \((57.36)\)A1
For using \(F = \mu R\): \([57.36 = \mu \times 122.9\) or \(100\sin 35° = \mu \times 150\cos 35°]\)M1 Allow \(\mu = 0.466\)
Coefficient of friction is \(0.467\) (ft for positive value from wrong \(X\)) \([(\frac{2}{3})\tan 35°]\)A1ft Total: 7
SR for candidate not using \(F\) explicitly and using \(F \leq \mu R\) implicitly (max 4 out of 7):
AnswerMarks
\(150\sin 35° - X \leq \mu R\) and \(5X - 150\sin 35° \leq \mu R\)A1
For eliminating \(X\) (not possible to eliminate \(\mu R\))M1
\(\mu R \geq 100\sin 35°\) or equivalentA1
## Question 5:

### Part (i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $R = 15 \times 10 \times \cos 35° = 123$ (AG) | B1 | Do not allow answers from $g = 9.81$ or $g = 9.8$; accept sin instead of cos for first M1 in (ii) |

### Part (ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| For resolving forces along the plane (either case) | M1 | |
| $150\sin 35° = X + F$ and $150\sin 35° = 5X - F$ | A1 | |
| For eliminating $F$ or $X$ | M1 | |
| $X = 28.7$ (ft from wrong $F$ or wrong positive $\mu$); ($28.1$ from $g = 9.81$ or $g = 9.8$) | A1ft | |
| $F$ or $\mu R = 10g\sin 35°$ or equivalent (may be implied) $(57.36)$ | A1 | |
| For using $F = \mu R$: $[57.36 = \mu \times 122.9$ or $100\sin 35° = \mu \times 150\cos 35°]$ | M1 | Allow $\mu = 0.466$ |
| Coefficient of friction is $0.467$ (ft for positive value from wrong $X$) $[(\frac{2}{3})\tan 35°]$ | A1ft | **Total: 7** |

**SR** for candidate not using $F$ explicitly and using $F \leq \mu R$ implicitly (max 4 out of 7):

| $150\sin 35° - X \leq \mu R$ and $5X - 150\sin 35° \leq \mu R$ | A1 | |
|---|---|---|
| For eliminating $X$ (not possible to eliminate $\mu R$) | M1 | |
| $\mu R \geq 100\sin 35°$ or equivalent | A1 | |

---
5

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{fcd2b219-d9b4-4972-b8fe-25cf543b9054-3_245_335_580_906}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}

A force, whose direction is upwards parallel to a line of greatest slope of a plane inclined at $35 ^ { \circ }$ to the horizontal, acts on a box of mass 15 kg which is at rest on the plane. The normal component of the contact force on the box has magnitude $R$ newtons (see Fig. 1).\\
(i) Show that $R = 123$, correct to 3 significant figures.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{fcd2b219-d9b4-4972-b8fe-25cf543b9054-3_369_1045_1247_555}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}

When the force parallel to the plane acting on the box has magnitude $X$ newtons the box is about to move down the plane, and when this force has magnitude $5 X$ newtons the box is about to move up the plane (see Fig. 2).\\
(ii) Find the value of $X$ and the coefficient of friction between the box and the plane.\\
(i) A particle $P$ of mass 1.2 kg is released from rest at the top of a slope and starts to move. The slope has length 4 m and is inclined at $25 ^ { \circ }$ to the horizontal. The coefficient of friction between $P$ and the slope is $\frac { 1 } { 4 }$. Find
\begin{enumerate}[label=(\alph*)]
\item the frictional component of the contact force on $P$,
\item the acceleration of $P$,
\item the speed with which $P$ reaches the bottom of the slope.\\
(ii) After reaching the bottom of the slope, $P$ moves freely under gravity and subsequently hits a horizontal floor which is 3 m below the bottom of the slope.\\
(a) Find the loss in gravitational potential energy of $P$ during its motion from the bottom of the slope until it hits the floor.\\
(b) Find the speed with which $P$ hits the floor.\\[0pt]
[1]\\[0pt]
[3]\\
$7 \quad$ A particle $P$ starts to move from a point $O$ and travels in a straight line. At time $t$ s after $P$ starts to move its velocity is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, where $v = 0.12 t - 0.0006 t ^ { 2 }$.
\begin{enumerate}[label=(\roman*)]
\item Verify that $P$ comes to instantaneous rest when $t = 200$, and find the acceleration with which it starts to return towards $O$.
\item Find the maximum speed of $P$ for $0 \leqslant t \leqslant 200$.
\item Find the displacement of $P$ from $O$ when $t = 200$.
\item Find the value of $t$ when $P$ reaches $O$ again.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2002 Q5 [8]}}