Answers
35 a The period of motion is 6 s. b 6.25 m s−2 c 0.68 m s−1 (2 s.f.)
T = _2 ω_π_ d As it passes through C, P is moving away from O
ω = _3π _ towards B.
Measuring the time, t seconds, from an instant
40 a When P is at the point X, AX = x metres and the
force of the spring be T newtons.
when C is at Q and the displacement from the Hooke’s Law:
centre of the oscillation, O: x = a cos ωt T = _λ _lx _ = _2_1_2._6 _x_ = 10.8x
_3π _ R(→): F = ma
The amplitude is 2l and ω =
After 0.75 s, C is at P:
__ − T = 0.3 x¨
x = 2l cos( _3π_ × 0.7_5_ )= (22l −co√s _2 _ ω4_ _ ) l = − 10.8x = 0.3 x¨
b = a_−_ x = 2l − √ 2 l = √ 2 l
b v = _√ _ 23_ _ π _l x¨ = − 36x = − 6 2 x
Comparing with the standard formula for simple
c 0.28 s (2 d.p.) harmonic motion, this is simple harmonic motion
36 a At C: v 2= ω 2( a 2− x 2) with ω = 6. c _1π_8_ s d 1.16 m (3 s.f.)
0 = ω 2( a 2 − 1.2 2) ⇒ a = 1.2 b 9T m= s_2 ω−_π1_ = _2 6_π_ = _ 3π_
At A: v 2= ω 2( a 2− x 2) 41 a 4 mg
( _ 13_0 √ _3_ ) 2= ω 2( 1.2 2 − 0.6 2) b Hooke’s Law:
ω 2 = _ 12_078_ = _ 41_ ⇒ ω = _21 _
_12 _ T = _λ _le _ = _ 4_m__g_( _41_l_ _l _+__x_) = mg + _4 _m_l_ g_x_
Checking a = 1.2 and ω = at B (1)
v 2= ω 2( a 2− x 2) Newton’s second law:
O==,__ 5114 __x (√1 _=5._ 2 0 2: − 0.8 2) = _ 51_ R(↓): F = ma
mg − t = m _dd_ 2t_ x2_
b v (2)
At
v 2 = _ 14_ ( 1.2 2 − 0 2) = 0.36
v = 0.6 ms−2 Substitute (1) into (2):
mg − (mg + _4 _m_l_ g_x_ ) = m _d d_ 2t_ x2_
c 0.15 m s−2 d 0.412 s (3 s.f.)
37 a _4π_ m h−1
b 4 hours − _4_m_l_ g_x_ = m _dd_ 2t_ x2_
38 a Let the piston be modelled by the particle P.
Let O be the point where AO = 0.6 m _d d_ 2t_ x2_ = − _4_gl_ x_
When P is at a general point in its motion. c _ 12 √_3_g__l_
Let OP = x metres and the force of the spring on P
be T newtons.
d First P moves freely under gravity until it returns
Hooke’s Law: to B. Then it moves with simple harmonic motion
T = _λ _lx _ = _40_8._6x_ = 80x
about O.
42 a _ 92_a_
R(→): F = ma b When P is at a general point X, let OX = x.
− T = 0.2 x¨ At this point the extension of the string is 0.5a + x.
− 80x = 0.2 x¨
x¨ = − 400x = − 20 2 x Hooke’s Law: (1)
Comparing with the standard formula for simple
harmonic motion, this is simple T = _λ _le _ = _ 8_m__g_(_ _12a_ _ a __+__x_) = mg + _2 _m_a_g _x_
harmonic motion with ω = 20. Newton’s second law:
T = _22 _0π_ = _ 1π_0_
c _1π_5_ s R(↓): F = ma (2)
b 6 m s−1 mg − t = m _d d_ 2t_ x2_
39 a Let E be the point where OE = 0.6 m Substitute (1) into (2):
When P is at a general point in its motion, let EP = x mg − (mg + _2 _m_a_g _x_ ) = m _dd_ 2t_ x2_
metres and the force of the
− _2_m_a_g _x_ = m _dd_ 2t_ x2_
spring on P be T newtons. _ dd_ 2t_ x2_ = − _2_ag_x _
Hooke’s Law:
