Chapter 5
A Hooke’s law: T = _ λ_lx _ Li nks λ is the modulus of elasticity of the spring
F = mm ax..
−T = and l is its natural length. ← FM1, Chapter 3
m x.. = − _ λ_lx _ λ, m and l are atlhlepofosritmivex.. constants, so the
x.. = − _ mλ__ l x equation is of = −ω 2x.
So P is moving with S.H.M.
The initial extension is the maximum value of x, so is the same as the amplitude.
When the particle is attached to an elastic spring, the particle will perform complete oscillations
because there will always be a force acting − a tension when the spring is stretched and a thrust when
the spring is compressed. The centre of the oscillation is where the tension is zero; that is the point
when the spring has returned to its natural length.
When the particle is attached to an elastic string, the particle will move with S.H.M. only while the
string is taut. Once the string becomes slack there is no tension and the particle continues to move
with constant speed until the string becomes taut again.
■■ For a particle moving on a smooth horizontal surface attached to one end of an elastic spring:
•• the particle will move with S.H.M.
•• the particle will perform complete oscillations.
■■ For a particle moving on a smooth horizontal surface attached to one end of an elastic string:
•• the particle will move with S.H.M. while the string is taut
•• the particle will move with constant speed while the string is slack.
■■ To solve problems involving elastic springs and strings:
•• use Hooke’s law to find the tension
•• use F = ma to obtain v
•• use information given in the question to obtain the amplitude.
Sometimes the particle is attached to two springs or strings which are stretched between two fixed
points. When this happens you will need to find the tensions in both the springs or strings.
Example 15
A particle P of mass 0.6 kg rests on a smooth horizontal floor attached to one end of a light elastic
string of natural length 0.8 m and modulus of elasticity 16 N. The other end of the string is fixed to
a point A on the floor. The particle is pulled away from A until AP measures 1.2 m and released.
a Show that, while the string remains taut, P moves with simple harmonic motion.
b Calculate the speed of P when the string returns to its natural length.
c Calculate the time that elapses between the point where the string becomes slack and the point
where it next becomes taut.
d Calculate the time taken by the particle to return to its starting point for the first time.
194
Dynamics
A x¨
A
TP
0.8 x Use Hooke’s law with λ = 16 and l = 0.8 to
a T = _ λ_lx _ find the tension.
T = _10_6._8x_ = 20x Use F = ma with F = T = 20x and m = 0.6.
F = ma Remember that the positive direction is
the direction of x increasing, and that the
20x = −0.6x.. acceleration acts in the opposite direction.
x¨ = − _02_.06_ x = − _103_0_ x
Tx..h=e−eωq u2xa,tisoonSr.eHd.Muc. eissptrootvheed.form
The particle is moving with S.H.M.
xC..o=m−pωar 2extthoefeinqduaωt 2i.on obtained in a with
b x¨ = − _103_0_ x
ω2 = _ 103_0_ The amplitude is the same as the initial
v 2 = ω 2(a2 − x2) extension and at the natural length x = 0.
v 2 = _ 103_0_ × 0.42
v = 2.309… The particle moves at a constant speed
At the natural length P has speed 2.31 m s−1 (3 s.f.). while the string is slack.
c The particle now moves a distance 1.6 m at The particle moves through a complete
2.309… m s−1. oscillation and two intervals of constant
Time taken = _ 2_.3_1_0.6_9_ …__ = 0.6928… speed (when the string is slack).
The string is slack for 0.693 s (3 s.f.).
____
√ d Period of the S.H.M. = _ 2ω_π_ = 2π × _10_3_0 _ = 1.088…
Total time = 1.088… + (2 × 0.6928…) = 2.473…
The time taken is 2.47 s (3 s.f.).
Example 16
A particle P of mass 0.8 kg is attached to the ends of two identical light elastic springs of natural
length 1.6 m and modulus of elasticity 16 N. The free ends of the springs are attached to two
points A and B which are 4 m apart on a smooth horizontal surface. The point C lies between A
and B such that ABC is a straight line and AC = 2.4 m. The particle is held at C and then released
from rest.
a Show that the subsequent motion is simple harmonic motion.
b Find the period and amplitude of the motion.
c Calculate the maximum speed of P.
195
Chapter 5
A x x¨
TA 0.8 kg TB Problem-solving
P
Use Hooke’s law to find the tensions in
A 2m 2m B each spring. Use your diagram to work
out the extensions.
a T = _ λ_lx _
TA = _ 16__(_0_1.4_.6 _ _+__x_) Use F = ma to form an equation of
TB = _ 16__(_0_1.4_.6 _ _−__x_)
xm.. =ot−ioωn 2xfotroPe.sRteadbulische this to the form
S.H.M.
F = ma 0.8x..
TB − TA =
_ 16__(_0_1.4_.6 _ _−__x_) − _ 16__(_0_1.4_.6 _ _+__x_) = 0.8x..
0.8x..
x.. −2 × _11_6.6_x_ =
= −2 × _ 1._6_1_×6__x0_ _.8_ = −25x
The motion is S.H.M. xC..o=m−pωa 2rxe the equation found in a with
to find ω.
b x.. = −25x
ω 2 = 25, ω = 5 You can give an exact value or a 3 s.f.
Period = _2ω_π_ = _ 25_π_ s answer (1.26 s) for the period.
Amplitude = 0.4 m
c v 2 = ω2(a2 − x2) As the springs are identical the centre of
v 2max = 25 × 0.42 the oscillation is at the midpoint of AB.
vmax = 5 × 0.4 = 2
P ’s maximum speed is 2 m s−1.
Exercise 5D
1 A particle P of mass 0.5 kg is attached to one end of a light elastic spring of natural length
0.6 m and modulus of elasticity 60 N. The other end of the spring is fixed to a point A on the
smooth horizontal surface on which P rests. The particle is held at rest with AP = 0.9 m and
then released.
a Show that P moves with simple harmonic motion.
b Find the period and amplitude of the motion.
c Calculate the maximum speed of P.
2 A particle P of mass 0.8 kg is attached to one end of a light elastic string of natural length 1.6 m
and modulus of elasticity 20 N. The other end of the string is fixed to a point O on the smooth
horizontal surface on which P rests. The particle is held at rest with OP = 2.6 m and then released.
a Show that, while the string is taut, P moves with simple harmonic motion.
b Calculate the time from the instant of release until P returns to its starting point for the first
time.
196
Dynamics
A 3 A particle P of mass 0.4 kg is attached to one end of a light elastic string of modulus of
elasticity 24 N and natural length 1.2 m. The other end of the string is fixed to a point A on
the smooth horizontal table on which P rests. Initially P is at rest with AP = 1 m. The particle
receives an impulse of magnitude 1.8 N s in the direction AP.
a Show that, while the string is taut, P moves with simple harmonic motion.
b Calculate the time that elapses between the moment P receives the impulse and the next time
the string becomes slack.
The particle comes to instantaneous rest for the first time at the point B.
c Calculate the distance AB.
P 4 A particle P of mass 0.8 kg is attached to one end of a light elastic spring of natural length 1.2 m
and modulus of elasticity 80 N. The other end of the spring is fixed to a point O on the smooth
horizontal surface on which P rests. The particle is held at rest with OP = 0.6 m and then released.
a Show that P moves with simple harmonic motion.
b Find the period and amplitude of the motion.
c Calculate the maximum speed of P.
5 A particle P of mass 0.6 kg is attached to one end of a light elastic spring of modulus of
elasticity 72 N and natural length 1.2 m. The other end of the spring is fixed to a point A on
the smooth horizontal table on which P rests. Initially P is at rest with AP = 1.2 m. The particle
receives an impulse of magnitude 3 N s in the direction AP. Given that t seconds after the
impulse the displacement of P from its initial position is x metres:
a find an equation for x in terms of t
b calculate the maximum magnitude of the acceleration of P.
P 6 A particle of mass 0.9 kg rests on a smooth horizontal surface attached to one end of a light
elastic string of natural length 1.5 m and modulus of elasticity 24 N. The other end of the string
is attached to a point on the surface. The particle is pulled so that the string measures 2 m and
released from rest.
a State the amplitude of the resulting oscillation.
b Calculate the speed of the particle when the string becomes slack.
Before the string becomes taut again the particle hits a vertical surface which is at right angles
to the particle’s direction of motion. The coefficient of restitution between the particle and the
vertical surface is _53 .
c Calculate:
i the period
ii the amplitude
of the oscillation which takes place when the string becomes taut once more.
P 7 A smooth cylinder is fixed with its axis horizontal. A piston of mass 2.5 kg is inside the
cylinder, attached to one end of the cylinder by a spring of modulus of elasticity 400 N and
natural length 50 cm. The piston is held at rest in the cylinder with the spring compressed to a
length of 42 cm. The piston is then released. The spring can be modelled as a light elastic spring
and the piston can be modelled as a particle.
a Find the period of the resulting oscillations.
b Find the maximum value of the kinetic energy of the piston.
197
Chapter 5
A 8 A particle P of mass 0.5 kg is attached to one end of a light elastic string of natural length
0.4 m and modulus of elasticity 30 N. The other end of the string is attached to a point on the
P smooth horizontal surface on which P rests. The particle is pulled until the string measures
0.6 m and then released from rest.
a Calculate the speed of P when the string becomes slack for the first time.
The string breaks at the instant when it returns to its natural length for the first time. When
P has travelled a distance 0.3 m from the point at which the string breaks the surface becomes
rough. The coefficient of friction between P and the surface is 0.25. The particle comes to rest
T seconds after it was released.
b Find the value of T.
E/P 9 A particle P of mass 0.4 kg is attached to two identical light elastic springs each of natural
length 1.2 m and modulus of elasticity 12 N. The free ends of the strings are attached to points
A and B which are 4 m apart on a smooth horizontal surface. The point C lies between A and B
with AC = 1.4 m and CB = 2.6 m. The particle is held at C and released from rest.
a Show that P moves with simple harmonic motion. (5 marks)
b Calculate the maximum value of the kinetic energy of P. (3 marks)
E/P 10 A particle P of mass m is attached to two identical light strings of natural length l and modulus
of elasticity 3mg. The free ends of the strings are attached to fixed points A and B which are 5l
apart on a smooth horizontal surface. The particle is held at the point C, where AC = l and A, B
and C lie on a straight line, and is then released from rest.
a Show that P moves with simple harmonic motion. (3 marks)
b Find the period of the motion. (2 marks)
c Write down the amplitude of the motion. (4 marks)
d Find the speed of P when AP = 3l. (2 marks)
E/P 11 A light elastic string has natural length 2.5 m and modulus of elasticity 15 N. A particle P of
mass 0.5 kg is attached to the string at the point K where K divides the unstretched string in the
ratio 2 : 3. The ends of the string are then attached to the points A and B which are 5 m apart
on a smooth horizontal surface. The particle is then pulled along AB and held at rest in contact
with the surface at the point C where AC = 3 m and ACB is a straight line. The particle is then
released from rest.
a Show that P moves with simple harmonic motion of period _π5 _ √ _2_ . (5 marks)
b Find the amplitude of the motion. (4 marks)
5.5 Vertical oscillation
You can investigate the motion of a particle which is attached to an elastic spring or string and is
oscillating in a vertical line.
A particle which is hanging in equilibrium attached to one end of an elastic spring or string, the other
end of which is fixed, can be pulled downwards and released. The particle will then oscillate in a
vertical line about its equilibrium position.
198
Dynamics
A W atch out The spring or string is not at its
l natural length in the equilibrium position. In this
position, the weight of the particle causes an
extension, e, in the spring or string.
e Equilibrium position
T x
P
mg x¨
Taking downwards as the positive direction, when the particle is a distance x below its equilibrium
position it has acceleration x¨ .
At the equilibrium position, the tension in the spring or string is mg.
Using Hooke’s law:
T = _ λ_×__e_x_tl _e _n_s_io_n_ λ is the modulus of elasticity and l is the natural
mg = _ λl_e length of the spring or string. e is the extension
e = _ m_λ_g _l of the spring or string in the equilibrium position.
Now consider the particle at a distance x below its equilibrium position.
T = _ λ_×__e_x_tl _e _n_s_io_n_ The particle is a distance x below the equilibrium
T = _λ_(x__l+ __e_) position, so the extension is x + e.
( )T = _ λ_ _x_+_l_ _ m__λ__g __l _ = _ λ_lx _ + mg
Using F = ma: W atch out When using F = ma the weight of the
particle must be included as well as the tension.
mg − T = m x..
( )mg _ λ_lx _ + m x..
− mg = m x..
− _ λ_lxx. _. ==
− _ mλ__ l x λ, m and l are atlhlepofosritmivex.. constants, so the
equation is of = −ω 2x. It is the same
The particle is moving with S.H.M. result as obtained for a horizontal oscillation.
As in the case of horizontal oscillations, a particle attached to one end of a spring will perform
complete oscillations. If the particle is attached to one end of an elastic string it will only move
with S.H.M. while the string is taut. If the amplitude is greater than the extension at the equilibrium
position the string will become slack before the particle reaches the upper end of the oscillation. Once
the string becomes slack the oscillatory motion ceases and the particle moves freely under gravity
until it falls back to the position where the string is once again taut.
199
Chapter 5
A ■■ For a particle hanging in equilibrium attached to one end of an elastic spring and displaced
vertically from its equilibrium position:
•• the particle will move with S.H.M.
•• the particle will perform complete oscillations
•• the centre of the oscillation will be the equilibrium position.
■■ For a particle hanging in equilibrium attached to one end of an elastic string and displaced
vertically from its equilibrium position:
•• the particle will move with S.H.M. while the string is taut
•• the particle will perform complete oscillations if the amplitude is no greater than the
equilibrium extension
•• if the amplitude is greater than the equilibrium extension the particle will move freely
under gravity while the string is slack.
A particle can be attached to two springs or strings which are hanging side by side or stretched in a
vertical line between two fixed points. The basic method of solution remains the same.
Example 17
A particle P of mass 1.2 kg is attached to one end of a light elastic spring of modulus of elasticity
60 N and natural length 60 cm. The other end of the spring is attached to a fixed point A on a
ceiling. The particle hangs in equilibrium at the point B.
a Find the extension of the spring.
The particle is now raised vertically a distance 15 cm and released from rest.
b Prove that P will move with simple harmonic motion.
c Find the period and amplitude of the motion.
d Find the speed of P as it passes through B.
e Find the speed of P at the instant when the spring has returned to its natural length.
a
A
0.6 m Change cm to m.
e T0 Use Hooke’s law to find the tension in terms
P of the extension.
mg At the equilibrium position the tension must
equal the weight.
T = _ λ_lx _
T0 = _ 60_0._6e_
1.2g = _ 60_0._6e_
e = _ 1._2__×__96_.8 _0_ ×__0__.6_ = 0.1176
The extension is 0.118 m (3 s.f.) or 11.8 cm (3 s.f.).
200
A b Dynamics
A On line Explore the simple harmonic
motion of a vertical spring using GeoGebra.
0.6 m
Draw a new diagram showing P at a distance
natural length x below the equilibrium level.
eT
Use Hooke’s law once more. This time the
equilibrium level extension is x + e.
x W atch out When you use F = ma you must
include the weight of the particle.
P Do not use an approximation for e. Instead,
use your work from a to replace mg with the
mg x¨ tension at the equilibrium level in terms of e.
When you simplify the equation e cancels.
T = _ λ_lx _ = _6_0_(_1x_.2_+ __e_)
F = mmxa.. P was raised 15 cm from its equilibrium level.
mg − T = x = 0 at B.
This is also the maximum speed of P.
mg − _6_0_0(_x_.6_+_ _e_) = mx.. At the natural length x = −e. Use at least
4 s.f. in your approximation for e.
_60_0._6e_ − _ 6_0_0(_x_.6_+_ _e_) = 1.2x..
x.. = − _1._2_6_×0__x0_ _.6_ = − _2_35 0_ x 201
P moves with S.H.M.
c x.. = − _2_35 0_ x
ω 2 = _ 2_35 0_
period = _ 2ω_π_ = √ _ _2__ 2___35π__ 0_ __ = 0.6882…
The period is 0.688 s (3 s.f.).
The amplitude is 15 cm.
d v 2 = ω 2(a2 − x2)
v 2 = _ 2_35 0_ (0.152 − 0)
B
vB = √ _ _2__35_ 0__ × 0.15 = 1.369…
The speed at B is 1.37 m s−1 (3 s.f.).
e v 2 = ω 2(a2 − x2)
v 2 = _ 2_35 0_ (0.152 − (−0.1176)2)
v = 0.8500…
The speed at the natural length is
0.850 m s−1 (3 s.f.).
Chapter 5
Example 18
A A particle P of mass 0.2 kg is attached to one end of a light elastic string of natural length 0.6 m
and modulus of elasticity 8 N. The other end of the string is fixed to a point A on a ceiling.
When the particle is hanging in equilibrium the length of the string is L m.
a Calculate the value of L.
The particle is held at A and released from rest. It first comes to instantaneous rest when the length
of the string is K m.
b Use energy considerations to calculate the value of K.
c Show that while the string is taut, P is moving with simple harmonic motion.
The string becomes slack again for the first time T seconds after P was released from A.
d Calculate the value of T.
A
T
P
0.2 g Use Hooke’s law.
a T = __ 0λ8__lx .__6e _
0.2g =
At the equilibrium position the tension must
e = _ 0_.2__×__9_8_. 8_ _×__0__.6_ = 0.147 equal the weight.
L = 0.6 + 0.147 = 0.747 The total length of the string is required.
b E.P.E. gained = _λ2_x_l2_
= _ 8_(2_K_×_−_ 0_0_..6_6_ )_2
G.P.E. lost = mgh = 0.2 × 9.8K Problem-solving
_ 8_(2_K_×_−_ 0_0_..6_6_ )_2 = 0.2 × 9.8K The question states that you must do this
part using conservation of energy. The
kinetic energy is zero at both points under
K2 − 1.2K + 0.36 = _ 2__×_0__.6__×__8 0 _ _.2__×_ _9_._8_K_ consideration, so the elastic potential energy
− 1.494K = 0.294K
gained is equal to the gravitational potential
+__0_.3__6__=_0_________
K2 energy lost. ← FM1, Section 3.4
K = _ 1._4_9_4__±__√_ 1 _._4_9_2 4_ _2_−__ 4__ ×__0_._3_6_
= 1.1919… or 0.3020…
∴ K = 1.19 (3 s.f.) The value of K must be greater than the
natural length of the string.
