CAIE M2 2002 November — Question 5 9 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2002
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHooke's law and elastic energy
TypeElastic string on smooth inclined plane
DifficultyStandard +0.8 This is a multi-step energy and dynamics problem requiring: (i) energy conservation with elastic potential energy on an incline (non-trivial setup), and (ii) applying Hooke's law with Newton's second law at maximum extension. Requires careful handling of the elastic string (only in tension when extended) and coordinate geometry on an incline, making it moderately challenging but still within standard M2 scope.
Spec6.02g Hooke's law: T = k*x or T = lambda*x/l6.02i Conservation of energy: mechanical energy principle

5 A light elastic string has natural length 2 m and modulus of elasticity 1.5 N . One end of the string is attached to a fixed point \(O\) of a smooth plane which is inclined at \(30 ^ { \circ }\) to the horizontal. The other end of the string is attached to a particle \(P\) of mass \(0.075 \mathrm {~kg} . P\) is released from rest at \(O\). Find
  1. the distance of \(P\) from \(O\) when \(P\) is at its lowest point,
  2. the acceleration with which \(P\) starts to move up the plane immediately after it has reached its lowest point.

(i)
AnswerMarks Guidance
\(\text{GPE} = 0.075g(d\sin 30°)\) or \(0.075g(d+x)\sin 30°\)B1
\(\text{EPE} = 1.5(d-2)^2/2x2\) or \(1.5x^2/2x2\)B1
Uses the principle of conservation of energy to form an equation with GPE and EPE termsM1*
\([\frac{3}{8}d = \frac{3}{8}(d-2)^2\) or \(\frac{3}{8}(2+x) = \frac{3}{8}x^2]\)
Attempts to solve a quadratic equation in \(d\) or attempts to solve a quadratic equation in \(x\) and uses \(d = x+2\)M1 dep
\([(d-1)(d-4) = 0]\) and \(d = 2 + 2]\)
Obtains distance as 4mA1 5 marks
(ii)
AnswerMarks Guidance
Obtains the tension at the lowest point as 1.5 NB1 ft
ft for 1.5\((d-2)/2\)
Uses Newton's 2nd law to obtain a 3 term equationM1
Obtains \(1.5 - 0.075g\sin 30° = 0.075a\)A1
Obtains acceleration as 15m\(s^{-2}\)A1 4 marks
**(i)**
$\text{GPE} = 0.075g(d\sin 30°)$ or $0.075g(d+x)\sin 30°$ | B1 |
$\text{EPE} = 1.5(d-2)^2/2x2$ or $1.5x^2/2x2$ | B1 |
Uses the principle of conservation of energy to form an equation with GPE and EPE terms | M1* |
$[\frac{3}{8}d = \frac{3}{8}(d-2)^2$ or $\frac{3}{8}(2+x) = \frac{3}{8}x^2]$ | |
Attempts to solve a quadratic equation in $d$ or attempts to solve a quadratic equation in $x$ and uses $d = x+2$ | M1 dep |
$[(d-1)(d-4) = 0]$ and $d = 2 + 2]$ | |
Obtains distance as 4m | A1 | 5 marks |

**(ii)**
Obtains the tension at the lowest point as 1.5 N | B1 ft |
ft for 1.5$(d-2)/2$ | |
Uses Newton's 2nd law to obtain a 3 term equation | M1 |
Obtains $1.5 - 0.075g\sin 30° = 0.075a$ | A1 |
Obtains acceleration as 15m$s^{-2}$ | A1 | 4 marks |
5 A light elastic string has natural length 2 m and modulus of elasticity 1.5 N . One end of the string is attached to a fixed point $O$ of a smooth plane which is inclined at $30 ^ { \circ }$ to the horizontal. The other end of the string is attached to a particle $P$ of mass $0.075 \mathrm {~kg} . P$ is released from rest at $O$. Find\\
(i) the distance of $P$ from $O$ when $P$ is at its lowest point,\\
(ii) the acceleration with which $P$ starts to move up the plane immediately after it has reached its lowest point.

\hfill \mbox{\textit{CAIE M2 2002 Q5 [9]}}