CAIE M2 2019 June — Question 5 8 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2019
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeElastic string – horizontal circle on surface
DifficultyStandard +0.3 Part (i) requires setting up two simultaneous equations using Hooke's law (T = λx/a), which is straightforward algebra. Part (ii) applies circular motion (T = mv²/r) with the found values. Standard multi-step application of formulas with no novel insight required, slightly above average due to the combination of elasticity and circular motion topics.
Spec6.02g Hooke's law: T = k*x or T = lambda*x/l6.05c Horizontal circles: conical pendulum, banked tracks

5 A light elastic string has natural length \(a \mathrm {~m}\) and modulus of elasticity \(\lambda \mathrm { N }\). When the length of the string is 1.6 m the tension is 4 N . When the length of the string is 2 m the tension is 6 N .
  1. Find the values of \(a\) and \(\lambda\).
    One end of the string is attached to a fixed point \(O\) on a smooth horizontal surface. The other end of the string is attached to a particle \(P\) of mass 0.2 kg . The particle \(P\) moves with constant speed on the surface in a circle with centre \(O\) and radius 1.9 m .
  2. Find the speed of \(P\).

Question 5(i):
AnswerMarks Guidance
\(4 = \frac{\lambda(1.6-a)}{a}\)B1 Use \(T = \left(\frac{\lambda x}{l}\right)\) twice
\(6 = \frac{\lambda(2-a)}{a}\)B1
\(1.5 = \frac{(2-a)}{(1.6-a)}\)M1 Attempt to solve the simultaneous equations
\(0.4 = 0.5(a),\ a = 0.8\)A1
\(\lambda = 4\)A1
Question 5(ii):
AnswerMarks Guidance
\(T = 4 \times \frac{1.1}{0.8}\ (= 5.5)\)B1 FT Use \(T = \frac{\lambda x}{L}\), ft candidates \(\lambda\) and \(a\)
\(5.5 = \frac{0.2v^2}{1.9}\)M1 Use Newton's Second Law horizontally
\(v = 7.23\ \text{ms}^{-1}\)A1
## Question 5(i):

| $4 = \frac{\lambda(1.6-a)}{a}$ | B1 | Use $T = \left(\frac{\lambda x}{l}\right)$ twice |
| $6 = \frac{\lambda(2-a)}{a}$ | B1 | |
| $1.5 = \frac{(2-a)}{(1.6-a)}$ | M1 | Attempt to solve the simultaneous equations |
| $0.4 = 0.5(a),\ a = 0.8$ | A1 | |
| $\lambda = 4$ | A1 | |

## Question 5(ii):

| $T = 4 \times \frac{1.1}{0.8}\ (= 5.5)$ | B1 | FT Use $T = \frac{\lambda x}{L}$, ft candidates $\lambda$ and $a$ |
| $5.5 = \frac{0.2v^2}{1.9}$ | M1 | Use Newton's Second Law horizontally |
| $v = 7.23\ \text{ms}^{-1}$ | A1 | |

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5 A light elastic string has natural length $a \mathrm {~m}$ and modulus of elasticity $\lambda \mathrm { N }$. When the length of the string is 1.6 m the tension is 4 N . When the length of the string is 2 m the tension is 6 N .\\
(i) Find the values of $a$ and $\lambda$.\\

One end of the string is attached to a fixed point $O$ on a smooth horizontal surface. The other end of the string is attached to a particle $P$ of mass 0.2 kg . The particle $P$ moves with constant speed on the surface in a circle with centre $O$ and radius 1.9 m .\\
(ii) Find the speed of $P$.\\

\hfill \mbox{\textit{CAIE M2 2019 Q5 [8]}}