202
Dynamics
A c
A
0.6 m Draw a diagram which shows the natural
length and the equilibrium level as well as
the distance of P from the centre of the
oscillation (x).
e = 0.147
xT Remember that x.. must be in the direction of
P increasing x.
0.2 g x¨
T = _ λ_×__e_x_tl _ e_n_s_io_n_ = _8_(_x0_._+6_ _e_) There is no need to use an approximation for
F = ma e as _0 8_.e6_ from part a.
0.2g − _8_(_x0_._+6_ _e_) = 0.2x.. xR.. e=d−uωce2xthtoe equation of motion to the form
_08_._6e_ − _ 8_(_x0_._+6_ _e_) = 0.2x.. establish S.H.M.
x.. = − _0_.6__8×_x_ 0__.2_ = − _81_020_ x
∴ S.H.M.
d Time to fall 0.6 m from rest: Until the string is taut, P is falling freely
s = ut + _ 21_ at2 under gravity.
0.6 = 0 + _ 21_ × 9.8t2 Problem-solving
t2 = _ 04_..69_
Because of the symmetry of S.H.M. there are
t = 0.3499... several ways to obtain the time for which
the string is taut. Whichever method you use
For S.H.M.: √ _1 0__0__×___ 182__ = 10√ __ 32__ = _ 10_3_√_ _6 __ you must show your working clearly.
√ ω = _ _81__02_0_ _ =
amplitude = K − L = 1.191 − 0.747 = 0.444 Using x = a cos ωt with the positive value
of x when the string is at its natural length
x = 0.444 cos ωt will give the time from the high point of the
when x = 0.147, 0.147 = 0.444 cos ωt oscillation (if it were complete) to the point
cos ωt = _00_..41_44_74_ where the string becomes taut.
t = ( ) _ω1_ arc cos _00_..41_44_74_ = 0.1510… Subtracting twice the time just found from
the period will give the time for which the
Period = _2ω_π_ = 2π × _ 10_3_√_ 6_ __ = 0.7695… string is taut in any one oscillation.
Time for which string is taut
= 0.7695 − 2 × 0.1510 = 0.4675… Finally, add the time taken while falling
freely under gravity to the time for which the
Total time = 0.4675… + 0.3499 = 0.8174… string is taut.
∴ T = 0.817 (3 s.f.)
203
Chapter 5
Exercise 5E
A Whenever a numerical value of g is required, take g = 9.8 m s−2.
E/P 1 A particle P of mass 0.75 kg is hanging in equilibrium attached to one end of a light elastic
spring of natural length 1.5 m and modulus of elasticity 80 N. The other end of the spring is
attached to a fixed point A vertically above P.
a Calculate the length of the spring. (3 marks)
The particle is pulled downwards and held at a point B which is vertically below A. (4 marks)
The particle is then released from rest. (2 marks)
b Show that P moves with simple harmonic motion.
c Calculate the period of the oscillations.
The particle passes through its equilibrium position with speed 2.5 m s−1. (4 marks)
d Calculate the amplitude of the oscillations.
E/P 2 A particle P of mass 0.5 kg is attached to the free end of a light elastic spring of natural length
0.5 m and modulus of elasticity 50 N. The other end of the spring is attached to a fixed point A
and P hangs in equilibrium vertically below A.
a Calculate the extension of the spring. (3 marks)
The particle is now pulled vertically down a further 0.2 m and released from rest. (2 marks)
b Calculate the period of the resulting oscillations. (2 marks)
c Calculate the maximum speed of the particle.
P 3 A particle P of mass 2 kg is hanging in equilibrium attached to the free end of a light elastic
spring of natural length 1.5 m and modulus of elasticity λ N. The other end of the spring is
fixed to a point A vertically above P. The particle receives an impulse of magnitude 3 N s in the
direction AP.
a Find the speed of P immediately after the impact.
b Show that P moves with simple harmonic motion.
The period of the oscillations is _π2 _ s.
c Find the value of λ.
d Find the amplitude of the oscillations.
E/P 4 A piston of mass 2 kg moves inside a smooth cylinder which is fixed with its axis vertical. The
piston is attached to the base of the cylinder by a spring of natural length 12 cm and modulus
of elasticity 500 N. The piston is released from rest at a point where the spring is compressed
to a length of 8 cm. Assuming that the spring can be modelled as a light elastic spring and the
piston as a particle, calculate:
a the period of the resulting oscillations (5 marks)
b the maximum speed of the piston. (2 marks)
204
Dynamics
A 5 A light elastic string of natural length 40 cm has one end A attached to a fixed point. A particle
P of mass 0.4 kg is attached to the free end of the string and hangs freely in equilibrium
E/P vertically below A. The distance AP is 45 cm.
a Find the modulus of elasticity of the string. (3 marks)
The particle is now pulled vertically downwards until AP measures 52 cm and then released
from rest.
b Show that, while the string is taut, P moves with simple harmonic motion. (4 marks)
c Find the period and amplitude of the motion. (3 marks)
d Find the greatest speed of P during the motion. (2 marks)
e Find the time taken by P to rise 11 cm from the point of release. (3 marks)
E/P 6 A particle P of mass 0.4 kg is attached to one end of a light elastic string of natural length
0.5 m and modulus of elasticity 10 N. The other end of the string is attached to a fixed point
A and P is initially hanging freely in equilibrium vertically below A. The particle is then pulled
vertically downwards a further 0.2 m and released from rest.
a Calculate the time from release until the string becomes slack for the first time. (4 marks)
b Calculate the time between the string first becoming slack and the next time it becomes
taut. (4 marks)
E/P 7 A particle P of mass 1.5 kg is hanging freely attached to one end of a light elastic string of
natural length 1 m and modulus of elasticity 40 N. The other end of the string is attached to
a fixed point A on a ceiling. The particle is pulled vertically downwards until AP is 1.8 m and
released from rest. When P has risen a distance 0.4 m the string is cut.
a Calculate the greatest height P reaches above its equilibrium position. (4 marks)
b Calculate the time taken from release to reach that greatest height. (3 marks)
E/P 8 A particle P of mass 1.5 kg is attached to the midpoint of a light elastic string of natural length
1.2 m and modulus of elasticity 15 N. The ends of the string are fixed to the points A and B
where A is vertically above B and AB = 2.8 m.
a Given that P is in equilibrium calculate the length AP. (3 marks)
The particle is now pulled downwards a distance 0.15 m from its equilibrium position and
released from rest.
b Prove that P moves with simple harmonic motion. (4 marks)
T seconds after being released P is 0.1 m above its equilibrium position.
c Find the value of T. (3 marks)
E/P 9 A rock climber of mass 70 kg is attached to one end of a rope. He falls from a ledge which is 8 m
vertically below the point to which the other end of the rope is fixed. The climber falls vertically
without hitting the rock face. Assuming that the climber can be modelled as a particle and the
rope as a light elastic string of natural length 16 m and modulus of elasticity 40 000 N, calculate:
a the climber’s speed at the instant when the rope becomes taut (3 marks)
b the maximum distance of the climber below the ledge (3 marks)
c the time from falling from the ledge to reaching his lowest point. (2 marks)
205
Chapter 5
Challenge
A A particle P of mass m kg is attached to one end of a light elastic string of natural length l m
and modulus of elasticity 5mg. the other end of the string is attached to a fixed point A on a
ceiling. The particle is pulled vertically downwards and released to oscillate with period T s.
A second particle Q of mass km kg is then also attached to the end of the string. The system
then oscillates with period 3T s.
Find the value of k.
Mixed exercise 5
Whenever a numerical value of g is required, take g = 9.8 m s−2.
E/P 1 A particle P of mass 0.6 kg moves along the positive x-axis under the action of a single force
which is directed towards the origin O and has magnitude _( x__+k__ 2_)_2 N where OP = x metres
and k is a constant. Initially P is moving away from O. At x = 2 the speed of P is 8 m s−1 and at
x = 10 the speed of P is 2 m s−1.
a Find the value of k. (6 marks)
The particle first comes to instantaneous rest at the point B.
b Find the distance OB. (4 marks)
E 2 A particle P of mass 0.5 kg is moving along the x-axis, in the positive x direction. At time
t seconds (where t . 0) the resultant force acting on P has magnitude _√ __ (_3___t5__+ ____4__) N and is
directed towards the origin O. When t = 0, P is moving through O with speed 12 m s−1.
a Find an expression for the velocity of P at time t seconds. (5 marks)
b Find the distance of P from O when P is instantaneously at rest. (5 marks)
E/P 3 A particle of mass m moves in a straight line on a smooth horizontal plane in a medium which
exerts a resistance of magnitude mkv2, where v is the speed of the particle and k is a positive
constant. At time t = 0 the particle has speed U.
Find, in terms of k and U, the time at which the particle’s speed is _43 U.
(5 marks)
E/P 4 A small pebble of mass m is placed in a viscous liquid and sinks vertically from rest through the
liquid. When the speed of the particle is v the magnitude of the resistance due to the liquid is
modelled as mkv 2, where k is a positive constant.
Find the speed of the pebble after it has fallen a distance D through the liquid. (5 marks)
5 A car of mass 1000 kg is driven by an engine which generates a constant power of 12 kW. The
only resistance to the car’s motion is air resistance of magnitude 10v2 N, where v m s−1 is the
E/P speed of the car.
Find the distance travelled as the car’s speed increases from 5 m s−1 to 10 m s−1. (8 marks)
206
Dynamics
A 6 A bullet B, of mass m kg, is fired vertically downwards into a block of wood W which is fixed
in the ground. The bullet enters W with speed U m s−1 and W offers a resistance of magnitude
E/P m(14.8 + 5bv 2) N, where v m s−1 is the speed of B and b is a positive constant. The path of B in
W remains vertical until B comes to rest after travelling a distance d metres into W.
Find d in terms of b and U. (8 marks)
E/P 7 A particle of mass m is projected vertically upwards, with speed V, in a medium which exerts a
resisting force of magnitude _m _cg_ 2v_ _2 , where v is the speed of the particle and c is a positive constant.
a Show that the greatest height attained above the point of projection is
( ) _2c_g 2_ ln 1 + _Vc_ 2 _2 (8 marks)
b Find an expression, in terms of V, c and g, for the time to reach this height. (3 marks)
E/P 8 A particle is projected vertically upwards with speed U in a medium in which the resistance is
proportional to the square of the speed. Given that U is also the speed for which the resistance
offered by the medium is equal to the weight of the particle, show that:
a the time of ascent is _π 4_Ug_ (6 marks)
b the distance ascended is _U 2_g _2 ln 2. (6 marks)
E/P 9 At time t, a particle P, of mass m, moving in a straight line has speed v. The only force acting is
a resistance of magnitude mk(V 02 + 2v2), where k is a positive constant and V0 is the speed of P
when t = 0.
a Show that, as v reduces from V0 to _21 V0, P travels a distance _l 4n_k_ 2_
(6 marks)
b Express the time P takes to cover this distance in the form _k _Vλ_ _0 , giving the value of λ to
two decimal places. (6 marks)
E/P 10 A car of mass m is moving along a straight horizontal road. When displacement of the car
from a fixed point O is x, its speed is v. The resistance to the motion of the car has
magnitude _m_k3__v _ 2 , where k is a positive constant. The engine of the car is working at a
constant rate P.
a Show that 3mv 2 _ dd_xv_ = 3P − mkv 3.
(6 marks)
When t = 0, the speed of the car is half of its limiting speed.
b Find x in terms of m, k, P and v. (6 marks)
E/P 11 A spacecraft S of mass m is moving in a straight line towards the centre of the Earth. When the
distance of S from the centre of the Earth is x metres, the force exerted by the Earth on S
ah asBmy magonditeulldineg_x k_t2 h , ewEhearrethk
is a constant, and is directed towards the centre of the Earth.
as a sphere of radius R and S as a particle, show that k = mgR2.
(2 marks)
The spacecraft starts from rest when x = 5R.
b Assuming that air resistance can be ignored, find the speed of S as it crashes onto (7 marks)
the Earth’s surface.
207
Chapter 5
A 12 A particle P is moving with simple harmonic motion between two points A and B which are
0.4 m apart on a horizontal line. The midpoint of AB is O. At time t = 0, P passes through O,
E moving towards A, with speed u m s−1. The next time P passes through O is when t = 2.5 s.
a Find the value of u. (4 marks)
b Find the speed of P when t = 3 s. (2 marks)
c Find the distance of P from A when t = 3 s. (5 marks)
E 13 A particle P of mass 1.2 kg moves along the x-axis. At time t = 0, P passes through the origin
O, moving in the positive x direction. At time t seconds, the velocity of P is v m s−1 and
OP = x metres. The resultant force acting on P has magnitude 6(2.5 − x) N and acts in the
positive x direction. The maximum speed of P is 8 m s−1.
a Find the value of x when the speed of P is 8 m s−1. (5 marks)
b Find an expression for v 2 in terms of x. (5 marks)
E/P 14 A particle P moves along the x-axis in such a way that at time t seconds its distance x metres
mOoivsegsivweinthbysimx p=le3 hsianr m _π4o_t n.ic
( )from theorigin motion. (4 marks)
that P
a Prove
b Write down the amplitude and the period of the motion. (3 marks)
c Find the maximum speed of P. (2 marks)
The points A and B are on the same side of O with OA = 1.2 m and OB = 2 m.
d Find the time taken by P to travel directly from A to B. (4 marks)
E/P 15 A particle P moves on the x-axis with simple harmonic motion such that its centre of
oscillation is the origin, O. When P is a distance 0.09 m from O, its speed is 0.3 m s−1 and the
magnitude of its acceleration is 1.5 m s−2.
a Find the period of the motion. (3 marks)
The amplitude of the motion is a metres. Find:
b the value of a (3 marks)
c tthheanto_ 2at_ aml teitmrees,. within one complete oscillation, for which the distance OP is greater
(5 marks)
E/P 16 A particle P of mass 0.6 kg is attached to one end of a light elastic spring of natural length
2.5 m and modulus of elasticity 25 N. The other end of the spring is attached to a fixed point A
on the smooth horizontal table on which P lies. The particle is held at the point B where
AB = 4 m and released from rest.
a Prove that P moves with simple harmonic motion. (4 marks)
b Find the period and amplitude of the motion. (3 marks)
c Find the time taken for P to move 2 m from B. (2 marks)
E/P 17 A particle P of mass 0.4 kg is attached to the midpoint of a light elastic string of natural length
1.2 m and modulus of elasticity 2.5 N. The ends of the string are attached to points A and B on
a smooth horizontal table where AB = 2 m. The particle P is released from rest at the point C
on the table, where A, C and B lie in a straight line and AC = 0.7 m.
a Show that P moves with simple harmonic motion. (4 marks)
b Find the period of the motion. (3 marks)
208
Dynamics
A The point D lies between A and B and AD = 0.85 m. (4 marks)
c Find the time taken by P to reach D for the first time.
E/P 18 A and B are two points on a smooth horizontal floor, where AB = 12 m.
A particle P has mass 0.4 kg. One end of a light elastic spring, of natural length 5 m and
modulus of elasticity 20 N, is attached to P and the other end is attached to A. The ends of
another light elastic spring, of natural length 3 m and modulus of elasticity 18 N, are attached
to P and B.
a Find the extensions in the two springs when the particle is at rest in equilibrium. (5 marks)
Initially P is at rest in equilibrium. It is then set in motion and starts to move towards B.
In the subsequent motion P does not reach A or B.
b Show that P oscillates with simple harmonic motion about the equilibrium position.
(4 marks)
c Given that P stays within 0.4 m of the equilibrium position for _31 of the time within each
complete oscillation, find the initial speed of P. (7 marks)
E/P 19 A particle P of mass 0.5 kg is attached to one end of a light elastic string of natural length
1.2 m and modulus of elasticity λ N. The other end of the string is attached to a fixed point A.
The particle is hanging in equilibrium at the point O, which is 1.4 m vertically below A.
a Find the value of λ. (3 marks)
The particle is now displaced to a point B, 1.75 m vertically below A, and released from rest.
b Prove that while the string is taut P moves with simple harmonic motion. (4 marks)
c Find the period of the simple harmonic motion. (4 marks)
d Calculate the speed of P at the first instant when the string becomes slack. (4 marks)
e Find the greatest height reached by P above O. (4 marks)
E/P 20 A particle P of mass m is attached to the midpoint of a light elastic string of natural length
4 l and modulus of elasticity 5mg. One end of the string is attached to a fixed point A and the
other end to a fixed point B, where A and B lie on a smooth horizontal surface and AB = 6 l.
_ 94_ l ,
The particle is held at the point C where A, C and B are collinear and AC = and released
from rest.
a Prove that P moves with simple harmonic motion. (4 marks)
Find, in terms of g and l:
b the period of the motion (2 marks)
c the maximum speed of P. (2 marks)
209
Chapter 5
A Challenge
The motion of a space shuttle which is launched from a point O on the surface of
the Earth can be modelled as a particle of mass m moving in a straight line, subject
to the universal law of gravitation using F = − _( Rm__M+__xG_)_ 2 where:
M is the mass of the Earth
m is the mass of the space shuttle
R is the radius of the Earth
x is the height of the space shuttle above the Earth
G is the universal constant of gravitation.
a Given a space shuttle is launched with initial velocity u m s−1, show that the
maximum height, H, above the Earth that the spaceship reaches can be
expressed as H = _ ( _ _2__MR__R__ G__u _− 2_ _u _ 2)_
The minimum velocity required to project the space shuttle into space is called the
escape velocity. This is the value of u for which H tends to infinity.
b Use
M = 5.98 × 1024
R = 6.4 × 106
G = 6.7 × 10−11
to work out the escape velocity for the space shuttle correct to 3 significant
figures.
