CAIE M2 2018 June — Question 2 6 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2018
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHooke's law and elastic energy
TypeVertical elastic string: released from rest at natural length or above (string initially slack)
DifficultyStandard +0.3 This is a standard energy conservation problem with elastic strings requiring students to set up and solve a quadratic equation. While it involves multiple energy forms (kinetic, gravitational potential, elastic), the approach is methodical and well-practiced in M2. The calculation is straightforward once the energy equation is established, making it slightly above average difficulty but routine for this level.
Spec6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle
One end of a light elastic string is attached to a fixed point \(O\). The other end of the string is attached to a particle \(P\) of mass 0.24 kg. The string has natural length 0.6 m and modulus of elasticity 24 N. The particle is released from rest at \(O\). Find the two possible values of the distance \(OP\) for which the particle has speed 1.5 m s\(^{-1}\). [6]
Question 2:
AnswerMarks Guidance
2EPE = 24 ( x−0.6 )2 /(2 × 0.6) B1
0.4 × 1.52/2 = 0.4gx – 24 ( x−0.6 )2 /(2 × 0.6)
AnswerMarks Guidance
[20x2 – 28x + 7.65 = 0 or equivalent]M1 Attempt to find a 3 term energy equation
M1Attempt to solve the 3 term quadratic equation
OP = 1.0279 m, 0.372 m (reject)A1 Correct answer chosen
0.4 × 1.52/2 = 0.4gxM1 Note the particle is moving upwards and the string is slack
OP = 0.1125 mA1
Total:6
QuestionAnswer Marks
2Alternative method
EPE = 24x2/(2 × 0.6)B1 x is the extension
0.4 × 1.52/2 = 0.4g(x + 0.6) – 24x2/(2 × 0.6)
AnswerMarks Guidance
[20x2 – 4x – 1.95 = 0 or equivalent ]M1 Attempt to find a 3 term energy equation
M1Attempt to solve the 3 term quadratic equation
[ x = 0.42787, – 0.22787 .reject] OP = 0.6 + 0.42787 = 1.0279A1
0.4 × 1.52/2 = 0.4g( x + 0.6) [x = – 0.4875]M1 Note the particle is moving upwards and the string is slack
OP = 0.6 – 0.4875 = 0.1125A1
Total:6 
Question 2:
2 | EPE = 24 ( x−0.6 )2 /(2 × 0.6) | B1 | Correct EPE term. Note x = OP
0.4 × 1.52/2 = 0.4gx – 24 ( x−0.6 )2 /(2 × 0.6)
[20x2 – 28x + 7.65 = 0 or equivalent] | M1 | Attempt to find a 3 term energy equation
M1 | Attempt to solve the 3 term quadratic equation
OP = 1.0279 m, 0.372 m (reject) | A1 | Correct answer chosen
0.4 × 1.52/2 = 0.4gx | M1 | Note the particle is moving upwards and the string is slack
OP = 0.1125 m | A1
Total: | 6
Question | Answer | Marks | Guidance
2 | Alternative method
EPE = 24x2/(2 × 0.6) | B1 | x is the extension
0.4 × 1.52/2 = 0.4g(x + 0.6) – 24x2/(2 × 0.6)
[20x2 – 4x – 1.95 = 0 or equivalent ] | M1 | Attempt to find a 3 term energy equation
M1 | Attempt to solve the 3 term quadratic equation
[ x = 0.42787, – 0.22787 .reject] OP = 0.6 + 0.42787 = 1.0279 | A1
0.4 × 1.52/2 = 0.4g( x + 0.6) [x = – 0.4875] | M1 | Note the particle is moving upwards and the string is slack
OP = 0.6 – 0.4875 = 0.1125 | A1
Total: | 6
One end of a light elastic string is attached to a fixed point $O$. The other end of the string is attached to a particle $P$ of mass 0.24 kg. The string has natural length 0.6 m and modulus of elasticity 24 N. The particle is released from rest at $O$. Find the two possible values of the distance $OP$ for which the particle has speed 1.5 m s$^{-1}$. [6]

\hfill \mbox{\textit{CAIE M2 2018 Q2 [6]}}
This paper (7 questions)
View full paper
Q1 6 Q2 6 Q3 5 Q4 7 Q5 7 Q6 9 Q7 10