T = _λ _lx _ = _10_2._6x_ = 20x
R(→): F = ma
− T = 0.8 x¨ Comparing with the standard formula for simple
− 20x = 0.8 x¨ harmonic motion, this is simple harmonic motion
x ¨ = − 25x = − 5 2 x ___
Comparing with the standard formula for simple 2_ 2πa_g _ __ 2a__g__ =
harmonic motion, this is simple √ √ √ with ω = ___
T = _2 ω_π_ = π
harmonic motion with ω = 5. _ 2_√1__2 __ a _2g_a_
T = _2 ω_π_ = _2 5_π_
c
244 On line Full worked solutions are available in SolutionBank.
Answers
d As a . _ 12 a, the string will become slack during its Exam-style practice: A level
motion. The subsequent motion of P will be partly
under gravity, partly simple harmonic motion. 1 a i 12.5 hours ii 5 m
b 2.46 m h–1
Challenge c 7.05 hours (3 s.f.)
1 a _dd_vt_ = − e 2kv 2 a i _11_1_2a_ ii _51_a2_
0∫ ∫ −2kv dv = − t b 24o
e dt 3 a _41_ ln( _ kk_ 2_ 2+_+_4_U_ U_ 2_ 2 ) m
u0
− _21_k_ [e 0
−2kv] − t
u =
t = _2 1_k_ ( 1 − e −2ku) U 2 U _k _ 2_+1__ v_ _2 dv = t
∫ ∫ b
− 2 dt
0
= _21_k_ ( _ e _ 2_ek_u 2_k−_u _1_ ) [ _ 1k_ arctan _kv_ ] U2U = − [2t] 0t
_ 1k_ arctan _2k_U_ − _ 1k_ arctan _Uk_ = − 2t
b _ _1_ _ − _ 1__+__2_k_u_ m
(4 k 2 4 k 2 e 2ku) So t = _2 1_k_ ( arctan _Uk_ − arctan _2k_U_ )
2 a = 8x _dd_xt_ 4 a 47.4 m s–1
Using a = v _dd_xv_ and v = _dd _xt_
v _dd_xv_ = 8xv b 23.4 , v , 77.4
∫ 5 a V= _4π_ = π[ _12_ sin 2x] _4π_ = π( _21_ − 0) = _ 2π_
_ dd_xv_ = 8x π c os 2x dx 0
0
b _4π_ − _ 21_
∫ dv = ∫ 8xdx c α = 74° (2 s.f.)
v = 4 x 2 + c 6 a At P, K.E. = _32 _ ag and P.E. = mg(a – a cos 60°) = _12 _ mg
so total energy is 2amg.
t = 0, x = 0, v = −k: −k = c
_dd_xt_ = 4 x 2 − k
Displacement When the string makes angle θ with OQ,
has maximum when _dd _xt_ = 0 total energy = _21 _ mv 2 + amg(1 – cos θ)
__ ⇒ v2 = 2ag(1 + cos θ)
2 − k = 0 ⇒ x = _√ 2_ k_
4 x maximum displacement is _21 _ √ _k_ T – mg cos θ = _ m_rv_ _2
So ⇒ T = mg cos θ + 2mg(1 + cos θ)
3 _2_1g_k _ ln( _44_+_+_4_k_k U_ U_ 2 _2 ) = mg(2 + 3 cos θ)
Exam-style practice: AS level String goes slack when T = 0, so 2 + 3 cos θ = 0
⇒ cos θ = – _23_
⇒ θ = –131.8°
sv(e o= _ 2t_ v–_ 1=_1e2_ )_ 2et_ _1 _ , + __2t(__2 e__–(1 __2et_(_ 1e __ + _2t_ _ 2_ t_ – 1_+_1) 1_) ),⇒1_ 2((ee_ m _ 2_ _t2t__ _+– s_−1_11_)) , 1 This corresponds to an angle of 48.2° with the
upward vertical.
1 a
b P does not make complete circles, as the string goes
slack before P reaches the top of the circle.
2 a R(→): T cos 60° = m × 2a cos 60° × _4k _ag_ ⇒ T = _m _2_k_ g_
__
b mg(1 − _√ _ 34_ k_ ) c k < _ 4____
√ 3
_3 a_g_
d R(→): T cos θ = m × 2a cos θ ×
⇒ T = 6mg
R(↑): T sin θ = mg ⇒ 6mg sin θ = mg
⇒ sin θ = _16 _ = _3a _
QX = 2a sin θ
__
∴QOQX= :2 QaO si=n 6 _031 _ °: √= _3_ a = √ 13 : 3 √ _3_
3 a i 1.1a ii 3.25 a
b a
c 20.5°
d Weight acts through midpoint.