Summary of key points
1 In forming an equation of motion, forces that tend to decrease the displacement are negative
and forces that tend to increase the displacement are positive.
2 Newton’s law of gravitation states that the force of attraction between two bodies of mass
M1 and M2 is directly proportional to the product of their masses and inversely proportional
to the square of the distance between them.
• F = _G_M_d_1 2_M __2 where G is a constant known as the constant of gravitation.
3 Simple harmonic motion (S.H.M.) is motion in which the acceleration of a particle is always
towards a fixed point O on the line of motion of P, and has magnitude proportional to the
displacement of P from O.
210
Dynamics
A 4 For S.H.M. of amplitude a defined by the equation x.. = –ω 2 x,
• v 2 = ω 2(a 2 − x 2)
• If P is at the centre of the oscillation when t = 0, use x = a sin ωt.
• If P is at an end point of the oscillation when t = 0, use x = a cos ωt.
• If P is at some other point when t = 0, use x = a sin (ωt + α).
• The period of the oscillation is T = _ 2ω_π_
5 For a particle moving on a smooth horizontal surface attached to one end of an elastic spring:
• the particle will move with S.H.M.
• the particle will perform complete oscillations.
6 For a particle moving on a smooth horizontal surface attached to one end of an elastic string:
• the particle will move with S.H.M. while the string is taut
• the particle will move with constant speed while the string is slack.
7 To solve problems involving elastic springs and strings:
• use Hooke’s law to find the tension
• use F = ma to obtain ω
• use information given in the question to obtain the amplitude.
8 For a particle hanging in equilibrium attached to one end of an elastic spring and displaced
vertically from its equilibrium position:
• the particle will move with S.H.M.
• the particle will perform complete oscillations
• the centre of the oscillation will be the equilibrium position.
9 For a particle hanging in equilibrium attached to one end of an elastic string and displaced
vertically from its equilibrium position:
• the particle will move with S.H.M. while the string is taut
• the particle will perform complete oscillations if the amplitude is no greater than the
equilibrium extension
• if the amplitude is greater than the equilibrium extension the particle will move freely
under gravity while the string is slack.
211
Review exercise 2
E 1 A particle P moves in a straight line. b Find the distance of P from O when P
At time t seconds, the acceleration of P is comes to instantaneous rest. (7)
e2t m s−2, where t > 0. When t = 0, P is at ← Section 4.1
rest. Show that the speed, v m s−1, of P at
time t seconds is given by
v = _12 (e2t − 1) (6) E/P 5 At time t = 0, a particle P is at the origin
O and is moving with speed 18 m s−1 along
← Section 4.1 the x-axis in the positive x direction. At
E 2 A particle P moves along the x-axis in the time t seconds (t . 0), the acceleration of
positive direction. At time t seconds, the P has magnitude
velocity of P is v m s−1 and its acceleration _ √_ _t__3+__ __4__ m s−2 and is directed towards O.
is _12 e− _16 t m s−2. When t = 0 the speed of
P is 10 m s−1. a Show that, at itsim(3e0t−se6c√o _tn_+d__s_4,_ t ) h me s−1. (6)
velocity of P
a Express v in terms of t. (6) b Find the distance of P from O when P
b Find, to 3 significant figures, the speed comes to instantaneous rest. (7)
of P when t = 3. (2) ← Section 4.1
c Find the limiting value of v. (1)
← Section 4.1 E/P 6 A particle moving in a straight line starts
from rest at a point O at time t = 0.
E 3 A particle P moves along the x-axis. At At time t seconds, the velocity v m s−1 is
time t seconds the velocity of P is v m s−1
and its acceleration is 2 sin _12 t m s−2, both given by
measured in the direction Ox. Given that
v = 4 when t = 0, {v = 3t(t − 4), 0<t<5
75t−1, 5 , t < 10
a find v in terms of t (6) a Sketch a velocity–time graph for the (3)
particle for 0 < t < 10.
b calculate the distance travelled by P
between the times t = 0 and t = _ π2_ (7) b Find the set of values of t for which
the acceleration of the particle is
← Section 4.1 positive. (2)
E 4 A particle P moves along the x-axis. c Show that the total distance travelled
At time t seconds its acceleration by the particle in the interval 0 < t < 5
is −4e−2t m s−2 in the direction of x is 39 m. (5)
increasing. When t = 0, P is at the origin
O and is moving with speed 1 m s−1 in the d Find, to 3 significant figures, the
direction of x increasing. value of t at which the particle returns
to O. (3)
a Find an expression for the velocity of ← Section 4.1
P at time t. (6)
212
Review exercise 2
E/P 7 A car is travelling along a straight A 11 A particle P is moving along the positive
horizontal road. As it passes a point O E x-axis in the direction of x increasing.
on the road, the engine is switched off.
At time t seconds after the car has passed When OP = x metres, the velocity of P is
O, it is at a point P, where OP = x metres, v m s−1 and the acceleration of P is
and its velocity is v m s−1. The motion of
the car is modelled by _ (x__4+_k_21 _)_2 m s−2, where k is a positive
v = _p_+1__ q_t
constant. At x = 1, v = 0.
a Find v 2 in terms of x and k. (5)
where p and q are positive constants. b Deduce that v cannot exceed 2k. (2)
a Show that, with this model, the ← Section 4.2
retardation of the car is proportional E 12 A particle P moves along the x-axis.
to the square of the speed. (3) At time t = 0, P passes through the
origin O, moving in the positive
When t = 0, the retardation of the car is x direction. At time t seconds, the
0.75 m s−2 and v = 20. Using the model, velocity of P is v m s−1 and OP = x metres.
find: The acceleration of P is _1 12_ (30 − x) m s−2,
measured in the positive x direction.
b the value of p and the value of q (2)
c x in terms of t. (5) a Give a reason why the maximum speed
← Section 4.1 of P occurs when x = 30. (2)
A 8 A particle P moves along the x-axis in Given that the maximum speed of P is
E such a way that when its displacement 10 m s−1,
from the origin O is x m, its velocity is b find an expression for v 2 in terms (5)
v m s−1 and its acceleration is 4x m s−2. of x.
When x = 2, v = 4.
← Section 4.2
Show that v 2 = 4x2. (4)
E/P 13 A particle P starts at rest and moves in
← Section 4.2 a straight line. The acceleration of P
initially has magnitude 20 m s−2 and, in
E 9 A particle P moves on the positive x-axis. a first model of the motion of P, it is
When OP = x metres, where O is the assumed that this acceleration remains
origin, the acceleration of P is directed constant.
away from O and has magnitude
( ) 1 − _ x4_2 m s−2. When OP = x metres, the
a For this model, find the distance
moved by P while accelerating from
velocity =of3√P _2_i ,ssvh mow s−t1h. Gatiwvehnenthat when rest to a speed of 6 m s−1. (4)
x = 1, v
x = _32 , v 2 = _ 439_ The acceleration of P when it is x metres
(5) from its initial position is a m s−2 and it is
then established that a = 12 when x = 2.
← Section 4.2 A refined model is proposed in which
a = p − qx, where p and q are constants.
E 10 A particle P is moving in a straight line.
When P is at a distance x metres from a b Show that, under the refined model, (5)
fixed point O on the line, the acceleration of p = 20 and q = 4.
P is (5 + 3 sin 3x) m s−2 in the direction OP.
Given that P passes through O with speed c Hence find, for the refinded model, the
4 m s−1, find the speed of P at x = 6. Give distance moved by P in first attaining a
your answer to 3 significant figures. (5) speed of 6 m s−1. (4)
← Section 4.2 ← Section 4.2
213
Review exercise 2
A 14 A particle P moves along the positive E 17 A particle P moves along the positive
E x-axis. When OP = x metres, the velocity x-axis. At time t seconds, the acceleration
of the particle is −(k + v) where v m s−1
of P is v m s−1 and its acceleration is is the velocity of the particle and k is a
positive constant. When t = 0, P is at O
_ (2_x_7_+_2_ 1_)_2 m s−2 in the direction of and v = U. The particle comes to rest at
the point A. Find:
x increasing. Initially x = 1 and P is
moving toward O with speed 6 m s−1. Find: a the distance OA (6)
a v2 in terms of x (5) b the time P takes to travel from O to A.
(5)
b the minimum distance of P from O. (3)
← Section 4.2 ← Section 4.3
E/P 15 A particle moves on the positive x-axis. E 18 A van starts from rest and moves along
The particle is moving towards the origin a straight horizontal road. At time t
O when it passes through t_h_e point A, seconds, the acceleration of the van
√ where x = 2c, with speed _ kc_ , where is _1 _6_19_0 _−0_ _v_ 2 m s−2 where v m s−1 is the
velocity of the van.
k is a constant. Given that the particle
experiences an acceleration of _2 k_x_ 2 + _ 4k_c_ 2 a Find v in terms of t (6)
acting in the direction of x increasing,
b Show that the speed of the van cannot
a show that it comes instantaneously to exceed 13 m s−1. (2)
rest at a point B, where x = c. (5)
← Section 4.3
As soon as the particle reaches B the E/P 19 A motorbike travels along a straight
acceleration changes to _2 k_x_ 2 − _ 4k_c_ 2 in a horizontal road. At time t seconds, the
direction away from O. speed of the motorbike is v m s−1 and
the acceleration is ( 6 – _5v _ ) m s−2. The
b Show that the particle next comes (5) motorbike starts from rest. Find:
instantaneously to rest at A.
← Section 4.2 a v in terms of t (6)
E 16 A particle moves along the x axis in the b the terminal speed of the motorbike. (3)
direction of x increasing. When the speed
of the particle is v m s−1, the acceleration ← Section 4.3
is − _1v_0 2_ m s−2. Initially the particle is at the
origin, O, and is moving with a speed of A 20 A particle, P, is moving along the x axis
12 m s−1. At time T seconds, the particle is E Ox in the direction of x increasing. At
at the point A with a velocity of 6 m s−1.
Find: time t seconds, the velocity of P is v m s−1
and the acceleration is –(a2 + v2) m s−2
a the value of T (6) where a is a constant. At time t = 0, P
is at O and its speed is 20 m s−1. At time
t = T, the particle is at point A and its
velocity is 12 m s−1. Find:
b the distance OA. (5) a T in terms of a (5)
← Section 4.3 b the distance OA. (6)
← Section 4.3
214
Review exercise 2
A 21 A particle P of mass 0.2 kg moves away A Find:
E from the origin along the positive x-axis.
a the value of t when the speed is (7)
It moves under the action of a force 8 m s−1
directed away from the origin O of
magnitude _x_+5__ 1_ N, where OP = x m. b the distance of P from O when (5)
Given that the speed of P is 5 m s−1 when t = 2.
x = 0, find the value of x, to 3 significant
figures, when the speed of P is 15 m s−1. (8) ← Section 5.1
← Section 5.1 E 25 A car of mass 800 kg moves along a
horizontal straight road. At time t
E 22 A particle P of mass 2.5 kg moves along seconds, the resultant force on the car has
the positive x-axis. It moves away from magnitude _(4t_8+_ 0_02_0)_2 N , acting in the
a fixed origin O, under the action of a direction of motion of the car. When
force directed away from O. When t = 0, the car is at rest.
OP = x metres, the magnitude of the
force is 2e−0.1x N and the speed of P is a Show that the speed of the car (7)
v m s−1. When x = 0, v = 2. Find: approaches a limiting value as t
increases and find this value.
a v2 in terms of x (4) b Find the distance moved by the car in
the first 6 s of its motion. (5)
b the value of x when v = 4. (2)
← Section 5.1
c Gnoivt eexacreeeadso√ _2n_0_w mh ys−t1h. e speed of P does
(2)
E/P 26 A particle P of mass _31 kg moves along
← Section 5.1 the positive x-axis under the action of a
E/P 23 A toy car of mass 0.2 kg is travelling in single force. The force is directed towards
a straight line on a horizontal floor. The the origin O and has magnitude _( x__+k__ 1_)_2 N,
car is modelled as a particle. At time t = 0 where OP = x metres and k is a constant.
the car passes through a fixed point O. Initially P is moving away from O.
After t seconds the speed of the car is At x = 1 the speed PofisP√ i _2s_ m4 m s− 1s.−1, and at
v m s−1 and the car is at a point P with x = 8 the speed of
OP = x metres. The resultant force on
the car is modelled as _1 1_0 x(4 − 3x) N Find:
in the direction OP. The car comes to
instantaneous rest when x = 6. Find: a the value of k (7)
b the distance of P from O when P first
comes to instantaneous rest. (5)
a an expression for v2 in terms of x (4)
← Section 5.1
b the initial speed of the car. (2)
← Section 5.1 E/P 27 A particle P of mass 0.7 kg falls vertically
from rest. A resisting force of magnitude
E 24 A particle P of mass 0.6 kg is moving 2.1v N acts on P as it falls where v m s−1 is
along the positive x-axis under the action the velocity of P after t seconds. Find:
of a force which is directed away from
the origin O. At time t seconds, the force a v in terms of t (8)
has magnitude 3e−0.5t N. When t = 0, the
particle P is at O and moving with speed b the distance that P falls in the first two
2 m s−1 in the direction of x increasing. seconds. (5)
← Section 5.1
215
Review exercise 2
A 28 A particle of mass m kg is projected A The kinetic energy of P at x = 2R is half
E/P vertically upwards with velocity U m s−1. of the kinetic energy at x = R.
The motion of the particle is subject to b Find c in terms of U and R. (3)
air resistance of magnitude _ mk_v_ N where
v m s−1 is the speed of the particle at time t ← Section 5.2
seconds and k is a positive constant.
E/P 31 A projectile P is fired vertically upwards
Find, in terms of U, g and k, the from a point on the Earth’s surface.
maximum height above O reached by the When P is at a distance x from the centre
particle. (8) of the Earth its speed is v. Its acceleration
is directed towards the centre of the
← Section 5.1 Earth
and has magnitude _x k_2 , where k is a
29 Above the Earth’s surface, the magnitude constant. The Earth is assumed to be a
E/P of the force on a particle due to the sphere of radius R.
Earth’s gravity is inversely proportional to a Show that the motion of P may be
the square of the distance of the particle modelled by the differential equation
from the centre of the Earth. Assuming
that the Earth is a sphere of radius R, v _ dd_xv_ = − _ gx_R_2_ 2 (3)
and taking g as the acceleration due to
gravity at the surface of the Earth, The initial speed of P is U, where
U 2 , 2gR. The greatest distance of P
a prove that the magnitude of the from the centre of the Earth is X.
gravitational force on a particle of mass
m when it is a distance x (where x > R)
_m _xg_R2_ _2 (3) b Find X in terms of U, R and g. (7)
from the centre of the Earth is
← Section 5.2
A particle is fired vertically upwards from E 32 A particle P moves in a straight line
the surface of the Earth with initial speed with simple harmonic motion about a
u, where u2 = _ 23 gR. Ignoring air resistance, fixed centre O with period 2 s. At time
b find, in terms of g and R, the speed of t seconds the speed of P is v m s−1.
When t = 0, v = 0 and P is at a point A
the particle when it is at a height 2R where OA = 0.25 m.
above the surface of the Earth. (7)
Find the smallest positive value of t for
← Section 5.2
E/P 30 A rocket is fired vertically upwards with which AP = 0.375 m. (5)
speed U from a point on the Earth’s ← Section 5.3
surface. The rocket is modelled as a E 33 A particle P of mass 0.2 kg oscillates
particle P of constant mass m, and the with simple harmonic motion between
Earth as a fixed sphere of radius R. At a the points A and B, coming to rest at
distance x from the centre of the Earth, both points. The distance AB is 0.2 m,
the speed of P is v. The only force acting and P completes 5 oscillations every
on P is directed towards the centre of the second.
_c x_m2_ ,
Earth and has magnitude where c is a a Find, to 3 significant figures, the
constant. maximum resultant force exerted
( )a Show that v 2 = U 2 + 2c _x1_ − _ R1_ on P.
(7) (3)
216
Review exercise 2
A When the particle is at A, it is struck a Ab
blow in the direction BA. The particle
now oscillates with simple harmonic A PQ B
motion with the same frequency as
previously but twice the amplitude. C
4l
b Find, to 3 significant figures, the speed For advanced players, the window PQ
of the particle immediately after it has is moved to the centre of AB so that
been struck. (3) AP = QB, as shown in the second figure
above.
← Section 5.3
c Find the time, in seconds to 2 decimal
places, taken for C to pass from Q to P
E/P 34 A piston P in a machine moves in a in this new position. (4)
straight line with simple harmonic motion
about a fixed centre O. The period of the ← Section 5.3
oscillations is π s. When P is 0.5 m from O,
its speed is 2.4 m s−1. Find: E 36 The points O, A, B and C lie in a straight
line, in that order, with OA = 0.6 m,
a the amplitude of the motion (4) OB = 0.8 m and OC = 1.2 m. A particle P,
moving in a straight line, has speed
b the maximum speed of P during its (2)
motion ( ) ( ) _13_0 √ _3_ m s−1 at A, _51 √ _5_ m s−1 at B and is
c the maximum magnitude of the instantaneously at rest at C.
acceleration of P during the
motion (2) a Show that this information is
consistent with P performing simple
d the total time, in seconds to 2 decimal harmonic motion with centre O. (4)
places, in each complete oscillation for
which the speed of P is greater than Given that P is performing simple
2.4 m s−1. (5) harmonic motion with centre O,
← Section 5.3 b show that the speed of P at O is (2)
0.6 m s−1.
