Edexcel M3 2002 January — Question 5 10 marks

Exam BoardEdexcel
ModuleM3 (Mechanics 3)
Year2002
SessionJanuary
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeBanked track – with friction (find maximum/minimum speed or friction coefficient)
DifficultyChallenging +1.2 This is a standard M3 banked circular motion problem requiring resolution of forces in two directions, application of friction laws, and solving simultaneous equations. While it involves multiple steps and careful handling of the friction inequality, it follows a well-established method taught in M3 with no novel insight required. The 10 marks reflect the algebraic manipulation rather than conceptual difficulty, placing it moderately above average.
Spec6.05c Horizontal circles: conical pendulum, banked tracks
A cyclist is travelling around a circular track which is banked at 25° to the horizontal. The coefficient of friction between the cycle's tyres and the track is 0.6. The cyclist moves with constant speed in a horizontal circle of radius 40 m, without the tyres slipping. Find the maximum speed of the cyclist. [10]
AnswerMarks Guidance
\(R(\angle): \text{Run } 25° - F \sin 15° = mg\)M1 A2
\(R(\angle): R \sin 25° + F \cos 15° = \frac{mv^2}{40}\)M1 A2
\(F = 0.6R\) usedM1
Eliminates \(R\)M1
Solves for \(v\)M1
\(v = 2.4 \text{ ms}^{-1}, 2.4 \text{ ms}^{-1}\)A1 (10)
Question 6(a):
AnswerMarks Guidance
\(\text{If } SHM, \omega = 1.2\)B1
Using \(v^2 = \omega^2(a^2 - x^2)\)M1
\(0.27 = \omega^2(1.2^2 - 0.4^2)\) or \(0.2 = \omega^2(3x^2 - 0^2)\)A1
Solve for \(\omega\) (\(E = 0.5\)) and use in other equationM1
Should be consistentA1
Show be lowestA1 e.s.o. (5)
Question 6(b):
AnswerMarks Guidance
\(V = 6w = 1.23\omega S = 0.6\) *M1 A1 (2)
Question 6(c):
AnswerMarks Guidance
\(x = \omega^2 x 0.6 = 0.15 \text{ ms}^{-1}\)
Question 6(d):
AnswerMarks Guidance
\(0.6 = a \sin \omega t\) or \(0.8 = a \sin \omega t\)M1
\(t = \frac{1}{\omega}\left(\tan^{-1}\frac{0.8}{a} - \sin^{-1}\frac{0.6}{a}\right)\)M1 A1 (V)
\(= 0.442s\) (3SF)A1 (4)
Question 7(a):
AnswerMarks Guidance
\(\frac{1}{2}m_1 \text{Tas} - \frac{1}{2}mv^2 = mgs\)M1 A1
\((\leftarrow), R = \frac{mv^2}{a} = \frac{3mg}{2}\)M1 A1 (4)
Question 7(b):
AnswerMarks Guidance
\(\frac{1}{2}m_1 \text{Tas} - \frac{1}{2}mv^2 = mgs(1 + \omega \theta)\)M1 A1
\((b'), mgs\omega = \frac{mv^2}{a}\)M1 A1
Eliminates \(v^2\)M1
Solving to give \(\omega \theta = K, \theta = 60°\) *M1 A1 (7)
Question 7(c):
AnswerMarks Guidance
\(V \cos 60° = t = a \sin 60°\)M1
\(v^2 = a g \cos 60°\)B1
Make it explicitM1
\(L = \sqrt{\frac{a}{5}}\)A1 (4) 
$R(\angle): \text{Run } 25° - F \sin 15° = mg$ | M1 A2 |
$R(\angle): R \sin 25° + F \cos 15° = \frac{mv^2}{40}$ | M1 A2 |
$F = 0.6R$ used | M1 |
Eliminates $R$ | M1 |
Solves for $v$ | M1 |
$v = 2.4 \text{ ms}^{-1}, 2.4 \text{ ms}^{-1}$ | A1 | (10)

# Question 6(a):

$\text{If } SHM, \omega = 1.2$ | B1 |
Using $v^2 = \omega^2(a^2 - x^2)$ | M1 |
$0.27 = \omega^2(1.2^2 - 0.4^2)$ or $0.2 = \omega^2(3x^2 - 0^2)$ | A1 |
Solve for $\omega$ ($E = 0.5$) and use in other equation | M1 |
Should be consistent | A1 |
Show be lowest | A1 e.s.o. | (5)

# Question 6(b):

$V = 6w = 1.23\omega S = 0.6$ * | M1 A1 | (2)

# Question 6(c):

$|x| = \omega^2 x 0.6 = 0.15 \text{ ms}^{-1}$ | M1 A1 (V) | (2)

# Question 6(d):

$0.6 = a \sin \omega t$ or $0.8 = a \sin \omega t$ | M1 |
$t = \frac{1}{\omega}\left(\tan^{-1}\frac{0.8}{a} - \sin^{-1}\frac{0.6}{a}\right)$ | M1 A1 (V) |
$= 0.442s$ (3SF) | A1 | (4)

# Question 7(a):

$\frac{1}{2}m_1 \text{Tas} - \frac{1}{2}mv^2 = mgs$ | M1 A1 |
$(\leftarrow), R = \frac{mv^2}{a} = \frac{3mg}{2}$ | M1 A1 | (4)

# Question 7(b):

$\frac{1}{2}m_1 \text{Tas} - \frac{1}{2}mv^2 = mgs(1 + \omega \theta)$ | M1 A1 |
$(b'), mgs\omega = \frac{mv^2}{a}$ | M1 A1 |
Eliminates $v^2$ | M1 |
Solving to give $\omega \theta = K, \theta = 60°$ * | M1 A1 | (7)

# Question 7(c):

$V \cos 60° = t = a \sin 60°$ | M1 |
$v^2 = a g \cos 60°$ | B1 |
Make it explicit | M1 |
$L = \sqrt{\frac{a}{5}}$ | A1 | (4)
A cyclist is travelling around a circular track which is banked at 25° to the horizontal. The coefficient of friction between the cycle's tyres and the track is 0.6. The cyclist moves with constant speed in a horizontal circle of radius 40 m, without the tyres slipping.

Find the maximum speed of the cyclist.
[10]

\hfill \mbox{\textit{Edexcel M3 2002 Q5 [10]}}
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