245
Index
Index
acceleration dynamics 171–211 Newton’s laws
circular motion 5–8, 19 horizontal oscillation 193–6 gravitation 181–3
horizontal circular path 5–8 motion in straight line with variable motion 172
straight line motion 147–52, 155–9, oscillation
force 172–8 horizontal 193–6
161–5, 172 Newton’s law of gravitation 181–3 vertical 198–203
varying with displacement 155–9 simple harmonic motion 184–91, 194, pendulums 11
varying with time 147–52 period, S.H.M. 185–91
varying with velocity 161–5 200 plane of symmetry 87
vertical circles 19 vertical oscillation 198–203 potential energy (P.E.) 19, 27
amplitude 185 elastic strings/springs 193–203 reference circle 187
angular speed 2–3 horizontal oscillation 193–6 resolving forces 11–15
axis of symmetry 48, 87 vertical oscillation 198–203 rigid bodies
calculus elemental strip 78 in equilibrium 103–7
acceleration 147–52, 155–9, 161–5 energy toppling and sliding 110–13
centres of mass 78–85 kinetic 19, 27 see also solids
centre of oscillation 184 potential 19, 27 rods, non-uniform 99–100
centres of mass 36–76, 77–123 equilibrium semicircles 81–2
calculus use to find 78–85 frameworks in 64–5 simple harmonic motion (S.H.M.)
composite lamina 47–51 laminas in 58–61, 103
framework 48, 54–6 rigid bodies in 103–7 184–91, 194, 200
frameworks in equilibrium 64–5 escape velocity 210 sliding 110–13
lamina in equilibrium 58–61, 103 frameworks 54 solids
non-uniform bodies 98–100 centres of mass 48, 54–6 non-uniform 98–100
non-uniform frameworks 68, 69–70 in equilibrium 64–5 of revolution 88–94
non-uniform laminas 68–9 non-uniform 68, 69–70 uniform 87–94
particles in plane 39–42 friction 7 see also rigid bodies
particles on straight line 37–8 frustum 94 speed
rigid bodies in equilibrium 103–7 hemispheres 90–1, 98–9, 104, 110–11 angular 2–3
standard uniform plane laminas 43–6 hemispherical shells 92–3 circular motion 2–3, 5–8, 11–15, 19–23
uniform bodies 87–94 Hooke’s law 194–6, 199–202 terminal (limiting) 162–3
uniform laminas 43, 78–85 horizontal circular motion 5–15 see also velocity
centroid 45 acceleration 5–8 spheres 87, 107
circles three-dimensional problems 11–15 springs see elastic strings/springs
arcs 84–5 integration straight line motion 146–70
sectors 83–4 constant of 147–8, 156, 157, 164, 172 simple harmonic motion 184–91, 194,
semicircles 81–2 straight line motion 147–52, 156–8,
circular motion 1–35 200
acceleration 5–8, 19 161–4 with variable force 172–8
angular speed 2–3 inverse square law 181 see also kinematics
horizontal 5–15 kinematics 146–70 strings see elastic strings/springs
object not staying on path 26–8 acceleration varying with displacement tension 194–5, 199–202
three-dimensional problems 11–15 time, acceleration varying with 147–52
vertical 19–23 155–9 toppling 110–13
coefficient of friction 7 acceleration varying with time 147–52 vector notation 19
cones 89–90, 98–9, 106 acceleration varying with velocity velocity
conical pendulum 11 acceleration varying with 161–5
conical shells 93–4 161–5 straight line motion 147–52, 155–9,
constant of gravitation 181–2 kinetic energy (K.E.) 19, 27
constant of integration 147–8, 156, 157, laminas 161–5
composite uniform 47–51 terminal (limiting) 149, 162–3
164, 172 in equilibrium 58–61, 103 see also speed
cubes 111–13 non-uniform composite 68–9 velocity–time graphs 148–9, 150–1, 152
cylinders 88–9, 99–100, 104–5, 110–11, 113 standard uniform 43–6 vertical circular motion 19–23
density 78 uniform 43, 78–85 acceleration 19
differentiation, straight line motion 147, mass, centre see centres of mass weight, line of action 110
medians 44 work–energy principle 19
149–50, 155–8, 161, 163–4 modulus of elasticity 194, 199
displacement
acceleration varying with 155–9
straight line motion 147–52, 155–9,
163–4
246246
Edexcel AS and A level Further Mathematics
Further Mechanics 2
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