E/P 35 b c Find the magnitude of the acceleration
P of P as it passes A. (4)
A QB
C d Find, to 3 significant figures, the time
4l taken for P to move directly from A
In a game at a fair, a small target C to B. (4)
moves horizontally with simple harmonic ← Section 5.3
motion between the points A and B,
where AB = 4l. The target moves inside
a box and takes 3 s to travel from A to E/P 37 The rise and fall of the water level in a
B. A player has to shoot at C, but C is harbour is modelled as simple harmonic
only visible to the player when it passes a motion. On a particular day the
window PQ where PQ = b. The window is maximum and minimum depths of the
initially placed with Q at the point shown water in the harbour are 10 m and 4 m
in the figure above. The target takes 0.75 s and these occur at 1100 hours and 1700
to travel from Q to P. hours respectively.
a Show that b = (2 − √ _2_ ) l. (3) a Find the speed, in m h−1, at which the
water level in the harbour is falling at
b Find the speed of C as it passes P. (3) 1600 hours on this particular day. (7)
217
Review exercise 2
A b Find the total time, between 1100 A 40 A particle P of mass 0.3 kg is attached
hours and 2300 hours on this E/P to one end of a light elastic spring. The
particular day, for which the depth in
the harbour is less than 5.5 m. (5) other end of the spring is attached to a
fixed point O on a smooth horizontal
← Section 5.3 table. The spring has natural length 2 m
and modulus of elasticity 21.6 N. The
E/P 38 A piston in a machine is modelled as a particle P is placed on the table at a point
particle of mass 0.2 kg attached to one A, where OA = 2 m. The particle P is now
end A of a light elastic spring, of natural pulled away from O to the point B, where
length 0.6 m and modulus of elasticity OAB is a straight line with OB = 3.5 m.
48 N. The other end B of the spring is It is then released from rest.
fixed and the piston is free to move in a
horizontal tube which is assumed to be a Prove that P moves with simple
smooth. The piston is released from rest harmonic motion of period _π 3_ s.
when AB = 0.9 m. (4)
b Find the speed of P when it reaches
A. (2)
a Prove that the motion of the piston is
simple harmonic with period _1 π_0_ s. (4) The point C is the midpoint of AB.
b Find the maximum speed of the
c Find, in terms of π, the time taken for
P to reach C for the first time. (4)
piston. (2)
c Find, in terms of π, the length of time Later in the motion, P collides with a
during each oscillation for which the particle Q of mass 0.2 kg which is at rest
length of the spring is less than at A.
0.75 m. (5) After impact, P and Q coalesce to form a
single particle R.
← Section 5.4
E/P 39 A particle P of mass 0.8 kg is attached d Show that R also moves with simple
to one end A of a light elastic spring OA, harmonic motion and find the
of natural length 60 cm and modulus of amplitude of this motion. (4)
elasticity 12 N. The spring is placed on a
smooth table and the end O is fixed. The ← Section 5.4
particle is pulled away from O to a point
B, where OB = 85 cm, and is released E/P 41 A light elastic string of natural length
from rest. l has one end attached to a fixed point
A. A particle P of mass m is attached to
the other end of the string and hangs in
a Prove that the motion of P is simple equilibrium at the point O, where
harmonic motion with period _2 5_π_ s. (4) AO = _54 l.
b Find the greatest magnitude of the a Find the modulus of elasticity of the
acceleration of P during the motion. (2) string. (3)
Two seconds after being released from The particle P is then pulled down and
rest, P passes through the point C. released from rest. At time t the length of
c Find, to 2 significant figures, the speed the string is _54_l + x.
of P as it passes through C. (2) b Prove that, while the string is taut,
d State the direction in which P is (2) _ dd_2tx_ 2 = − _4_gl_ x_ (5)
moving 2 s after being released.
← Section 5.4
218
Review exercise 2
A When P is released, AP = _ 74 l. The point B Challenge
is a distance l vertically below A.
A 1 A particle moves in a straight line with an initial
c Find the speed of P at B. (2) velocity of u m s−1. When the particle is moving
with a velocity of v m s−1 the acceleration is –e2kv
d Describe briefly the motion of P after where k is a constant.
it has passed through B for the first
time until it next passes through O. (2) a Show that the time taken for the particle to
← Section 5.5 come to rest is _2 1_k_ ( e_ _2ek_u 2_ k−u_ 1_ )
E/P 42 A light elastic string, of natural length b Find the distance the particle moves before
4a and modulus of elasticity 8mg, has coming to rest.
one end attached to a fixed point A. A ← Section 4.3
particle P of mass m is attached to the
other end of the string and hangs in 2 A particle P travels on the x-axis, passing
equilibrium at the point O.
the origin at time t = 0 with velocity –k m s–1,
a Find the distance AO. (4) where k is a positive constant. At time t the
The particle is now pulled down to a particle is a distance x m from the origin and
point C vertically below O, where OC = d.
It is released from rest. In the subsequent its acceleration, a m s–2, is given by a = 8x _dd_xt_
motion the string does not become slack. Show tnheavtetrheexdceisetdans c_12 e √o _k_f the particle from the
origin metres.
√ b ← Section 4.2
Show that P moves with simple _ _2_g_a__
harmonic motion of period π (5) 3 A particle of mass m is projected vertically
upwards at time t = 0, with speed U. The particle
mThoetigorneaiste_12 s √t _gs_ap_ . eed of P during this moves against air resistance of magnitude
c Find d in terms of a. (2) mgkv2, where v is the speed of the particle at
time t and k is a constant. Find in terms of U, g
Instead of being pulled down a distance and k the height of the particle at the first point
d, the particle is pulled down a distance a. where its speed is half of its initial speed.
Without further calculation,
← Section 5.1
d describe briefly the subsequent motion
of P. (2)
← Section 5.5
219
Exam-style practice
Further Mathematics
AS Level
Further Mechanics 2
Time: 50 minutes
You must have: Mathematical Formulae and Statistical Tables, Calculator
1 A particle P moves along the positive x axis. At time t seconds, the acceleration of the particle
is _ 1_4_44_ 8_−_ _v_ 2 where v m s−1 is the velocity of the particle. The particle starts from rest.
a Find v in terms of t. (6)
b Show that the speed of the particle cannot exceed 12 m s−1. (2)
2 Q
2a
O 60° P
A particle P of mass m is attached to one end of a light inextensible string of length 2a.
The other end of the string is fixed to a point Q which is vertically above the point O on a
smooth horizontal table. The particle P remains in conta_c_t_with the surface of the table and
√ moves in a circle with centre O and with angular speed _ k4_ag_ , where k is a constant.
Throughout the motion the string remains taut and ∠QPO = 60°, as shown in the figure above.
a Show that the tension in the string is _m _2k_ g_
(3)
b Find, in terms of m, g and k, the normal reaction between P and the table. (3)
c Deduce the range of possible values of k in order for the particle to remain in contact with
√ the table. (2)
of is to _ _3a__g__ . The above
The angular speed
P changed particle P now moves in a horizontal circle
the table. The centre of this circle is X. (7)
d Show that QX : QO = 1 : 3√ _3_
220220
Exam-style practice
3 D C (4m)
2a
A 5a B (2m)
The figure above shows four uniform rods joined to form a rigid rectangular framework ABCD,
where AB = CD = 5a, and BC = AD = 2a. The mass of each rod is _ma _ per unit length.
Particles, of masses 2m and 4m, are attached to the framework at points B and C respectively.
a Find the distance of the centre of mass of the loaded framework from:
i AB
ii AD (8)
A new uniform rod, PQ, of mass 10m is added to the loaded framework to join AB and CD. P
lies on AB and Q lies on DC such that PQ is parallel to AD. Given that the centre of mass of the
new framework is a distance 2.5a from AD,
b find the distance AP. (5)
The loaded framework is freely suspended from D and hangs in equilibrium.
c Find, correct to three significant figures, the angle which DC makes with the vertical. (3)
d State how in your calculations you have used the assumption that the rods are uniform. (1)
221
Exam-style practice
Further Mathematics
A Level
Further Mechanics 2
Time: 1 hour and 30 minutes
You must have: Mathematical Formulae and Statistical Tables, Calculator
1 In a harbour, sea level at low tide is 10 m below the level of the sea at high tide. At low tide the
depth of the water in the harbour is 8 m. On a particular day, low tide occurs at 1 p.m. and
the next low tide occurs at 1.30 a.m. A ship can remain in the harbour safely when the depth
of water is at least 12 m. The sea level is modelled as rising and falling with simple harmonic
motion.
a Write down the
i period
ii amplitude of the motion. (2)
A ‘safe mooring height’ marker is attached to the harbour wall at a depth of 12 m.
b Find the speed, in metres per hour, at which the water level is rising when it passes this (4)
marker.
c Find the total length of time between two consecutive low tides for which the water in the (4)
harbour is at a safe mooring depth.
2 A uniform rectangular piece of card ABCD has D E
AB = 2a and AD = a. Corner C is folded down to C
meet side AB as shown in the diagram. 2a
a Find the distance of the centre of mass of the a
lamina from
i AD ii AB (7)
The lamina is freely suspended by a string A B
attached to the point A and hangs at rest.
(4)
b Find, to the nearest degree, the angle between DE and the vertical.
3 A particle P of mass m moves along the positive x-axis. At time t seconds, the acceleration of the
particle is −2(k2 + v2) m s−2 where v m s−1 is the velocity of the particle and k is a positive constant.
When t = 0, P is at O and v = 2U. The particle passes through the point A with velocity U.
a Find the distance OA. (6)
(5)
b Show that the time P takes to travel from O to A is _ 21_k_ ( arctan _ 2k_U_ − arctan _ Uk_ )
222222
Exam-style practice
4 A circular track is banked at an angle α to the horizontal, where tan α = _ 34 . A car moves round
the track at constant speed in a horizontal circle of radius 306 m. The car is modelled as a
particle and the track is modelled as being smooth and any non-gravitational resistance to
motion is ignored.
a Find the speed of the car. (6)
The model is refined to account for the roughness of the track, and the coefficient of friction
between the car and the track is taken to be 0.5. Any non-gravitational resistances to motion are
still ignored.
b Given that the car does not skid either up or down the banked surface of the track, find the (8)
range of possible speeds of the car.
5 The region R is bounded by the curve with equation y
y = √ _c_o_s_ 2_x_ , the positive x-axis and the line x = π_4_ .
The unit of length on both axes is the metre. (4) y = cos 2x
A uniform solid S is formed by rotating R through 2π (7) R
about the x-axis.
a Show that the volume of S is _ π2_ m3. O π x
b Find, in terms of π, the x-coordinate of the centre 4
of mass of S. You must use algebraic integration
to obtain your answer.
The solid S is placed on an inclined plane, rough enough to prevent slipping, with its circular
face on the plane. The plane is slowly tilted until the solid S is about to topple.
c Find the inclination of the plane to the horizontal when the solid S is about to topple. (4)
6 A particle P of mass m is attached to one end of a light inextensible O
string of length a. The other end of the string is attached to a fixed 60°
point O. The particle is held with the string taut so that OP makes an a
angle of 60° with the √d _3o_gw_a_n iwn aarddivreecrttiiocanl,wOhQic.hTishepeprapretnicdleiciusltahretno OP P
projected with speed
so that the particle moves in the vertical plane OPQ.
Show that P does not make complete circles about O and find the Q
angle that OP makes with the upward vertical when the string first
goes slack. (14)
223
Answers
Answers
CHAPTER 1 Challenge y
Prior knowledge check a y = _ pq_ _2 x 2
b 3.05 kg (3 s.f.)
1 a 21.9 N (3 s.f.) b 58.6 J (3 s.f.)
2 a 10.6 m (3 s.f.)
Exercise 1A
1 a 0.524 rad s−1 (3 s.f.) b 12.6 rad s−1 (3 s.f.)
c 38.2 rev min−1 (3 s.f.) d 1720 rev h−1 (3 s.f.)
2 a 80 m s−1 b 83.8 m s−1 (3 s.f.) x
3 a 8 rad s−1 b 76.4 rev min−1 (3 s.f.)
4 a 2 m s−1 b 2.09 m s−1 (3 s.f.) b Acceleration is 2q in the positive y-direction. Speed at
5 a 44.9 s (3 s.f.) b 0.14 rad s−1 (3 s.f.) the origin__i_s_p_.__
6 a 0.628 rad s−1 (3 s.f.) b 0.0754 m s−1 (3 s.f.)
c 0.0503 m s−1 (3 s.f.) c y = R − √ R 2− x 2
7 a 0.279 rad s−1 (3 s.f.) b 39.8 m (3 s.f.) d R = _2p _q 2_
8 3.14 m s−1 (3 s.f.), 5.24 m s−1 (3 s.f.) e 2q
f The acceleration of P and Q are equal.
9 a 0.242 rad s−1 (3 s.f.) b 0.362 m s−1 (3 s.f.) Exercise 1C
10 0.056 rad s−1 (3 s.f.)
11 a 0.000145 rad s−1 (3 s.f.), 0.00175 rad s−1 (3 s.f.)
b 1.45 × 10−5 m s−1 (3 s.f.), 2.62 × 10−4 m s−1 (3 s.f.) 1 18.4 N (3 s.f.), 4.52 rad s−1 (3 s.f.)
12 62.8 m s−1 (3 s.f.) 2 10.3 N (3 s.f.), 4.43 rad s−1 (3 s.f.)
13 a 4.71 rad s−1 (3 s.f.) b 2.55 cm (3 s.f.) 3 23.7 N (3 s.f.), 60° (nearest degree)
14 29 900 m s−1 (3 s.f.) 4 73.5 N, 0.6 m
15 r . 5 5 T = mlω 2
Challenge 6 Let the tension in the string be T.The angle between
_1π_9_ rad s−1 the string and the vertical is θ, and the radius of the
circle is r.
Exercise 1B R(→): T sin θ = mr ω 2
R(→): T cos θ = mg
1 4 m s−2
2 20.8 m s−2 (3 s.f.) Dividing the horizontal component by the vertical
3 a 5 rad s−1 b 15 m s−1 component
4 a 12.9 rad s−1 (3 s.f.) b 7.75 m s−1 (3 s.f.) tan θ = _m _m_r_ gω _ _2
5 2.14 m s−2 (3 s.f.) _ xr_ = _r _ gω_ _2 so, ω 2 x = g
6 0.283 rad s−1 (3 s.f.)
7 0.72 N √ 7 ω = __ 3g__
8 48.6 N
9 a 0.588 N (3 s.f.) b 4.5 N 8 d = 5 cm
10 a 0.24 m s−1 b 0.0072 N
11 0.0294 (3 s.f.) 9 ω = 1.8 rad s−1, v = 5.4 m s−1
12 3.13 rad s−1 (3 s.f.)
13 0.157 (3 s.f.) 10 18.1 rad s−1 (3 s.f.)
14 0.233 rad s−1 (3 s.f.)
11 9.5°
15 a 320 N (3 s.f.) b 0.000153 (3 s.f.) 12 22.8 rad s−1 (3 s.f.)
16 a 2.42 rad s−1 (3 s.f.) 13 9.44 rad s−1 (3 s.f.)
b No, because it is the minimum possible value for W.
If the speed or the coefficient of friction reduced at 14 a R(→): R cos α = mg
all, the people would slip down the cylinder. R(↑): R sin α = _m _r_v_ _2
17 1.4 m s−1 Dividing the horizontal component by the vertical
18 a μ . _ gv_R 2_
component to eliminate R:
tan α = _rm _m_v_g _2 = _vr_g 2_
b Model assumes that the tyres all experience the So, v 2 = rg tan α = √ _rg__ t_a_n_ α_
same friction.
19 0.322_ m___(3 s.f.) b This model assumes there is no friction between the
20 ω < _√_ 11_90_7 _ vehicle and the road.
15 2.22 m s−1 (3 s.f.)
16 0.4
17 42° (nearest degree)
18 29.5 m s−1 (3 s.f.), 9.94 m s−1 (3 s.f.)
224 On line Full worked solutions are available in SolutionBank.
Answers
19 a 20° (nearest degree), 20 800 N (3 s.f.); 5 a a_4_ √b _g__4_a__
68° (nearest degree), 52 700 N (3 s.f.) √ c _ _9__4g__a __ , 64° (nearest degree)
b To turn in a shorter time the aircraft will need to 6 a 49° (nearest degree)
decrease the radius of the circular arc in which it
turns. Thus the angle to the horizontal and lift force b The particle will fall through a parabolic arc
must both increase. (projectile motion) towards the surface in the
positive x-direction.
20 R(↑): T sin θ + R sin θ = mg ⇒ T + R = _ sm_in_g_ θ_
R(→): T cos θ − R cos θ = ml cos θ ω2 ⇒ T − R = ml ω2 7 a 10.3 m s−1
b At R: _ 21_ × 2 × v 2= 2g(12 − 5 cos 70° − 7 cos 40°)
Eliminating R: So, v 2= 96.58
2T = _ s_min_g_ θ _ + ml ω2 Resolving towards B: mg cos θ − R = _m _7_v_ _2
T = _21 _ m(l ω2 − g cosec θ) R = 2g cos 40° − _2 _7v_ _2 = −12.6
21 2.20 rad s −1(3 s.f.) R,0
Exercise 1D This is impossible, so the particle must have lost
contact with the chute.
1 a 3.13 m s−1 (3 s.f.) b 17.6 N (3 s.f.) c In reality, energy is lost due to friction between the
la_u__n_d_ry bags and the chute.
2 a 3.43 m s−1 (3 s.f.) b 19.6 N
√ 8 a 1_7_3_g_ a_
3 a 2.97 m s−1 (3 s.f.) b 15.7 N (3 s.f.)
4 a 2.21 m s−1 (3 s.f.) b 5.88 N (3 s.f.) b Energy would be lost due to the frictional force
5 a 8.52 m__ _s_−1__(3__s_._f._)_ ______ b 46.9 N__(_3__s.f.) acting on the marble, requiring a larger initial speed
6 a v = √ u 2− 1.4g( 1 − cos θ) b u > √ 2.8g for the marble to leave the bowl.
