OCR Further Mechanics 2020 November — Question 2 6 marks

Exam BoardOCR
ModuleFurther Mechanics (Further Mechanics)
Year2020
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHooke's law and elastic energy
TypeBungee jumping problems
DifficultyStandard +0.8 This is a multi-step energy conservation problem requiring careful consideration of three phases (free fall, elastic extension, and air resistance throughout). Students must set up and solve a quadratic equation involving gravitational PE, elastic PE, and work against resistance. While the method is standard for Further Mechanics, the bookkeeping and algebraic manipulation elevate it above routine difficulty.
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 principle6.02j Conservation with elastics: springs and strings

2 A bungee jumper of mass 80 kg steps off a high bridge with an elastic rope attached to her ankles. She is assumed to fall vertically from rest and the air resistance she experiences is modelled as a constant force of 32 N . The rope has natural length 4 m and modulus of elasticity 470 N . By considering energy, determine the total distance she falls before first coming to instantaneous rest.

Question 2:
AnswerMarks
2Initial energy = 80g(x + 4) where x is the final
extension of the rope
470x2
Elastic PE =
2×4
Energy lost (work done by air resistance) =32(x+4)
470x2
80g(x+4)= +32(x+4)
2×4
235x2−3008x−12032=0 soi
x = 16
AnswerMarks
She falls 20 m before first coming to restM1
M1
M1
A1
M1
A1
AnswerMarks
[6]3.3
3.3
3.3
3.4
1.1
AnswerMarks
2.3Attempt GPE
Attempt EPE
Attempt Work done
Initial energy = Final Elastic
PE + energy lost
Can be multiple of this
Only from W-E principle
with three terms
AnswerMarks
–3.2 rejectedOr 80gd where d is the distance
below the bridge
470(d−4)2
2×4
470(d−4)2
80gd = +32d
8
235d2−4888d+3760=0
d = 20 (and 0.8 rejected)
Question 2:
2 | Initial energy = 80g(x + 4) where x is the final
extension of the rope
470x2
Elastic PE =
2×4
Energy lost (work done by air resistance) =32(x+4)
470x2
80g(x+4)= +32(x+4)
2×4
235x2−3008x−12032=0 soi
x = 16
She falls 20 m before first coming to rest | M1
M1
M1
A1
M1
A1
[6] | 3.3
3.3
3.3
3.4
1.1
2.3 | Attempt GPE
Attempt EPE
Attempt Work done
Initial energy = Final Elastic
PE + energy lost
Can be multiple of this
Only from W-E principle
with three terms
–3.2 rejected | Or 80gd where d is the distance
below the bridge
470(d−4)2
2×4
470(d−4)2
80gd = +32d
8
235d2−4888d+3760=0
d = 20 (and 0.8 rejected)
2 A bungee jumper of mass 80 kg steps off a high bridge with an elastic rope attached to her ankles. She is assumed to fall vertically from rest and the air resistance she experiences is modelled as a constant force of 32 N . The rope has natural length 4 m and modulus of elasticity 470 N .

By considering energy, determine the total distance she falls before first coming to instantaneous rest.

\hfill \mbox{\textit{OCR Further Mechanics 2020 Q2 [6]}}