Challenging +1.3 This is a multi-part SHM question requiring equilibrium analysis, derivation of SHM conditions using Hooke's law with two strings, and energy conservation. While it involves several steps and careful bookkeeping of two elastic strings, the techniques are standard for Further Maths mechanics: finding equilibrium position, showing restoring force proportional to displacement, and applying energy methods. The conceptual framework is well-established for FM students, though the algebra requires care.
\(A\) and \(B\) are two fixed points on a smooth horizontal surface, with \(AB = 3a\) m. One end of a light elastic string, of natural length \(a\) m and modulus of elasticity \(mg\) N, is attached to the point \(A\). The other end of this string is attached to a particle \(P\) of mass \(m\) kg. One end of a second light elastic string, of natural length \(ka\) m and modulus of elasticity \(2mg\) N, is attached to \(B\). The other end of this string is attached to \(P\). Given that the system is in equilibrium when \(P\) is at \(M\), the mid-point of \(AB\), find the value of \(k\). [3]
The particle \(P\) is released from rest at a point between \(A\) and \(B\) where both strings are taut. Show that \(P\) performs simple harmonic motion and state the period of the motion. [5]
In the case where \(P\) is released from rest at a distance \(0.2a\) m from \(M\), the speed of \(P\) is \(0.7\) m s\(^{-1}\) when \(P\) is \(0.05a\) m from \(M\). Find the value of \(a\). [3]
Find k by equating equilibrium tensions: mg (a/2)/a = 2mg (3a/2 – ka) / ka M1 A1
(vertical motion can earn M1 only) ½ = 3/k – 2, k = 6/5 or 1⋅2 A1
2 2
Apply Newton’s law at general point, e.g.: m d x/dt = – mg (a/2 + x)/a
(lose A1 for each incorrect term) + 2mg (3a/2 – ka – x) / ka
2 2
or m d y/dt = + mg (a/2 – y)/a
– 2mg (3a/2 – ka + y) / ka M1 A2
2 2
Simplify to give standard SHM eqn, e.g.: d x/dt = – (1 + 2/k)gx / a
S.R.: B1 if no derivation (max 2/5) = – 8gx/3a A1
State or find period using 2π/ω with ω = √(8g/3a): T = 2π√(3a/8g) or π√(3a/2g)
(√ on ω) or 3⋅85√(a/g) or 1⋅22√a [s] B1
Substitute values in v 2 = ω 2 (x0 2 – x 2 ): 0⋅7 2 = (8g/3a){(0⋅2a) 2 – (0⋅05a) 2 } M1 A1
Answer
Marks
Solve to find numerical value of a: 0⋅49 = (8g/3) × 0⋅0375a, a = 0⋅49 A1
3
5
Answer
Marks
Guidance
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11
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Mark Scheme
Syllabus
Cambridge International A Level – October/November 2015
9231
21
Question 3:
3 | Find k by equating equilibrium tensions: mg (a/2)/a = 2mg (3a/2 – ka) / ka M1 A1
(vertical motion can earn M1 only) ½ = 3/k – 2, k = 6/5 or 1⋅2 A1
2 2
Apply Newton’s law at general point, e.g.: m d x/dt = – mg (a/2 + x)/a
(lose A1 for each incorrect term) + 2mg (3a/2 – ka – x) / ka
2 2
or m d y/dt = + mg (a/2 – y)/a
– 2mg (3a/2 – ka + y) / ka M1 A2
2 2
Simplify to give standard SHM eqn, e.g.: d x/dt = – (1 + 2/k)gx / a
S.R.: B1 if no derivation (max 2/5) = – 8gx/3a A1
State or find period using 2π/ω with ω = √(8g/3a): T = 2π√(3a/8g) or π√(3a/2g)
(√ on ω) or 3⋅85√(a/g) or 1⋅22√a [s] B1
Substitute values in v 2 = ω 2 (x0 2 – x 2 ): 0⋅7 2 = (8g/3a){(0⋅2a) 2 – (0⋅05a) 2 } M1 A1
Solve to find numerical value of a: 0⋅49 = (8g/3) × 0⋅0375a, a = 0⋅49 A1 | 3
5
3 | 11
Page 4 | Mark Scheme | Syllabus | Paper
Cambridge International A Level – October/November 2015 | 9231 | 21
$A$ and $B$ are two fixed points on a smooth horizontal surface, with $AB = 3a$ m. One end of a light elastic string, of natural length $a$ m and modulus of elasticity $mg$ N, is attached to the point $A$. The other end of this string is attached to a particle $P$ of mass $m$ kg. One end of a second light elastic string, of natural length $ka$ m and modulus of elasticity $2mg$ N, is attached to $B$. The other end of this string is attached to $P$. Given that the system is in equilibrium when $P$ is at $M$, the mid-point of $AB$, find the value of $k$. [3]
The particle $P$ is released from rest at a point between $A$ and $B$ where both strings are taut. Show that $P$ performs simple harmonic motion and state the period of the motion. [5]
In the case where $P$ is released from rest at a distance $0.2a$ m from $M$, the speed of $P$ is $0.7$ m s$^{-1}$ when $P$ is $0.05a$ m from $M$. Find the value of $a$. [3]
\hfill \mbox{\textit{CAIE FP2 2015 Q3 [11]}}