7 a T = 4.5g cos θ + 3_ 4_u_ 2 − 3g
8 a 0.27 ____ Mixed exercise 1
b u . √ 10g 1 R(→): R cos θ = mg
b 0.26
9 a 0.30 b 0.28 R(↑): R sin θ = _m _r_v_ _2 = _ 2_3m__au_ _2
Dividing the horizontal component by the vertical
c The particle will continue in a parabolic arc
(projectile motion) in the negative x-direction, component gives tan θ = _32 _au_g _2
initially increasing in y before decreasing in y.
10 a 6.26 m s −1(3 s.f.), 25.3 N (3 s.f.)
b 5.6 m s −1, 18.8 N (3 s.f.) b g m s−2 From the problem’s geometry, tan θ = _ (_ _3__2_a_ _ )_ = _ 3____
11 a 9.66 m s−1 (3 s.f.) √ 7
(_ √ _72 _ a )
c 45.8 N (3 s.f.)
12 3 9.6° (3 s.f.) to the upward vertical,
v__=_2.74 m s −1(3 s.f.) So, _√ 3_ _7__ =__3 _2_au_g _2
13 √ 8gr _ ____ b 1_1_5_m_ g_ 9ag = 2 √ 7 u 2
√14 a 1_6_5g_ _r 2 a 3_m_2_ g_ b m_2_g_
15 a T = mg (1 + 3 sin θ) b 19.5° (3 s.f.)
16 a At point S: G.P.E. = 0.4 × g × 3.8 = 1.52g, K.E. = 0 3 a 1_3_5_m_ g_ _____
4 108 m (3 s.f.)
At point P: G.P.E. = 0.4 × g × 4 sin θ = 1.6 gsin θ, b √ 60gl
K.E. = _21 _ × 0.4 × v 2= 0.2v 2
By conservation of energy: 1.52g = 1.6 gsin θ + 0.2v2 5 a T is the tension in string AP. S is the tension in
string BP.
⇒ 0.2v 2= 1.52g − 1.6g sin θ
The triangle is equilateral.
⇒ v 2= _7_._6_g__−__8_g_ s_i_n_ θ_
⇒ v = √ 7 .6g − 8 g sin θ R(→): T cos 60° = mg + S cos 60°
b 3.8 m T − S = 2mg
c In reality there will be frictional forces acting on the R(↑): T cos 30° + S cos 30° = mrω 2
handle so the height will be less than 3.8 m.
(T + S)cos 30° = m(l cos 30°)ω 2
T + S = mlω 2
Exercise 1E Eliminating S to find T:
1 a mg + 3mg cos θ b 4_3_l c 4_2_07_l 2T = 2mg + mlω 2
b m_2_T (l=ω _2m 2_− (22gg+)
2 a 9g cos θ − 6g b 48° (nearest degree) lω 2)
c Both strings
c 6.7 m b _34 ____ taut ⇒ lω2 – 2g . 0, ω2 . _ 2_lg _
3 a 9_r4_g_ − 2rg cos θ
d _3_√2_ _ r_ _g__ √c 3_r4_g_ 6 a Let T be the tension in the string.
e 64° (nearest degree) R(↑): T cos _4_5° = mg
4 a 4_8_°_ (nearest degree) √ T____=___√ 2 mg
b √ 8g , 74° (nearest degree)
b ω = _ √_ 2l_ g_
225
Answers
7 a 6.48 N b 25° (nearest degree) ______ ______
8 a 4mg
√b 3π _ r_g_ √ c _1_95_g_ r _ √d _7_13_5g_ r_
9 a _m_r_v_ _2 = μR = μmg 16 a _u_2_3+_a_2g_a _g_ b 34° (nearest degree)
_ vr_g 2_ = μ
Challenge (x, x2), _dd_yx_ = 2x
a At point
_1_0_0_2 _1×_ _29 _._8_ = μ R(↑): R cos θ = mg (1)
μ = 0.45 R(→): R sin θ = mxω2 (2)
b _13_3_5_6_ (2) ÷ (1): tan θ = _x _gω_ _2 (3)
10 a _√_ 34_ m_ _ (rω2 + 2g) tan θ = _dd _yx_ = 2x
b Maximum speed gives the shortest time. At the ⇒ 2x = _x _gω_ _2 ⇒ 2g = ω 2 ⇒ ω = ___
maximum speed with the rod still on the s__urface of
the sphere, R = 0. Radius of the circle is _√ _ 23_ r_ . √ 2g
Hence ω is independent of the vertical height.
From (3), ω 2 = _ g_ t_ax_n _ θ_
_c_mo_sg_ α _ = _ 2_√m_ _3__g _ b _g_ t_ax_n _ θ_ = k
For ω to be independent of
When R = 0, T cos α = mg ⇒T = ⇒ tan θ = ax for constant a x ⇒ for constant k
__
T sin α = m × _√_ 23_ r_ = ω2
__ _dd_yx_ = tan θ = ax ⇒ y = _21 _ ax 2 + b
So _2_√m_ _3__g _ × _12_ = m × ω2 Hence f(x) = px 2 + q for for constants p and q
_ √ _ 23_ r_ =
___
√ ω2 = _ 23_gr_
______ ___ CHAPTER 2
√ √ Time for one revolution = _2 ω_π_ = π _4_2×__g3_ _r = π _6g_r_ Prior knowledge check
c i The minimum period decreases. 1 x = 3, y = 2
ii The minimum period increases. 2 3.6 m
3 144 cm2
11 a Let F be the force due to friction. Exercise 2A
Let R be the normal reaction force. 1 (3.2, 0)
2 (0, 2.5)
R(→): R = mg 3 (1.1, 0)
R(↑): F = mrω 2 4 2_ 31_ m
5 m = 6
If P does not slip, then μR > mrω 2 6 0.7 kg
7 6.5
_73_ mg > _35_a_ mω 2 ⇒ ω 2 < _ 75_ag_ 8 (0, −2)
b 4_5_2g_a_ < ω2 < 64__52_ga_ 9 m1 = 2, m2 = 3
10 1 × (2m + 5) = ( (m − 1) × −1) + ((5 − m) × 1) + (2 × m)
12 a _43 ga + 2ga sin θ b mg( _43 + 3 sin θ )
+ ((m + 1) × 0)
c 206° 2m + 5 = 6
d v = 0 before the particle reaches the top of the 2m = 1
13 √2__.6c_ig_r c le. m = 0.5 kg
14 a g_2_a_ (5 − 4 cos θ)
b Resolving towards the centre O: Challenge _32 _ PQ))=
mg cos θ − R = _m _r_v_ _2
(2 × PQ) + ( 3 × (PQ + (6 × PG)
_12_9 PQ = 6PG
mg cos θ − R = _m _2_g _ (5 − 4 cos θ)
19PQ = 12PG
Substituting cos θ = 0.9 PQ: PG is 12: 19
R = 0.9mg − _m _2_g _ (5 − 3.6) Exercise 2B
R = 0.2mg
1 (3, 2)
R . 0 ⇒ P is__s_t_ill on the hemisphere 2 (0.5, −0.75)
√ c i _ 5_6___ _ ii 5__g6_a _ 3 (4.6, 4.2)
√ d 5__g2_a _ 4 3i + 2.5j
e 61° (nearest degree) 5 (2.1, 0.3) b 1.5
6 a 1
15 a u 2 − 1_ 5_6_ gr b m_5_g_ 7 p = 1, q = −2
8 (1, 3)
226 On line Full worked solutions are available in SolutionBank.
Answers
9 3_ 34_ cm from AB and 4_31 _ cm from AD. Challenge _ 3_2√_ _3 __ x
10 a 3 g b 3.2 cm Area of hexagon
ABCDEF = 2
Challenge Using Pythagoras, find the height of the midpoint:
0.2 kg _ √2_ _3__
x 2 = ( _x2_ ) 2 h 2 ⇒ h =
So
+
Exercise 2C the centre of mass of ABCDEF is _x2_
( _ √ _3_2_ _x )
1 a (2, 3) b (3, 4) __
c (_ 31_ , 1) d ( _83_a_ , 3a) Area of triangle to be removed = _12 _ × __ × _x2 _ = _ √4_ 3_ x 2
√ 3 x
2 Centre of mass is on the axis of symmetry at a (x _ √ _3+_2_ _x_ 6x _ )
distance _ 13_6π_ cm from the centre. So the centre of mass of DEF is
3 a = 3, b = 3 a from AB, Distance _2 3_a_ from BC. _y = __ to symmetry
4 a Distance
_ √ _ 32_ x_ due
__ x3_ 2√=_ _3 __ _ 1 x3_1 20_(x_ _,√ _3 __22xs__ _ xo ) N − ( _ √ _14_3 _3_1__ 0_ (x_ _ √ ,_ 7 _3_26___ x√ _x _ _3 2) __ x_= ) _ 5_ 4√ _ _3 __ ( _ √ __3x_2_ _ x )
b Distance _a3_ from BC, Distance _4 3_a_ from AB.
c On the line of symmetry, _4 3_a_ from the line AB.
d (3a, _23_a_ ) with A as the origin and AC as the x-axis.
MN = _2x _ − _ 13_10_x_ = _ 12_5_ x
5 (2, 7) and (−13, 4)
6 a B has coordinates (_ 15_8 , _45_1 ) , D has coordinates ( _15_2 , − _ 51_ ) Exercise 2E
b
(3, 4)
7 a (3, 5), (3,−3) 1 a ( _1463 , _1261 ) b (_ 1583 , _290 )
b (3, _37_ ) , ( 3, − _13 _ ) c (2.67, 2.26) (3 s.f.) d (3.73, 3.00) (3 s.f.)
__
8 y¯ = _3y _ = y− 2 Centre of mass is on line of symmetry through O, and a
_4 _ 3√ _ 3 _ __
__ _ 9_(_6√ _ 3+_ _+π_ 2__) from
_23_ y = _4 _ 3√ __ 3 __ distance of O.
3 _17_1 a
y = 2√ 3 4 a 5.83 cm (3 s.f.)
Centre of mass is at x = 2 b i 817 g (3 s.f.) ii 4.41 cm (3 s.f.)
So, using Pyth_a_g_o_r_a_s_:_____ 5 _15_2 m horizontally to the right of A and _11 _90 m vertically
A B 2 = BC 2 = √ 2 2 + (2 √ 3 )2 below A.
AB = BC = 4 6 Centre of mass is on the line of symmetry at a distance
So the triangle is equilateral. of _ 23_π_ below the line AB.
Exercise 2D 7 a 1.75 m 19_y ⇒ _y =
b
1 a ( _52 , _1143 ) b (1.7, 2.6) c ( _13103 , _3409 ) d ( _73 , 2) (20 × 1.75) − (1 × 0.5) = _ 63_89
△y = _63 _89 − 1.75 = _7 5_6
e ( _392 , _1538 ) f (2, _1673 ) g (_ 285 , 3)
2 _92_ a
3 2.89a (3 s.f.) _ C15_ √h _5a_ lclmenge
4 a 10 cm
b 12 cm horizontally to the right of A and 4.5 cm Exercise 2F
vertically above A.
1 a 20.4° (3 s.f.) b 24.4° (3 s.f.). c 56.8° (3 s.f.)
5 a 4.5 cm horizontally to the right of O, and 3.5 cm
vertically above O. 2 63.0° (3 s.f.)
b 5.83 cm horizontally to right of O, and 4.32 cm 3 80.5° (3 s.f.)
vertically above O.
4 33.1° (3 s.f.)
6 a 12(13) + 24( 54 ) + 16( 92 ) = 52( xy ¯¯ ) 5 81.0° (3 s.f.)
( xy ¯¯ ) = ( __4161_193_3 )
6 ab __217768 ccmm
c α = 22.2° (3 s.f.)
( _619_3 , _ 41_13 ) 7 67.1° (3 s.f.)
8 65.3° (3 s.f.)
b Since y ¯ = _14 _13 for original plate, holes must be 9 a 3.5 m
symmetrically placed about the line y = _ 411_3
b At A: 5g N; At B: 7g N
c _790_
7 x = a c 149°
10 _8_−π__ π_
227
Answers
11 a For a rectangular lamina: Area = 2a × 3a = 6a 2 Challenge
a Let width = 2, then height = _t a__n_21_2_ _.5__° = _ √_ _2__ 1_−_ _1_ = √ _2_ + 1
a b On central axis, a distance 0.54x from the bottom edge
Centre of mass ( _32a_ )
For the quarter circle: Area = _14 _ × π × a 2 = _ π_4a_ _2 of the paper.
Centre of mass: Mixed exercise 2
d = 2_ _r_ 3s_iαn_ _ α_ = _2 _a3_ s_×i_n__4π( __ _ 4π __ ) = _4_√3_ _2π__ a _
This distance is the hypotenuse of a right-angled 1 a 0.413 m (3 s.f.) b 12° (nearest degree)
c 0.275 (3 s.f.)
triangle, with the other sides equal. 2 θ = 36.9° (3 s.f.)
d 2= 2l 2 ⇒ ( _ 4_√3_ _2π__ a _ ) 2 = 2l 2 ⇒ l 2 = _ 19_6_πa_2_2 ⇒ l = _43_aπ_ 3 (− _17_ , _23_ ) b _49_a_ c 45° d m = _5 _9M_ _
(2 − _43a_π_3 4 _π ) a ) 4 a _1_39_a _ ii _a2_
( 5 a i _45_a_
So, centre of mass of the quarter circle b θ = 58°
Centre of mass of the lamina: 6 a 1.7a b 1.1a
a _ π_4a_ _2 ( (2 − _43a_π_3 4 _π )a )= _4π _ ) a 2( __xy ) 7 59.1° (3 s.f.)
6a 2( _ 32a_ )
− ( 6 − 8 a 39.0° (3 s.f.) b 7.8° (3 s.f.)
9 AB = 0.25M, BC = _32 _ M , CD = 0.5M, AD = M
(6 − _4π _ ) a 2 _x = 6a 3 − _ π_2a_ _3 + _ a3_ 3_ ⇒ _x = _ 2_a3_((3_2_84_ _−−_3_π_π) _)
(6 − _4π _ ) a 2 _y = 9a 3 − _a3_ 3_ ⇒ _y = _3_(12__04_4_−a__ π_) 0.25M(33 ) + _ 32_ M( 55 ) + 0.5M( 83) + M( 61) = _ 12_29_ M( xy)
( yx) = ( _1_ 722_6_9999_ )
b _3_(2_3_44_W_−_ _π_) and W_ _3 _((32_8_4_ −_−_3_π_π) _) tan θ = _( (__ 152__699__ − _−__27 __993_ ) )_ = _ (( __ _2682_69__29 ) )_ = _ 34_31_
θ = arctan( _ 34_31_ )
c arctan( _1_1_2_6_−8_ _9_π_ ) (39.1° (3 s.f.))
Challenge 10 a T1 = 1.2Mg and T2 = 0.8Mg b 45°
0.0343M (3 s.f.) 11 a T1 = Mg and T2 = 2Mg b 47.3° (3 s.f.)
Exercise 2G Challenge
13.6° (3 s.f.)
1 21.8° (3 s.f.)
2 104° (3 s.f.) CHAPTER 3
3 a x = 8 cm
4 30.6° (3 s.f.) b 0.2 Prior knowledge check
5 28.5° (3 s.f.) __ __
6 19.4° b 78.7° (3 s.f.) _ √_ _3__√ _ 2+_ _1_ mg, _√ _ _3__ _−__1_ mg
7 a T1 = _85 _ W , T2 = _38 _ W b 25.5° (3 s.f.) 1 T1 = T2 = √ 3 + 1
8 a T1 = _ 17_8 W , T2 = _ 19_1 W
2 A = 8 and B = 4
Challenge
θ = 26.6° (1 d.p.) 3 _ 29_612_
Exercise 3A
1 ( _23 , 2 )
2 (1.5, 3.6)
3 (2.4, 0.75)
( ) 4 _124_5 , _ 23_35
Exercise 2H ( _ _532π_ a , ,_ 8π_0 )
1 _37_ c m horizontally to the right of AD and 2 cm vertically ( ) 5
below AB.
6
2 2.4 cm vertically below AB and 5.6 cm to the right of
(original) AD. ( ) 7 _1_−l_n___ l2_n_ 2_ , _ 4_ l1_n_ 2_
3 20.1° (3 s.f.) ( ) 8 _ 4_√3_2π_ _r , 0
4 ( _13_15 , _ 11_12 ) ( ) 9 _116_5 , _ 26_14
5 (2.89, 1.31) (3 s.f.) 10 (2_ 55_54 , 1_ 15_1 )
6 50.2° (3 s.f.) 11 (1.01, 0) (3 s.f.)
7 a 0.07 m 12 (1.34, 0.206) (3 s.f.)
b C, as the centre of mass of the composite lamina is Challenge ___ ___
between O and C.
c At B: 17.2g; At C: 2.8g √ √(0, 0), (4, 0), (4, 2), (4, −2), ( _11_09_ , _11_09_ )and ( _ 11_09_ , − _11_09_ )
8 a 61.2° (3 s.f.) b 27.6° (3 s.f.)
228 On line Full worked solutions are available in SolutionBank.
Answers
Exercise 3B 3 a A square based pyramid has base area A and height
h, the centre of mass is on the axis of symmetry.
1 (2, 0) 2 (1, 0)
Volume = _13 _ Ah
3 (π, 0) 4 (0, −3) Mass = _31 _ Ahρ
5 ( _56 , 0 ) Take a slice of thickness δx at a distance xi from
6 ( 2 _32 , 0 ) vertex. The base of the slice is an enlargement of
7 ( _345_8 , 0 ) _xh _ i .
the base of the pyramid with scale factor
8 (1.65, 0) (3 s.f.)
