Edexcel M3 — Question 2 9 marks

Exam BoardEdexcel
ModuleM3 (Mechanics 3)
Marks9
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TopicHooke's law and elastic energy
TypeParticle attached to two separate elastic strings
DifficultyStandard +0.3 This is a standard M3 equilibrium problem with elastic strings requiring resolution of forces in two directions and application of Hooke's law. While it involves multiple steps (finding extensions, applying Hooke's law to both strings, resolving horizontally and vertically, solving simultaneous equations), the method is routine and well-practiced. The geometry is straightforward (3-4-5 triangle), and the question follows a standard template for this topic, making it slightly easier than average.
Spec6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^2
The figure shows a particle \(P\), of mass 0·8 kg, attached to the ends of two light elastic strings. \(AP\) has natural length 20 cm and modulus of elasticity \(\lambda\) N. \(BP\) has natural length 20 cm and modulus of elasticity \(\mu\) N. \(A\) and \(B\) are fixed to points on the same horizontal level so that \(AB = 50\) cm. When \(P\) is suspended in equilibrium, \(AP = 30\) cm and \(BP = 40\) cm. Calculate the values of \(\lambda\) and \(\mu\). \includegraphics{figure_2} [9 marks]
AnswerMarks Guidance
\(APB = 90°\), \(\sin \theta = 0.6\), \(\cos \theta = 0.8\)B1 M1
\(T \cos \theta + S \sin \theta = 0.8g\)A1 B1
\(4T + 3S = 39.2\)
Solve: \(S = 4.704 \text{ N}\), \(T = 6.272 \text{ N}\)M1 A1 (both)
Horiz: \(S \cos \theta = T \sin \theta\), so \(4S = 3T\)
Now \(S = \frac{\mu}{0.2} \times 0.2 = \mu\)M1 A1 A1
\(T = \frac{\lambda}{0.2} \times 0.1 = \frac{\lambda}{2}\)
So \(\lambda = 12.5\), \(\mu = 4.70\) Total: 9 marks 
$APB = 90°$, $\sin \theta = 0.6$, $\cos \theta = 0.8$ | B1 M1 |
$T \cos \theta + S \sin \theta = 0.8g$ | A1 B1 |
$4T + 3S = 39.2$ | |
Solve: $S = 4.704 \text{ N}$, $T = 6.272 \text{ N}$ | M1 A1 (both) |
Horiz: $S \cos \theta = T \sin \theta$, so $4S = 3T$ | |
Now $S = \frac{\mu}{0.2} \times 0.2 = \mu$ | M1 A1 A1 |
$T = \frac{\lambda}{0.2} \times 0.1 = \frac{\lambda}{2}$ | |
So $\lambda = 12.5$, $\mu = 4.70$ | | **Total: 9 marks**
The figure shows a particle $P$, of mass 0·8 kg, attached to the ends of two light elastic strings. $AP$ has natural length 20 cm and modulus of elasticity $\lambda$ N. $BP$ has natural length 20 cm and modulus of elasticity $\mu$ N. $A$ and $B$ are fixed to points on the same horizontal level so that $AB = 50$ cm. When $P$ is suspended in equilibrium, $AP = 30$ cm and $BP = 40$ cm. Calculate the values of $\lambda$ and $\mu$.

\includegraphics{figure_2}
[9 marks]

\hfill \mbox{\textit{Edexcel M3  Q2 [9]}}
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