Edexcel M3 2024 January — Question 4 9 marks

Exam BoardEdexcel
ModuleM3 (Mechanics 3)
Year2024
SessionJanuary
Marks9
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Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeParticle on cone surface – no string (normal reaction only)
DifficultyStandard +0.8 This is a challenging M3 circular motion problem requiring resolution of forces in 3D (normal reaction, friction, weight) on a conical surface, application of F=mrω² for circular motion, and careful handling of limiting friction. The geometry requires finding the cone's semi-vertical angle (tan α = 4/3), then solving simultaneous equations to eliminate the normal reaction and derive the given expression. While systematic, it demands strong spatial reasoning and algebraic manipulation beyond routine circular motion questions.
Spec3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks
4. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{92131234-bfc1-4e0e-87d4-db9335fbf343-12_760_1212_294_429} \captionsetup{labelformat=empty} \caption{Figure 3}
\end{figure} Figure 3 shows a thin hollow right circular cone fixed with its circular rim horizontal.
The centre of the circular rim is \(O\). The vertex \(V\) of the cone is vertically below \(O\).
The radius of the circular rim is \(4 a\) and \(O V = 3 a\).
A particle \(P\) of mass \(m\) moves in a horizontal circle of radius \(r ( 0 < r < 4 a )\) on the inner surface of the cone. The coefficient of friction between \(P\) and the inner surface of the cone is \(\frac { 1 } { 4 }\) The particle moves with a constant angular speed.
Show that the maximum possible angular speed is \(\sqrt { \frac { 16 g } { 13 r } }\)
Question 4:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Resolve vertically: \(R\sin\theta = mg + F\cos\theta\)M1, A1, A1 Need all terms. Dimensionally correct. Condone sign errors and sin/cos confusion. Unsimplified equation with at most one error; then correct unsimplified equation.
Equation for horizontal motion: \(R\cos\theta + F\sin\theta = mr\omega^2\)M1, A1, A1 Need all terms. Dimensionally correct. Condone sign errors and sin/cos confusion. Accept any form of acceleration for the method mark only. Direction of \(F\) consistent with vertical resolution. Incorrect form of acceleration is one error.
Use of \(F = \mu R\): \(F = \dfrac{1}{4}R\)M1 Used, not just quoted.
Substitute for trig and solve for max \(\omega\); \(\left(R = \dfrac{20mg}{13},\, F = \dfrac{5mg}{13}\right)\)DM1 Dependent on all preceding M marks. If more than two equations are produced, the correct two must be used.
\(\Rightarrow \omega = \sqrt{\dfrac{16g}{13r}}\) *A1* Obtain given answer from correct working
[9]
ALT1 (using N2L parallel and perpendicular to the incline):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Perpendicular: \(R - mg\sin\theta = mr\omega^2\cos\theta\)M1, A1A1 Need all terms. Dimensionally correct. Condone sign errors and sin/cos confusion. Note that the acceleration must have a sin/cos component. Accept any form of acceleration for the method mark only. Mark A's as above.
Parallel: \(F + mg\cos\theta = mr\omega^2\sin\theta\)M1, A1A1 A1A0 unsimplified equation with at most one error. A1A1 correct unsimplified equation. 
# Question 4:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolve vertically: $R\sin\theta = mg + F\cos\theta$ | M1, A1, A1 | Need all terms. Dimensionally correct. Condone sign errors and sin/cos confusion. Unsimplified equation with at most one error; then correct unsimplified equation. |
| Equation for horizontal motion: $R\cos\theta + F\sin\theta = mr\omega^2$ | M1, A1, A1 | Need all terms. Dimensionally correct. Condone sign errors and sin/cos confusion. Accept any form of acceleration for the method mark only. Direction of $F$ consistent with vertical resolution. Incorrect form of acceleration is one error. |
| Use of $F = \mu R$: $F = \dfrac{1}{4}R$ | M1 | Used, not just quoted. |
| Substitute for trig and solve for max $\omega$; $\left(R = \dfrac{20mg}{13},\, F = \dfrac{5mg}{13}\right)$ | DM1 | Dependent on all preceding M marks. If more than two equations are produced, the correct two must be used. |
| $\Rightarrow \omega = \sqrt{\dfrac{16g}{13r}}$ * | A1* | Obtain given answer from correct working |
| | [9] | |

**ALT1 (using N2L parallel and perpendicular to the incline):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Perpendicular: $R - mg\sin\theta = mr\omega^2\cos\theta$ | M1, A1A1 | Need all terms. Dimensionally correct. Condone sign errors and sin/cos confusion. Note that the acceleration must have a sin/cos component. Accept any form of acceleration for the method mark only. Mark A's as above. |
| Parallel: $F + mg\cos\theta = mr\omega^2\sin\theta$ | M1, A1A1 | A1A0 unsimplified equation with at most one error. A1A1 correct unsimplified equation. |
4.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{92131234-bfc1-4e0e-87d4-db9335fbf343-12_760_1212_294_429}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}

Figure 3 shows a thin hollow right circular cone fixed with its circular rim horizontal.\\
The centre of the circular rim is $O$. The vertex $V$ of the cone is vertically below $O$.\\
The radius of the circular rim is $4 a$ and $O V = 3 a$.\\
A particle $P$ of mass $m$ moves in a horizontal circle of radius $r ( 0 < r < 4 a )$ on the inner surface of the cone.

The coefficient of friction between $P$ and the inner surface of the cone is $\frac { 1 } { 4 }$\\
The particle moves with a constant angular speed.\\
Show that the maximum possible angular speed is $\sqrt { \frac { 16 g } { 13 r } }$

\hfill \mbox{\textit{Edexcel M3 2024 Q4 [9]}}
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