Ratio of areas is ( _ xh_ i )2
( ) 9 _12_ _((ee_22__+−__11_)) , 0
10 (0, 1.34) (3 s.f.) Area of base of slice is _hx _ i22_ A
11 (1.01, 0) (3 s.f.) Mass of slice mi = δx
12 1.46 (3 s.f) δlxi→m0 ∑x=h0 m ix i= ∫0 h_ xh_ 23_ Aρ dx = _14 _ h 2Aρ
13 3.04 cm _x = _ ∑__Mm__ i_x _ i = __ 14 _ _31_ _ h A_ 2_hA_ρρ_ = _34 _ h
14 _3r_ above the base
15 _56_ c m above the base The centre of mass lies on the line of symmetry at a
16 The arc of the circle x2 + y2 = a2, a – h < x < a is distance _ 43_ h from the vertex, or _4h _ from base.
b 3.39 cm (3 s.f.) below O
rotated about the x-axis.
x̅ = π ∫ a (a 2− x 2)x dx _ [ _12_[ _ a _a_ 22_xx _ 2−_ −_ _12 _ _14x_ x _3]_ 4 aa]_ − aa_h − _h 4 ab _ _ 4h83__ √ w _6_ hm erbeelhowis
_ _a_−h_____ _______ =
π ∫ h (a 2− x 2)dx the height of the tetrahedron.
O
a−h
= _ _14_ _2a3_ a4_ _3−_−_ _12_ aa_ 22 _( (a _a_−_−_hh_)_) 2+_ _+ _13 _ __1(4a _ (a_−_−_h_)h 2_) _ 4 5 a 1.96 cm (3 s.f.) b 0.48 cm below O (3 s.f.)
= _ _13_ (2_a__ 2_− __14 _ (3a_a _2 _2−(_a_( a−_ _−h__)h _+ ) _2)(_ a2_ _−_h__) 3_)
= _43_ _(23_ah_h _2a_−_−_h_h 2_) 3_ 2 6 a 11.4 kg
b 0.589 m (3 s.f.) from its base.
c Mass of post will decrease as cross-section of
cylinder is less than 1. Position of centre of mass
will be the same.
= _ 43_ _((23_a_a_−−__hh_)_) 2 7 a Shape Mass Vertical
distance from
O to CoM
∴ Distance of centre of mass from base of cap Cylinder 32πr3 r
(i.e. x = a – h) is
Hemisphere _ 2_π3_r_ _3 _ 38_r_
_34_( (2_3_aa__−−__hh_)) _2 − (a − h) = _ 3_( 2_a__−__h_)_ 2 4_−(_3_4a_(_3−_a_h_−)_ _h_)(_a__−__h_)
= _44_(a_3h_a_−_−_h_h 2_) Composite 32πr3 − _2_π3_r_ _3 = _9_4_3π_ r_ _3 x _
body
17 a y _ = 3 = _h4_((_43_aa__−−_h_h_)) 32πr 3( r) − _2 _π3_r_ _3 ( _ 38_r_ ) = _ 9_4_3π_ r_ _3 (x _ )
b 4.18 cm _ 1_2_74_π_ r_ _4 = _9_4_3π_ r_ _3 x _ ⇒ x _ = _3 3_87_1_6r_
18 k = _57 _
19 a _24_3_4h_ b i _33_89_7_2r_ ii _4r_9_
8 a _81_91 m
b On the common axis of symmetry of the cone and c 0.727 m (3 s.f.) b 4.76 m (3 s.f.)
d 3.45 m (3 s.f.)
ocvyef lrtithniecdaeclry,alaixniddsieosrtfaatnhncedeca_2y 4_l3di4_nhi_ sd tbaeern.lcoew_1 tr_1h_ etototpheplleafnteoffatchee
9 a 824 kg (3 s.f.) b 8.73 m
10 a _23_ b _76_ m
Challenge
20 a 4.07 cm (3 s.f.) b 4.17 cm (3 s.f.) a = –2.5 and b = 3
Challenge Exercise 3D
_R2_ from the plane face
1 θ = 68° (to the nearest degree)
2 θ = 40° (to the nearest degree)
Exercise 3C 3 θ = 63° (to the nearest degree)
1 The centre of mass lies on the axis of symmetry at a 4 a 2.78 cm from large circular face
point 0.42 cm below the base of the cone. b α = 51° (to the nearest degree)
2 Centre of mass is on the axis of symmetry at a distance 5 T1 = _1_3.3 N, T2 = 6.34 N b T1 = 38.3 N, T2 = 42.9 N
5.33 cm away from base of the cylinder. 6 a _√3_ 2_ m
229
Answers
7 a At P reaction is _1 _94_5_ g_ N and at Q reaction is _4 _44_5_ g_ N. 6 a Consider the cube in equilibrium, on the point of
b 390 kg toppling, so the reaction force acts through the
bottom corner A.
8 a Mass of elemental disc = ρπr 2h = 100π e 0.1h δh R(→): P − F = 0 ⇒ F = P
Total mass = 100π ∫ 30 e 0.1h dh= 100π [10e 0.1h] 30 0 R(↑): R − W = 0 ⇒ R = W
0 Moments about A: P × 4a = W × 3a
= 100π(10 e 3− 10)= 1000π(e 3− 1) ⇒ P = _34 _ W
1000π(e 3− 1) y _ = 100π ∫ 30 he 0.1hd h ⇒ y _ = ∫ 30 he 0.1hd h If equilibrium is broken by toppling:
_ 01__0_(e_ _3 _−__1_)_
0 P = _43 _ W ⇒ F = _43 _ W
But F , μR
∫ 30 he 0.1hdh = [ 10he 0.1h] 30 0− ∫ 30 10e 0.1hdh I⇒sfliμp _34_p , Win _34g_, twhμehWnent⇒hFeμ=c.uμbR_ 43e_. iws itlhl ebecoonndtihtieonpofoinrttoopf pling.
0 0
= [ 10he 0.1h − 100e 0.1h] 03 0= 100(2e 3+ 1)
∴ y _ = _ 1 _1 0_00_((_e2_ 3e _ 3−__+1__)1 _)= _ 1_0_e (_2 3_e_− 3__+1_ _1_)
b 40°
9 a T1 = 1.07 × 10−3 N, T2 = 1.21 × 10−3 N b 2a b 0 , k , 6
b 46.7° (3 s.f.) 7 a 25.6°
10 a In equilibrium the centre of mass G lies below the 8 16.1°
point of suspension S. Let distance SG = x. 9 a Shape
O is the centre of the base of the cone and V is its Mass Mass Distance
ratio of CoM
vertex. from O
A is the point on the base connected to the string
_atA3xa_nrl_ ns d=o θBt_ x=a_ins−_r_3 x_ _ θrt_r _h =(efr_xp o_om−_ri _nr_tt r(oifarnnotgmhleetrVliinSaenGg)SlGe a distance r from G. Hemisphere _23_ πρ r 3 h + _38 _ r
ABS) 2r
Cylinder πρr2h 3h _h2_
x = 3x − 3r Composite πρr 2( _23_ r + h) 2r + 3h _x
solid
2x = 3r ⇒Mo82mrrhhen++ts3_34 :_ r r2 22r +(+h6_ 23_+ h h 2 _832 _= =r)4(2+(2r3r+h+3×3hh_2h) __x) _x = (2r + 3h)_x
x = _3 2_r_ _21 _ ⇒ _x = _ 6_h_ 4 2_(_+3_h8__h +_r _2+_r_)3_ _r _2
tan θ =
__
b _√ _ 5_2 m_ _g_
11 22.8° (3 s.f.) b α = 29° (to the nearest degree)
Exercise 3E c μ . _ 84_14_
10 a Shape
1 4.77 cm (3 s.f.) b μ = _35 Mass Mass Distance
b 35 cm (2 s.f.) ratio of CoM
2 a 31° (to the nearest degree) 8 from O
3 a 30° _24_h_
4 a _ 2_2_+M__√g__3 __
b _12 Large cone _13_ πρ(2r 2)2h
5 a Let the height of the small cone be h.
Using similar triangles:
_ h_+_h_ 2_r_ = _ 2r_r_ Small cone _13_ πρr 2h 1 h + _h4 _
Frustum _13_ πρ(8r 2h − r 2h ) 7 _x
2h = h + 2r
h = 2r The centre of the base is the point O.
Shape Mass Mass Distance of CoM The radius of the smaller cone is obtained by
ratio from centre axis similar triangles.
7_x
Large cone ρ _13_ π(2r) 2 4r 8 r Moments: 8 ×_x _2= 4_h_ _ 21 _−81_ h1 × _5 4_h_ =
Small cone ρ _13_ πr 2× 2r 1 _1_14_h _ = 7_x ⇒
2r + _2 4_r_ = _52_r_
_x
Frustum ρ _13_ π × 14r 3 7 b θ = 38° (to the nearest degree)
11 a P . μMg b P . _ 83_ Mg
Take moments about centre axis: c i Slide ii Topple iii Slide and topple
_5 2__xr_ == 7_x
8r − _ 11_14_r_ 12 a P = _49 _0g_
b i Yes ii No b P = _13 _0g_
c As the angle of the slope is 40°, limiting friction
would imply μ = tan 40°. c _49_0g_ , _ 13_0g_
No slipping implies that μ > 0.839 (3 s.f.)
230 On line Full worked solutions are available in SolutionBank.
Answers
Challenge b Shape Mass Mass Distance of
ratio CoM from AB
a Shape Mass Mass Distance of
ratio CoM from O
Large 2πρa2 4 _83_aπ_
Cone _13_ πρr 2h h _ 4h_ semicircle
Hemisphere _ 32_ πρr 3 2r _ −_83_ r_
Toy _31_ πρ(r 2h + 2r 3) h + 2r _x Semicircle _12_ πρ a 2 1 _43_aπ_
diameter AD
Moments: (h + 2r)_x = h × _4h _ + 2r( −__ 83_ r_ ) Semicircle _12_ πρa 2 1 _43_aπ_
(_xh =+_4 h2_( _h2r_)−_+x__ 3=2_r_r _ h2_)4_ 2_ − _ 3_4r_ _2 diameter OB πρa2 2 _x
Remainder
Moments: 4 × _83 _aπ_ − 1 × _43 _aπ_ − 1 × _43 _aπ_ = 2_x
b i Fall over _23_4_πa_x_ == 2_x
ii Return to vertical position _4π_a_
iii Remain in new position
c The distance from OC is a
The distance from OB is _2 π_a_
Mixed exercise 3 d 78° (to the nearest degree)
1 a V = ∫ π y 2 dx= π ∫ 44 x dx 8 a Shape Mass Mass Distance
ratio of CoM
= π [2 x 2] 04 0 Cylinder πρr2h from O
Hemisphere _ 32_ πρ (3r) 3 h −_2_ h_
= 32π Mushroom π ρr 2(h + 18r) 18r _83_ (3r)
h + 18r
b x ̅ = _38_ π ∫1 2 _x1_ 2_ dx 0
2 a V = ∫ π y 2 dx=
= π [ −_x_ 1_ ] 12
= _ 2π_ m 2 Moments: − h × _2h _ + 18r × _83 _ × 3r = 0
_ h2_ 2_ = _ 8_14__r_ _ _2_
b 39 cm to the nearest cm
√ h = r _ 82_1_
3 34 cm
4 9_1_r0_ b θ = 9° (to the nearest degree)
5 a 0.71 (2 s.f.) 9 a V = π ∫ a4ax dx
0
b The centre of mass of the body is at C which is
always directly above the contact point. = π [2ax 2] 0a
6 a _y = _ ρ_ ρ∫_ ∫ _21__ y_y_ d2_ dx_x _ = _ _ 12 __ ∫0_ 4_ _1x_ _ _416_ 2__ ∫0 ( _ 1 4 _46_x_−_−_8_xx _ 2 _+ d_x_x _ 2_)d_x_ = 2πa 3
= 2 _ [ _13__ _x_ [ 32_ −_ x__ 812__ x_− _4_ 13__+ _ x__ 831__0] 04_ x _ 5_]_ 04 = _ 25_ b _x = π ∫ a4 ax 2 dx
_ _0__2_π_ a_ _3 ___
b θ = 79° (to the nearest degree)
= π _[ 2_ 4__πa_3_x _a_ 3_ ] 3_ 0a
7 a Take the diameter as the y-axis and the midpoint of
TthheedniaMm _xe =terρ ∫a s2ythxe dxorwighine.re = _23_ a
c ρ1: ρ2 = 6 : 1
2Mρπ=a _21 _2 ρ _xπ = (2ρa ∫)2 2π2axn√d _4_wa_ h2__e−_r_xe_ 2_x d 2x+ y 2 = (2a) 2 d As centre of mass is at centre of hemisphere this
will always be above the point of contact with the
plane. (Tangent–radius property).
0 10 a Shape Mass Mass Distance
ratio of CoM
= −__ 32_ ρ_ [( 4a 2− x 2) _ 23_ ] 20π from AB
_2x ρ =πa_ 13_6 2 _ xa 3=÷_ 23_2ρ_ π ×a 8a 3 Cylinder πρ (2r) 2× 3r 12r _ 32_r_
2
Cone _ 13_ πρr 2 × h _ 31_ h _41_ h
= _83_aπ_ Remainder 1 2r − _13 _ h _x
π ρ(12r 3 − _ 13_ r 2h)
Moments: ( 12r − _31 _ h)_x = 12r × _3 2_r_ − _ 3h_ × _ 4h_
231
Answers
(12r − _31 _ h)_x = 18r 2 − _ 1h_2 2_ 7 a α = 70.5° (3 s.f.) √ b l= 2_ 3_ _2_kr___ g _ _
8 a 2mg 2π
_x = _ 1_18_2r_r 2_ _−−__3h__ 1 h_ _2_ 2 b
_x = _ 24_1(_36_6_rr_ 2 _−−__hh_) _2 __
9 a tan 60° = _ _rh r__ ⇒ r = _2h _ tan 60° = _√2_ 3_ h
b θ = 38° (to the nearest degree) b Tension in AP = mg + _12 mh 2 and tension in
BP = _12 mh 2 − mg
11 a 0.265 m (3 s.f.) b 0.478mg (3 s.f.) ___
12 a 1.11 (3 s.f.) b 14.4° (3 s.f.) √ c BP 0 _2h_g_
9 Tension in . ⇒ω .
13 a ∫ 9 (1000 + 400x )_ 21_ d x = [ 1000x + ]_8 _30 0_ x _ 32_
0 0 ___ ___
√ √ As T = _2 ω_π_ , T , 2π _2h_g_ ⇒ T , π _2g_h_
= (9000 + 7200)− (0)
= 16 200 kg 10 a R(↑): T cos θ − mg = 0 ⇒ T = _ cm_o_sg_ θ_
R(←): T + T sin θ = mrω 2
b 4.9 m
14 a 40 cm ⇒r =ωh 2 t a=n_ hg _θ ( _⇒1_+_s_ ci_mn_so__i _θsg_n_ θ_ θ_ ( ) 1 + sin θ) = mh _cs_oi_ns_ θ θ_ ω 2
b i e.g. Suitable if uniform across cross-section, or
suitable as height ≫ diameter, or unsuitable as
may be non-uniform across cross-section.
ii Unsuitable as rod has no width so will never be b ω 2 = _ hg_ ( _ s_i1n_ θ_ + 1)and sin θ , 1 so _s _i1n_ θ_ . 1
stable.
___
c 5.71° (3 s.f.) √ ⇒ _ hg_ × _2h_g_
15 TA = 48g N, TB = 120g N ω 2 . 2 ⇒ ω .
16 60 m
__
c T = 2_ _√3_ 3 _ mg or 1.15 mg
Challenge ( )11 2a
a Mass of elemental disc =ρπ (h − x) 2 δx 12 a R(←): T cos 30° = m(2a cos 30°) _ 3k_ag_ ⇒ T = _2 _k_3m_ _g_
b mg 1 − k_3 _ c k , 3
= π (x + 1)(h 2− 2hx + x 2)δx
= π(h 2 x − 2hx 2+ x 3+ h 2− 2hx + x 2)δ x
Mass = π∫ h h 2x − 2hx 2+ x 3+ h 2− 2hx + x 2 dx d PX = 2a cos θ
0 R(←): T ′ cos θ = m × 2a cos θ × _2 a_g_ ⇒ T ′ = 4mg
= 2 − _ 32_ h x 3 + _ 14_ x 4+ h 2x − hx 2 + _ 31_ x 3] 0h R(↑): T ′ sin θ − mg = 0 ⇒ sin θ = _m T_g _′ = _4m_m_g_g_ = _ 14_
π[ _21_ h 2x AX = 2a sin θ = _21 _ a, AO = 2a sin 30 = a
So AX = _21 _ AO
= π(( _21_ h 4 − _ 23_ h 4 + _ 14_ h 4+ h 3− h 3 + _ 31_ h 3)− (0)) 13 a R(↑): T cos θ − S cos θ − mg = 0
= π(_ 11_2 h 4 + _ 31_ h 3)
= _ 11_2 πh 3 (h + 4) ⇒ T − S = _c m_o_sg_ θ_ = _4 _m3__ g_ (1)
b 5 m
=y_ =_ _ 6 1 _ 1__01 _2k _ π π_hh_h f_4 o3 _( r(2 _ h_hs_o+_+m_45_e)_) constant
c y _ R(←): T sin θ + S sin θ = mrω 2 = ml sin θ ω 2
If
k ⇒ T + S = mlω 2 (2)
⇒ k = _ _ 6 1 _ 1__01 _2 _ π π_h_h _3 3 _( (2 _ h_h_+_+_45_)_) = _ 52_(_hh_+_+_45__) Solving (1) and (2) simultaneously
=_ 2_ (_5h_(_h+_ +_4_)4_−_) _3_ = _ 52_ − _ 5_(_h_3+_ _4__) ⇒ T = _16 _ m(3lω 2+ 4g) c As S > 0, ω 2 > _ 43_gl_
As h → ∞, k → _ 25_
Hence as h varies the height of the centre of mass of the b _16 m (3lω 2 − 4g)
cone above its base cannot exceed _25 _ h.
