AQA M2 2010 June — Question 9 8 marks

Exam BoardAQA
ModuleM2 (Mechanics 2)
Year2010
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeElastic string – horizontal circle on surface
DifficultyStandard +0.3 This is a straightforward application of circular motion with elastic strings requiring students to equate centripetal force to elastic tension using Hooke's law. The setup is standard M2 material with clear given values, requiring only substitution into F=mv²/r and T=λx/l, then solving a simple equation. Slightly easier than average due to the direct single-concept application with no complications.
Spec6.02h Elastic PE: 1/2 k x^26.05b Circular motion: v=r*omega and a=v^2/r
9 A particle, of mass 8 kg , is attached to one end of a length of elastic string. The particle is placed on a smooth horizontal surface. The other end of the elastic string is attached to a point \(O\) fixed on the horizontal surface. The elastic string has natural length 1.2 m and modulus of elasticity 192 N . \includegraphics[max width=\textwidth, alt={}, center]{3ffa0a2b-aa7d-46eb-b92b-3e3ee59f235c-18_165_789_571_630} The particle is set in motion on the horizontal surface so that it moves in a circle, centre \(O\), with constant speed \(3 \mathrm {~ms} ^ { - 1 }\). Find the radius of the circle. \includegraphics[max width=\textwidth, alt={}, center]{3ffa0a2b-aa7d-46eb-b92b-3e3ee59f235c-19_2349_1691_221_153} \includegraphics[max width=\textwidth, alt={}, center]{3ffa0a2b-aa7d-46eb-b92b-3e3ee59f235c-20_2505_1730_212_139}
Question 9:
A particle of mass 8 kg on a smooth horizontal surface, attached to elastic string (natural length 1.2 m, modulus 192 N), moves in a circle of centre O with speed 3 m s⁻¹. Find the radius.
AnswerMarks Guidance
Working/AnswerMark Guidance
Let radius = \(r\), extension = \(x = r - 1.2\)
Tension = \(\frac{\lambda x}{l} = \frac{192(r-1.2)}{1.2}\)M1 Correct use of Hooke's Law with extension \((r - 1.2)\)
\(T = 160(r - 1.2)\)A1 Correct expression for tension
Newton's second law for circular motion: \(T = \frac{mv^2}{r}\)M1 Correct equation of motion for circular motion
\(160(r - 1.2) = \frac{8 \times 9}{r}\)A1 Correct equation
\(160r(r - 1.2) = 72\)M1 Rearranging to form quadratic
\(160r^2 - 192r - 72 = 0\)A1 Correct quadratic
\(20r^2 - 24r - 9 = 0\)
\((10r + 3)(2r - 3) = 0\)M1 Attempt to solve quadratic
\(r = \frac{3}{2} = 1.5\) mA1 Correct positive root, rejecting negative
Radius = 1.5 m
## Question 9:

A particle of mass 8 kg on a smooth horizontal surface, attached to elastic string (natural length 1.2 m, modulus 192 N), moves in a circle of centre O with speed 3 m s⁻¹. Find the radius.

| Working/Answer | Mark | Guidance |
|---|---|---|
| Let radius = $r$, extension = $x = r - 1.2$ | | |
| Tension = $\frac{\lambda x}{l} = \frac{192(r-1.2)}{1.2}$ | M1 | Correct use of Hooke's Law with extension $(r - 1.2)$ |
| $T = 160(r - 1.2)$ | A1 | Correct expression for tension |
| Newton's second law for circular motion: $T = \frac{mv^2}{r}$ | M1 | Correct equation of motion for circular motion |
| $160(r - 1.2) = \frac{8 \times 9}{r}$ | A1 | Correct equation |
| $160r(r - 1.2) = 72$ | M1 | Rearranging to form quadratic |
| $160r^2 - 192r - 72 = 0$ | A1 | Correct quadratic |
| $20r^2 - 24r - 9 = 0$ | | |
| $(10r + 3)(2r - 3) = 0$ | M1 | Attempt to solve quadratic |
| $r = \frac{3}{2} = 1.5$ m | A1 | Correct positive root, rejecting negative |

**Radius = 1.5 m**
9 A particle, of mass 8 kg , is attached to one end of a length of elastic string. The particle is placed on a smooth horizontal surface. The other end of the elastic string is attached to a point $O$ fixed on the horizontal surface.

The elastic string has natural length 1.2 m and modulus of elasticity 192 N .\\
\includegraphics[max width=\textwidth, alt={}, center]{3ffa0a2b-aa7d-46eb-b92b-3e3ee59f235c-18_165_789_571_630}

The particle is set in motion on the horizontal surface so that it moves in a circle, centre $O$, with constant speed $3 \mathrm {~ms} ^ { - 1 }$.

Find the radius of the circle.

\includegraphics[max width=\textwidth, alt={}, center]{3ffa0a2b-aa7d-46eb-b92b-3e3ee59f235c-19_2349_1691_221_153}\\
\includegraphics[max width=\textwidth, alt={}, center]{3ffa0a2b-aa7d-46eb-b92b-3e3ee59f235c-20_2505_1730_212_139}

\hfill \mbox{\textit{AQA M2 2010 Q9 [8]}}
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