CAIE FP2 2019 June — Question 2 8 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2019
SessionJune
Marks8
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Mark schemeDownload PDF ↗
TopicSimple Harmonic Motion
DifficultyChallenging +1.2 This is a Further Maths SHM problem requiring application of standard formulas (v² = ω²(a² - x²)) with algebraic manipulation to find amplitude and ω from given conditions. The multi-step nature, need to set up simultaneous equations from the speed ratio condition, and integration for time make it moderately challenging, but it follows standard SHM techniques without requiring novel insight.
Spec4.10f Simple harmonic motion: x'' = -omega^2 x
A particle \(P\) moves on a straight line in simple harmonic motion. The centre of the motion is \(O\). The points \(A\) and \(B\) are on the line on opposite sides of \(O\) such that \(OA = 3.5\) m and \(OB = 1\) m. The speed of \(P\) when it is at \(B\) is twice its speed when it is at \(A\). The maximum acceleration of \(P\) is 1 m s\(^{-2}\).
  1. Find the speed of \(P\) when it is at \(O\). [4]
  2. Find the time taken by \(P\) to travel directly from \(A\) to \(B\). [4]
Question 2:
AnswerMarks Guidance
2(i)v = 2v , v 2 = 4 v 2 , ω2 (a2 – 12) = 4ω2 (a2 – 3⋅52)
B A B AM1 Find amplitude a: allow M1 for v = 2 v
A B
AnswerMarks Guidance
3 a2 = 48, a = [±] 4 [m]A1
1 = aω2, ω = 1/√a [ = ½]B1 Find ω from given maximum acceleration
v = aω = √a = 2 [m s-1]
AnswerMarks Guidance
OB1 Find speed v at O
O
4
AnswerMarks
2(ii)ω t = sin-1 (3.5/a) + sin-1(1/a)
AB
AnswerMarks Guidance
= sin-1 0⋅875 + sin-1 0⋅25M1 A1 Find equation (AEF) for ω t , combining t and t ,
AB A B
using for example x = a sin ωt allow sign errors for the M1
= 1⋅065 + 0⋅253 (or t = 2⋅131 + 0⋅505)
AnswerMarks Guidance
ABA1 or x = a cos ωt
t = 2 × 1⋅318 = 2⋅64 [s]
AnswerMarks Guidance
ABA1 Hence find t
AB
Alternative method for question 2(ii)
ω t = cos-1 (–1/a) – cos-1(3.5/a)
AB
= cos-1 (–0⋅25) – cos-1 0⋅875
AnswerMarks Guidance
or π – cos-1 0⋅25 – cos-1 0⋅875 (AEF)M1 A1 or x = a cos ωt
= 1⋅823 – 0⋅505 (or t = 3⋅647 – 1⋅011)
AnswerMarks
ABA1
t = 2 × 1⋅318 = 2⋅64 [s]
AnswerMarks Guidance
ABA1 Hence find t
AB
4
AnswerMarks Guidance
QuestionAnswer Marks 
Question 2:
--- 2(i) ---
2(i) | v = 2v , v 2 = 4 v 2 , ω2 (a2 – 12) = 4ω2 (a2 – 3⋅52)
B A B A | M1 | Find amplitude a: allow M1 for v = 2 v
A B
3 a2 = 48, a = [±] 4 [m] | A1
1 = aω2, ω = 1/√a [ = ½] | B1 | Find ω from given maximum acceleration
v = aω = √a = 2 [m s-1]
O | B1 | Find speed v at O
O
4
--- 2(ii) ---
2(ii) | ω t = sin-1 (3.5/a) + sin-1(1/a)
AB
= sin-1 0⋅875 + sin-1 0⋅25 | M1 A1 | Find equation (AEF) for ω t , combining t and t ,
AB A B
using for example x = a sin ωt allow sign errors for the M1
= 1⋅065 + 0⋅253 (or t = 2⋅131 + 0⋅505)
AB | A1 | or x = a cos ωt
t = 2 × 1⋅318 = 2⋅64 [s]
AB | A1 | Hence find t
AB
Alternative method for question 2(ii)
ω t = cos-1 (–1/a) – cos-1(3.5/a)
AB
= cos-1 (–0⋅25) – cos-1 0⋅875
or π – cos-1 0⋅25 – cos-1 0⋅875 (AEF) | M1 A1 | or x = a cos ωt
= 1⋅823 – 0⋅505 (or t = 3⋅647 – 1⋅011)
AB | A1
t = 2 × 1⋅318 = 2⋅64 [s]
AB | A1 | Hence find t
AB
4
Question | Answer | Marks | Guidance
A particle $P$ moves on a straight line in simple harmonic motion. The centre of the motion is $O$. The points $A$ and $B$ are on the line on opposite sides of $O$ such that $OA = 3.5$ m and $OB = 1$ m. The speed of $P$ when it is at $B$ is twice its speed when it is at $A$. The maximum acceleration of $P$ is 1 m s$^{-2}$.

(i) Find the speed of $P$ when it is at $O$. [4]

(ii) Find the time taken by $P$ to travel directly from $A$ to $B$. [4]

\hfill \mbox{\textit{CAIE FP2 2019 Q2 [8]}}
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Q1 4 Q2 8 Q3 10 Q4 10 Q5 12 Q6 7 Q7 8 Q8 8 Q9 10 Q10 11 Q11 24