14 a R(↑): R − mg = 0 ⇒ R = mg
R(←): F = mrω 2 = m( _34_ a) ω 2
As P remains at rest F < μR
_ 29_0_ga_
⇒ m(_ 43_ a) ω 2 < _ 35_ mg ⇒ ω 2 <
Review exercise 1 b 2_ _g0_ a_
ω 2max = 1 2__90_ga_ and ω 2min =
15 a ⇒Ti s2iv.n52 θ _g=_ =N_9 1__ mg _6_ xr__v _ ⇒2_ ⇒ vii( = _5a2_g_43r_ _ )c √c( _g_ 53o_x__s ) ( _ =45_ )o_ (2 _43_r_ v _x3 _2)6 .9°
1 3.67 m s−1 (3 s.f.) b
2 a T = 5.5 N (2 s.f.) b θ = 26° (nearest degree)
_c o_m_s _6g_ 0_° = 2mg
√ 3 a R(↑): _T_ c_os 60° − mg = 0 ⇒ T =
ω = _2L_g_
b √16 a u = g_2_l
b 5_6_l
4 190 m (2 s.f.)
5 24 m s−1 (2 s.f.) 17 a Use conservation of energy:
6 a _5_m4_ g_ ____ _12_ m(u 2− v 2) = mgl(1 − cos θ)
⇒ v 2= u 2− 2gl(1 − cos θ)= 3gl − 2gl + 2gl cos θ
b v = √ 6gl
c The tensions could not be assumed to have the v 2 = gl + 2gl cos θ
same magnitude.
232 On line Full worked solutions are available in SolutionBank.
Answers
Resolve along the string: b c = −_13
T − mg cos θ = _m __lv _ 2_ = _ m_g_l__+_2__ml _g_l_ c_o_s_ θ_ 32 a 10.7 cm (3 s.f.)
⇒ T = mg(1 + 3 cos θ) b 25° (nearest degree)
___
√ b v_=___ _g3_l 33 a The area of rectangle ABDE is 6a × 8a = 48a 2
c 4_2_07_l The area of △BCD is _12 _ × 6a × 4a = 12a 2
The area of lamina ABCDE is 48a 2+ 12a 2= 60a 2
√ 18 a _2_3g_ _l Shape Lamina Rectangle Triangle
b Resolve along the string: T − mg cos θ = _m __lv _ 2_ (1) Mass ratios 60a 2 48a 2 12a 2
Conservation of energy:
_12_ mu 2 − _ 12_ mv 2 = mgl(1 − cos θ) (2) Displacement GX 4a − _34_ a
from X
Solving (1) and (2) simultaneously: T _ m3_g_ ( 9 cos θ
c 2_m_3_ g_ < T < 5_m_3_ g_ = − 4) M(X): 60a 2 × GX = 48a 2× 4a + 12a 2× (− _43_ a)
= 192a 3− 16a 3= 176a 3
GX = _1 6_7_0_6a_a _2 _3 = _ 414_5 a
19 a _ 43 or 0.75 the radius OB: mg cos θ = _m __lv _ 2_ = _ m0__.v8_ 2_ b _11_51
b Resolve along
34 a 6_5_l
So v 2= 0.8g cos θ = 0.6g = 5.88 b l c 51° (nearest degree)
c u = 1.4 35 a 11_9_5 a b 4_7_5 M
20 a _21_ mv 2 = mg(a cos α − a cos θ) ⇒ v 2_=__2_ga(cos α − cos θ) 36 a 6 cm b 22.6° (1 d.p.)
√ 2 1 abb __θR3221_ mme=sgo6×l0v_7 °i_2ngo_a g_r a−_3π _l o_ 12r_n amgdVira 2an d=si umsg:ma(g1 c+ocsc θos= θ_ 7_)m_ 2g__a_a V_ (_ 1 _2 ) (2) 37 a 6.86 cm (2 d.p.) b 32.1° (1 d.p.)
Eliminate V 2 from equations (1) and (2):
⇒___c_os θ = _12 _ ⇒ θ = 60° 38 a 25 cm b _13_1 m
39 a ( _92 , _23 )
√ c _6g_a_
b The centre of mass of the lamina is (3, 0)
Let the centre of mass of the combined system of the
lamina and the three particles be at the point G.
The total mass of the system is 12m + km = (12 + k)m
Shape Combined Particles Lamina
system
22 a v = 8 b m = 6 c 1300 N (2 s.f.) Mass (12 + k)m 12m km
23 a 30 m s−1 (2 s.f.) b 1900 N Distances (x) 4 _92_ 3
Distances (y) λ _32_ 0
c 28 m s−1 (2 s.f.) d 560 N (2 s.f.)
e Lower speed at C ⇒ the normal reaction is reduced.
24 a v 2 = ag (1 − 2 cos θ) M(Oy): ( 12 + k)m × 4 = 12m × _ 92_ + km × 3
⇒ 48 + 4k = 54 + 3k ⇒ k = 6
b R(↙) along radius:
_m _a_v_ 2_ ⇒
T+ mg cos θ = T = (1 − 3 cos θ) mg c _94 d 83.7° (1 d.p.)
c _34 a d _24_7 a or 0.148a
40 a Let the distances of the centre of mass of L, say G,
25 a _23 ga + 2ga sin θ b _3_m2__ g_ (1 + 2 sin θ) from AD and AB be x¯ and y ¯ respectively.
The mass of L is 3m + 4m + m + 2m = 10m
c Put T = 0, then sin θ = − _21 _ so θ = 210°
d No. v = 0 when θ = 229° (3 s.f.), so P changes Shape L ABCD A B C
direction before it reaches the top of the circle.
Mass 10m 3m 4m m 2m
e Consider motion at start and at A:
no change i_n__P._E. ⇒ no change in K.E. Distances (x) x ¯ 2.5a 0 5a 5a
Distances (y) ¯ y a 0 0 2a
√ so v = u = _ 3_2g_a _
f 73.2° (3 s.f.) b 0.7a c 20° (nearest degree)
26 a _23 + _ 3u_a_ 2g_ b 34° (nearest degree) d M(O): P × 2a = 10mg × (2.5a − x ̅)= 10mg × 0.25a
27 a u 2 − 3ga b 5_m_2_ g_ _2 _._52_ma__ g_a_ = _45_ mg
⇒ P=
_____ ____
___
√ c _7_2g_a _ d u = √ 5ag e 5__√4_6_ 5_ mg
28 a Taking moments about the x-axis: 41 a 3 b 2 c 37° (nearest degree)
42 a i _ 52 a ii _43 a
2(8 + λ) = 3 × 4 + 5 × 0 + 4λ so λ = 2
b 15° (nearest degree)
b k−1.1
29 x = y = _12
30 (3i + 2.5j) m 43 a i 1_5_2l_ ii _3l_
31 a The total mass is 2M + M + kM = (3 + k)M b 39° (nearest degree)
M(Oy): ( 3 + k)M × 3 = 2M × 6 + M × 0 + kM × 2 44 a d b d c 63° (nearest degree)
b 0.211 (3 s.f.)
⇒ 9 + 3k = 12 + 2k ⇒ k = 3 45 a 3.28 cm (3 s.f.)
233
Answers
46 a 43.8 cm (3 s.f.) b Shape Mass Mass Position
ratio of CoM
b T1 = 0.452 N (3 s.f.), T2 = 0.548 N (3 s.f.)
c k = 7.45 (3 s.f.) Cone _ 31_ πρ a 3 k k _ 43_ ka
47 _83 from O Hemisphere _ 32_ πρ a 3 2 ka + _83 _ a
48 a The centre of mass lies on the x-axis from Top _ 31_ πρ a 3(k + 2) k+2 x _
symmetry. Taking moments about V:
An elemental strip of area is 2yδx k(_ 34_ ka__)+ 2(ka + _83 _ a)= (k + 2) x_ ⇒ x _ = _ (3_ _k_ 24 _(+_k_8_ +_k_ 2+_)_ 3__)a_
The boundary of the semicircle has equation c k = √ 3
x 2+ y 2= a 2, 0 < x < a
x_ = _ ρ_ ∫ _ 2__x_ ρ (a_×_ 2_ _−π_ _ 2a_ _x _2 _2_) _ 21__ d_x_= _π_2a_ _2 [− _32_ (a 2− x ]2) _32_ a = _ 43_aπ_
0
b Shape Mass Mass Distance of 52 a Shape Mass Mass Distance of
ratio CoM from O
ratio CoM from O
Large
Semicircle _12_ πρ a 2 a 2 _ 43_aπ_ hemisphere _ 23_ πρ a 3 8 _ 38_ a
radius a
Small _32_ πρ ( _a2_ ) 3
Semicircle _12_ πρ b 2 b 2 _34_πb_ hemisphere 1 _ 13_6 a
radius b _ 12_ πρ(a 2− b 2) a 2− b 2 x _
Remainder Remainder _ 32_ πρ _7_ 8a_ _3 7 x _
Moments about O : a 2 × _ 43_aπ_ − b 2 × _ 34_πb_ = (a 2− b 2) x _ Taking moments about O: 8⇒× x__38 _ =a − 1 × _1 3_6 a = 7 x_
so x_ = _ 34_π_ _((aa_ _23_−−__bb_ 23_)) = _ 34_π__ _(a__−_( a_b_−)_(a_b _2)( _ a+__ +a_b_b_)+ __b _ 2_) _k 85_π_= _27 _ 41_51_a2_
b
=_ 4_(_a_3 _2π_+(_a_a _+b_ _b+_) _b _ 2_) 53 a
c _2π_a_
_ π_ _0 _2 __41__ x_ (__ x8_5 _π _− __2_)_ 4_ d_x_ = _85_π_ ×
_ 35_2 ∫− 20 u 5+ 2u 4 du = _3 5_2 [ _61_ u 6
∫ ∫ b x_= _4π _ 0 (u+ 2) u 4 du
=
49 a Let the equation of the line AC be y = c – mx
−2
_ _ 12__ ∫∫ _0 _m_ c_ _ m c _( _ cc_−−__ mm_ _xx_d)_ 2x_ d_ x_= _ _ 12_[ _ [−_− __2 _3_1_m__1m_ __ _ _ ( c (_c_− −__mm_x_x)_ )2_ 3]_] _0mc_ _ _ 0 m c_ + _ 25_ u 5] 0
0
y _
−2
= = _35_2 (0 − (− _36_04 ) ) = _ 13_
= _ 6c_m _3_ ÷ _ 2c_m _2_ = _ 31_ c c _5_19_2W_ _
54 a Shape Mass Mass Distance of
ratio CoM from O
b, c Shape Mass Position of CoM
Cylinder πρ r 2 h 1 _h2_
Rectangle 2a 2ρ ( _a2_ , a)
Triangle a 2 ρ ( _43_a_ , _23_a_ ) Cone _ 13_ πρ r 2( _h2_ ) _61_ h − _41 _ (_ h2_ )
Lamina 3a 2 ρ ( x_ , y_ ) Ornament _ 56_
_ 65_ πρ r 2 h x _
Taking moments: Taking moments about O, centre of plane base:
⎜ ⎟ 2a 2 ρ( _ aa2_ )+ a 2 ρ⎛⎝_ _ 4233__aa__ ⎞⎠ = 3a 2 ρ( xy __) 1 × _h2 _ − _16_ × _ 78_h_ = _56_ x _ ⇒ _ 56_ x _ = _ 14_7_8h_ ⇒ x_ = _ 14_70_h_
b α = 66.5° (1 d.p.)
⎜ ⎟ 50 a ⎛⎝x_ 2 a a=++ _∫∫_4_ π_23_ _πa3__xa _ _ y _ y⎞⎠ 2_ =2 d_ d_3xx_ ( yx __=) _ __ 1441⇒___ ∫∫_11 _22 x _ xx_ _ =−−_34_ _ 97dd__ axx_, = y __ [[ =−−__ __3211_ 89____ xxa__ −−_23_]]_ 1221 = _ 79_ 55 a The centre of mass lies on the y-axis from
symmetry. An elemental strip of area is 2x δx
hy2x dy hy(h − y) dy
_ 0__h______ = _ _0 _h____ ______
=∫ ∫ y_
∫ ∫ 2x dy (h − y) dy
00
The centre of mass is ( _79_ − 1) = _ 72_ m from the larger = _ [ [ _ _yy__22h__h __ −_− __y 2_ _y3_2__ 3 ] _]h 0h_ 0 = _h3_
plan face.
b _32_90 m or 0.967 m _ [ _π_R_ 2_ _x__ 232__ _ 2_π _ R−_ _3_ π __4x__ __4_ ]_ R0 = _ _ _ 4123___ ππ_RR_ _43
∫ R πx(R 2− x 2) dx
51 a x_ = _38_ _ 0 _R_____ 23__ π_ _R _ 3_ _____ =
=
234 On line Full worked solutions are available in SolutionBank.
Answers
b Shape Mass Distance of CoM from A H × _H4 _ − h × _h4 _ = (H − h)x _ ⇒ x _ = _ 14_(H_(H_ 2__−−__hh_) 2_ ) = _ 41_ (H + h)
△ ABC 1 2ρ a 2 2a Distance from the vertex = H − _14 _ (H + h) = _41 _ (3H − h)
c _3H_H__+−_h_h_
△ BDC 4ρ a 2 2 a + _2 3_a_
Remainder 8ρ a 2 x_ 60 a Shape Mass Mass Position
ratio of CoM
Taking moments about A:
_58_r_
12ρa 2× 2a − 4ρa 2(_ 83_a_ )= 8ρa 2 x _ ⇒ _ 4_03_a _ = 8 x_ Hemisphere 4 πρ r 3 4r
⇒ x _ = _ 53_a_ _ h2_ + r
Cylinder πρ r 2 h h x _
c Taking moments about C: Toy π ρ r 2 (4r + h) 4r + h
Mg × 8a sin θ = Mg( _43_a_ cos θ − 4a sin θ)
Taking moments about O:
⇒ 12Mga sin θ = _43 _ Mga cos θ
⇒ tan θ = _91 _ 4r × _5 8_r_ + h( _h2_ + r)= (4r + h) x _
So CB makes an angle arctan( _91_ ) with the vertical. x _ _ _ 5__2 r__ _2_ 4 _+r__ _h+ 2__ 2_ _h _+ _r_h_ so d _ h _ 2_2 +__(2_h_r+_h _ _4+_r_5)_ r_ _2
56 a Shape
Mass Distance of ⇒ = =
CoM from O __
b h = √ 3 r
Circular disc π ρ a 2 0 61 a Shape Mass Distance of CoM
from AB
Hemispherical bowl 2πρ a 2 _21_ a
Closed container 3 πρ a 2 x_ Hemisphere M _ 38_r_
Taking moments about O: Cone m − _34_r_
0 + 2πρ a 2 × _ a2_ = 3πρ a 2 x _ ⇒ x _ = _ a3_ x _
b 56° (nearest degree) Toy m + M
57 θ = 53.1° (1 d.p.) Taking moments about AB:
(m + M) x _ = _ 3_M8__ r_ − _3_m4_ _r
58 a Shape Mass Distance of CoM from C ⇒ x _ = _ 3_8(M_(_M_−_ +_2_mm__)) r_
H 8M _ 38_a_
− _31_a6_ b tLNaeont αeαq==ui_3 alr_ir_nb gr⇒ilue mx _b e.⇒tw_ 13_ x e_ re .n OA and axis of cone
x _ r tan α
KM
S 9M So _ 3 _8(M_(_M_−_ +_2_mm__)) r_ . _ 31_ r
Taking moments about C:
8M × _3 8_a_ − M × _31 _a6_ = 9M x_ ⇒ 3a − _31 _a6_ = 9 x_ 9(M − 2m) . 8(M + m) ⇒ M . 26m
⇒ x _ = _ 41_56_a_ ÷ 9 = _51 _a6_ 62 a x_ = _ ∫∫0_ 0 r_ rr_ r x_x _2 d_ dx_x_ = _ [[ _ __3121___ rr_ xx_ _23_]] _r00r = _ __ 1132___ rr_ 43_ = _ 32_ r
∫ b _41_65
_ π_ 0_ _h x__( _r _2_ −_ _ _13__ 2 _π_ _rh _r _2_ 2_ x_ h_ _+ _ _ r__ _h2__ x 2__ _2_ )_d _x_ b α = 72° (nearest degree)
59 a x_
= 63 a x_ = _ _ 12__ ∫ ∫_0 π π_ ss_ii_nn _ x2_ x_d_ xd_ x_= _ _ 41__ ∫_0 π_∫ 1_ π_ −s_i_nc _o x_s _d 2_xx_ _ d_x_
00
_r _ 32_ h _ 0 h _ 2_ r_ h_2_ x_ 2_ + r _2 xh 32 dx
∫ = 2 − = _ _ 41__ [_x_[ −−__ c_12_ o_ _ss_i x n_] 2_ 0π _x_]_ 0π = _2 _4π __ = _ 8π_
x r
b 76° (nearest degree)
= _r _ 32_ h _ [_ x_22_ r _ 2 − _ 2_ r_3_ 2h_ x _ _3 + _r4_ 2 _h x_ 2 _4 ] h0 = _ r_ 32_ h _ × _ h_1 2_2 r_ _2 = _ h4_ 64 a _56_1_4a_ or 0.797a (3 s.f.)
b α = 70.6° (3 s.f.)
b Shape Mass Mass Position
ratio of CoM
Large cone _13_ πρ r 3 H H _ H4_ c β = 38.7° (3 s.f.)
Small cone _ 13_ πρ r 3 h h _ 4h_
Remainder _ 31_ πρ r 2(H − h) H–h x _
235
Answers
65 a Shape Mass Mass Distance b α = 74° (nearest degree)
ratio of CoM
from O 68 a Shape Mass Mass Position of
ratio CoM from O
Large _ 32_ πρ (6a) 3 27 _83_ × 6a Cylinder πρ r 2 h 3 + _h2_
hemisphere − _h4_
x_
Small _23_ πρ (2a) 3 1 _38_ × 2a Cone _13_ πρ (2r) 2 h 4
hemisphere x _
Remainder _23_ πρ (6 3 − 2 2) a 3 26 Tree π ρ r 2 h(1 + _43 _ ) 7
Taking moments about O: Taking moments about O:
26 x _=
⇒ x_ = 27 × (_ 83_ × 6a)− 1 × ( _38_ × 2a)= 60a 3 × _h2 _ − 4 × _h4 _ = 7 x_ ⇒ _ 32_h_ − h = 7 x_
_ 31_0_3a_ ⇒ x _ = _ 1h_4_
b Shape Mass Mass Distance b r = _14 h
ratio of CoM
from O
69 a Shape Mass Distance of CoM from O
_4_136 _ πρ a 3 52 _31_03_a_
Bowl B Hemisphere 2M h + _3 8_r_
Cylinder 2 4πρ a 3 9 9a Cylinder 3M _h2_
x _
Combined solid _4_83 8_ πρ a 3 61 y_ Combined solid 5M
Taking moments about O: Taking moments about O:
61 y _= _3 1_0_3a_ +
⇒ y _ = 52 × 9 × 9a = 120a + 81a = 210a 2M(h + _3 8_r_ ) + 3M × _h2 _ = 5M x_
_ 26_011_ a
⇒ _72_h_ + _34_r_ = 5 x _
c S will not topple. ⇒ x_ = _ 1_4_h_2_0+_ _3_r_
7 0 ab __672r_ r below O
66 a Shape Mass Mass Distance of b 40.6°
ratio CoM from O
Hemisphere _23_ πρ a 3 2 _ 38_ a c The toy will not topple.
Cone _31_ πρ a 3 1 _ 4a_
Remainder _ 31_ πρ a 3 1 x _ 71 a 1.07 kg (2 d.p.) b 0.57 m (2 d.p.)
Taking moments about O: 72 a 3.41 kg (2 d.p.) b 0.58 m (2 d.p.)
2 × _38 _ a − 1 × _a4 _ = 1 x_ ⇒ x _ = _ a2_
73 a e.g. the density of the rod increases as the distance
from A increases.
b Let N be the point of contact between the solid and b _14_1
the plane. sin α = _a_ a2 __ = _ 12_ c __1__91 2_ m
From △OGN 74 4 √ 3 m
⇒ α = _6π _ Challenge
1 e = _21 _
c _√1_3___ 2 a Reaction of ring on ball = _m _R_ v_ 2_
67 a Shape Mass Mass Distance of Using F = ma with frictional force:
ratio CoM from O _m_μ_R_ v _ _2 = –m_ dd_vt_
_ v1_ 2_ _dd_vt_ = _−R_ μ_
Cone _ 31_ πρ (3r) 2 h 3h _34_ h
Cylinder π ρ (4r) 2 r 16r h + _2r _ _−v_ 1_ = _−_R μ_ _t + c
Bollard πρ(16r + 3h) r 2 16r + 3h
x _ When t = 0, v = u ⇒ c = _− u_ 1_
Taking moments about O: _−v_ 1_ = _−_R μ_ _t − _u1_ ⇒ v = __ μ R__t _1+_ __ u1__ = _ R_+_u_Ru_ μ__t
3h × _3 4_h_ + 16r(h + _2r _ ) = (16r + 3h) x _ b _e _ π_22 _ −_0_ 1_ s or 0.191 s (3 s.f.)
⇒ _ 9_ 4h_ _2 + 16rh + 8r 2= (16r + 3h) x _
⇒ x_ = _ 3_2_ r_4 2_(_+1_6_6r_4 _+r_h _3_+h_)_9 _ h_ _2
3 a T1 = 4Mg, T2 = T3 = 0 b 6.72°
4 4 cm
236 On line Full worked solutions are available in SolutionBank.
Answers
CHAPTER 4 b 2.5 , t < 6 c (_ 1_34 2_ + 72 ln(2)) m
Prior knowledge check c − c_o__s5_ 5π_π _x_ + c Challenge
_dd_vt_ = _k6_t_0 _2
1 a _3_ (_2__−4_ _3_x_)_ 2 + c b _4_3e__ 3_x + c Integrating gives:
2 y = 3e _ 3_(xx−_+1_2 _)
3 ln _43_ v = − _6k_0t_ + c
Solving for k and c using v = 0 when t = 2, and v = 9 when
Exercise 4A
t = 5 gives v = 15 − _3 _t0 _ . As t > 2, _3 _t0 _ . 0 so v , 15 and the
1 (16 − 12e−0.25t) m s−1 speed of the car never reaches 15 m s−1
2 v = t sin t, x = ∫ t sin t dt. Using integration by parts and the Exercise 4B
initial condition x = 0 at t = 0, we get x = sin t − t cos t. 1 v 2 = _x 2_2 + 4x + 25
At t = _2π _ , x = − _2π _ cos _2π_ + sin _2π _ = 0 + 1 = 1. 2 v = ± √ _(8_0__−_4_x_2_)
3 _51
3 2ln 3 m 4 16 m
5 a _ 1_265 _ ___
4 11.5 m__(3 s.f.) b x = _43 sin 3t c t = _3π _ b ±4√14 m s−1 as the particle will pass through this
5 a 6√ 2 m s−2
6 v = _23 _ − _ 2__+3_ _t _2 position in both directions.
7 a v m s–1
4.5 6 4 36 cos _3x _
4
7 a v 2 = 52 −
①
② ___
b 2√22 m s−1 (≈ 9.38 m s−1)
8 4.72 m s−1 (3 s.f.), in the direction of x increasing.
O 3 6 t(s) 9 a 1.95 m s−1 (3 s.f.) b 26._8_ (3 s.f.)
10 a v = x + _2x _ b 2√2 m s−1
b (27 − 3 ln 2) m __ 11 10 m b x = (t + 2)3
8 a 4 m s−1 12 a v = 3x _23
b (π − 2√2 ) m
9 a v = 40 − 20 e0.2t m s−1 b (200 ln 2 − 100) m
( )b 3200 ln _ 8_08_0_+_ _t
10 a c = 80, d = 1 Challenge
11 a t = ln 2.5, ln 3 ( )b 18 − _1 _5_(_l2n_ _ 3_) _2 m v 2 = _15_ (25x − _x 2_ 2_ ) + _1_62_3 _
12 a t = 3 b (12 + 3 ln 12) m Exercise 4C
13 a t = 1, t−=_ 52_ _ t23_2 b a = 25 m s−2 c s = _3_59_41 7_ m 1 a v = ln (t + 1) b v = ln (11)
d x = t3 2t = t(t2 − _ 25_ t + 2)
+ 2 0.137 s (3 d.p.)
For t . 0, when t2 − _ 52_ t + 2 = 0 3 a 2.04 s (3 s.f.) b 9.80 m (3s.f.)
The discriminant of the quadratic equation is
4 a _dd_vt_ = g − 2v, so _( _g_d−__v2_ _v_) = dt.
_24_5 − (4 × 2) , 0.
The equation has no real roots. Therefore, P never Integrating both sides and using initial conditions:
returns to the origin for any t . 0.
− _12_ ln ( g − 2v) = t − _ 12_ lng.
14 a velocity (m s–1) ⇒ 1 − _2 g_v_ = e −2t ⇒ g(1 − e −2t)= 2v.
b _4g_ (1 − e −4) m
12
5 a v = 12 − _1 e _2 _4t__ b 12 m s−1
6 6 1.16 m (3 s.f.)
7 _21_ ln( _u_ 2_k+_ _k_ ) m
O 2.5 5 6 12 time (s) ∫ ∫ 8 a _dd_vt_ = − (a2 + v2) ⇒ _ a_2_d+_v_ v__2 = − dt
so _ 1a_ arctan _av_ = − t + c
Using t = 0, v = U and t = T, v = _ 21_ U gives
T = _1a_ [arctan _Ua_ − arctan _2U_a_ ]
–12.5 b _12_ ln _ 44_(_aa _ 22_++__uu_ _ 22 _) m
237
Answers
∫ ∫ 9 _dd_vt_ = _1_6_0_60_4_−v_ _v_ _2 ⇒ _1_6_0_60_4_v−_ _v_ _2 dv = 1dt 17 a 6.04 m s−1 (3 s.f.) b 2.56 (3 s.f.)
⇒ − 32 ln(1600 − v 2) = t + c
T = − 32 ln(1600 − 2 0 2)− (− 32 ln(1600 − 1 0 2)) 18 a 7 m s−1 b x = 7.56 m (3 s.f.)
= − 32 ln 1200 + 32 ln 1500 = 32 ln _ 11_25_0000_ = 32 ln _45_ b x = _32 (e2t − 1)
___v_d_v_ __ = − kdx
(U 2 + v2)
∫ ∫19 a v = 2x + 3
20 v _dd_xv_ = –k(U 2 + v2) ⇒
Mixed exercise 4 so _12_ ln (U 2+ v 2) = − kx + c _2 1_k_ ln _58_
using distance = 0 when v
1 a 8(e0.5t − 1) b 6 m s−2 when v = _ 12_ U = U gives distance as
2 a v = 18 − 10e −t 2 b 18 m s−1 21 209 m (3 s.f.)
3 6 Challenge
4 a v = 10 − _2 _t5_+_0_ 5_ 11 776 km
b (100 − 25 ln 5) m ≈ 59.8 m
5 a _2π_ m s−1 CHAPTER 5 b 675 kJ
b From part a, v = _ 41_ sin 2t + _ 12_ t b 1.68 m
Prior knowledge check
x = ∫vdt = − _81_ cos 2t + _ 41_ t2 + c
Using inital conditions gives c = _ 18_ , hence 1 a 11170 N
x = − _18_ cos 2t + _ 14_ t2 + _ 18_ 2 a 33.9 m s−1
Substituting t = _ 4π_ gives: 3 13 _89_ N
Exercise 5A
x = − _81_ cos _2π_ + _ 41_ (_ 4π_ ) 2 + _ 81_ = _ 61_4 ( π2 + 8) 1 a 9.09 m s−1 (3 s.f.) b 1.41 m s−1 (3 s.f.)
6 a 2.5 m s−2 in the direction of x increasing.
c P first comes to rest when t = π.
d 14.2 m (3 s.f.) e OP = 20 m
b 8 e−1 m s−2 in the direction of x decreasing. 2 a 10
( )c _ 53_6 − 8 e−2 m ≈ 17.6 m (3 s.f.) b The van moves 10.6 m in the first 4 seconds (3 s.f.)
7 a v = 2t + ln(t + 1) b (2 + 3 ln 3) m 3 a Maximum speed occurs when acceleration is zero,
8 a T = 3√ _1_0_ s i,e. when force is zero. ⇒ _ 61_ ( 15 − x)= 0 ⇒ x = 15
b velocity b 6.85 m s−1
4 a 6.79 m s−1 (3 s.f.) b 8.23 m s−1 (3 s.f.)
13 c 8.10 N
8 5 x = 0.677 (3 s.f.)
6 a 0.25 _dd_vt_ = − _( t__+8__1 _) _2 ⇒ ∫ 0.25 dv= − ∫ _(t__+8__1 _) _2 dt
⇒ 0.25v = _( t__+8_ _1_) + c ⇒ v = _( t_3_+_2_1 _) + d where d = 4c
When t = 0, v = 10: 10 = _3 1_2 + d ⇒ d = − 22
⇒ v = 2( _(t_1_+_6_1 _) − 11)
b x = 32ln6 − 132
O2 4 5 3 10 time 7 k = 66
3
8 _14 ln 4 s = 0.347 s (3 d.p.)
9 2 ln 2 m = 1.39 m (3 s.f.)
10 a ln 2.5 s = 0.916 s (3 d.p.)
b (12 − 8 ln 2.5) m = 4.67 m (3 s.f.)
( )11 2__1k_ _g ln _ m__+m_k_ _u_2
c _73_ , t , 5 m _dd_vt_ ⇒ − _m k_ ∫ _a_ 2_1+__ v_ _2 dv
d 59.2 m (3 s.f.) 12 a R(→): −k(a 2+ v
b _92_071_ m 2) = ∫ 1 dt =
9 a a = 6t − 20
⇒ t = A − _a m_k_ arctan( _av_ )
c The discriminant of the equation When t = 0, v = U: A = _a m_k_ arctan( _ Ua_ )
x = t(t2 − 10t + 32t) = 0 is less than 0 for t . 0 and
thus x can never be 0 for t . 0.
10 _ 2_k1_U _ s When t = T, v = _U2 _ : T = _ am_k_ ( arctan( _ Ua_ ) − arctan( _ 2U_a_ ) )
11 a_t_=___ 32__ _o_r_t__= 2 b _22_478_ m ( ) ( ) b 2_m_k_ ln 4_4_aa_2_2+_+_4_UU_ 2 _2
√ ( )12 20 − _1 x_2_ 13 a 12( 1 − e− _ 4t ) b 12 m s−1
14 a m _dd_vt_ = − mg − _m _k_g_ v_ ⇒ _ dd_vt_ = − g(1 + _kv _ )
13 x = 8
14 x = _51 b 5 ∫ T − g dt ∫ 0 _k_+k__ v_ dv − g [t]0 T k [ln(k v)] U0
15 a v 2 = 16 + 6x − x2 0 U
16 a v 2 =___5 _3x_ _3 − _ x4_4 + _ 2_53_0_ 0_ ⇒ = ⇒ = +
b _5_0_3√_ _3_ m s−1
⇒ − gT = k(ln k − ln(k + U)) ⇒ T = _kg _ ln(_ k__+k_ _U_ )
238 On line Full worked solutions are available in SolutionBank.
Answers
b H = _ kg__ (_U___−__k_ l_n( _ k__+k_ _U_ ) ) √ __ b 4, _22_π_ = π s c 8 m s−1 d _6π_ e _1π_2_
14 a x = 3 sin(4t + _12 _ ) ⇒ x˙ = 12 cos( 4t + _21 _ )
√15 a v = _ 1_−__ke_ −_2_g k_x b _1k_ + _21 _ ) ⇒
⇒ x¨ = − 48 sin(4t x¨ = 16x ∴ S.H.M.
c This model has the particle rapidly approaching b Amplitude Period = _2π _ sd 0.660 (3 s.f.)
terminal velocity, within two metres of release the c x = 1.44 (3 = 3 m,
exponential term is of the order 10−19. s.f.)
Challenge 15 a 11.51 a.m. (nearest minute)
b b b 8.39 p.m. (nearest minute)
∫ a 3 x 2− x _31_ dx [ x 3 _ 43_ x ]_ 34_ 16 P takes 0.823 s to travel directly from B to A (3 s.f.)
Work done = = −
a
a Challenge
x¨ = − ω 2 x v 2= ω 2 (a 2− x 2 )
= b 3 − _ 34_ b −_ 34_ a 3 + _ 43_ a _43_ v 1 2= ω 2 (a 2 − x 1 2 ) (1)
v 2 2= ω 2 (a 2 − x 2 2 ) (2)
Hence work done is independent of the initial velocity. (2) − (1): v 2 2− v 1 2= ω 2( a 2 − x 2 2) − ω 2 (a 2 − x 1 2 )
v 2 2− v 1 2= ω 2( a 2 − x 2 2− a 2 + x 1 2 )
b 208 J (3 s.f.) Rearranging gives ω 2= _ xv_ 21_ 22_ −−__vx_ _12 _22 so ω = ( _xv_ 12_ 22_ −−__vx_ _12 _22 ) _ 21_
Exercise 5B T = _2 ω_π_ = 2π( _ xv _ 21_ 22_ −−__vx_ _12 _22 ) _ 21_
1 F = _d k_ 2_ , where d = distance from centre
distance (x − R) above surface
⇒ distance x from centre ⇒ F = _x k_ 2_
On surface F = mg, x = R ⇒ mg = _x k_ 2_ ⇒ k = mg R 2
∴ Magnitude of the gravitational force is _m _xg__ 2R_ _2
Exercise 5D
2 For a particle of mass m, distance x from the centre of
the earth. 1 a F = ma ⇒ −T = =0._λ5 _lx x_¨ =
Hooke’s law: T _60_0._6x_ =
F = ma ⇒ _ xk_ 2_ = mA 100x
On the surface of the earth, x = R, A = g b _⇒ 1π_0_ −√ 12__0 s0 x = 0.5x¨ ⇒ x ¨ = −200x ∴ S.H.M.
0.3 m c 4.24 m s−1
⇒ mg = _R k_ 2_ ⇒ k = mg R 2 ⇒ mA = _m _xg__ 2R _ _2 ⇒ A = _g x_ R _2 _ 2 (3 s.f.)
_____ 2 a F = ma ⇒ −T = 0.8x¨
Hooke’s law: T = _λ _lx _ = _21_0._6x_ = _ 2_25_ x_
3 √ 2g R ⇒ − _2_25_ x_ = 0.8x¨ ⇒ x¨ = − _1_28_5_ x_ ∴ S.H.M.
__________________
√ 4 __ U__ 2__X___+__UX _ 2_+R _ _R−_ _ 2_g_ _R_X_ b 3.21 s (3 s.f.)
√ 5 _ 7_g5___R _ _
√ 6 2 _ g3_R_ 3 a F = ma ⇒ −T = 0.4x¨
_x k_ 2_ when x = R, F = mg somg = _R k_ 2_ ⇒ k = mg R 2 Hooke’s law: T = _λ _lx _ = _21_4._2x_ = 20x
7 a F = ⇒ − 20x = 0.4x¨ ⇒ x¨ = −50x ∴ S.H.M.
_m _xg__ 2R _ _2 b 0.489 s (3 s.f.) c 1.84 m (3 s.f.)
gravitational force on S = 4 a F = ma ⇒ −T = 0.8x¨
____ Hooke’s law: T = _λ _lx _ = _81_0._2x_
√ b speed = _7_g2_R _
⇒ − _81_0._2x_ = 0.8x ¨ ⇒ x ¨ = − _11_0_.02_x_ ∴ S.H.M.
Challenge
a 5.98 × 1024 kg b 5500 kg m−3 5 a x = 0.5 sin 10t b 50 m s−2
Exercise 5C 6 a 0.5 m b 2.11 m s−1 (3 s.f.)
1 a _2π_ s b 1.83 m s−1 (3 s.f.) c i 1.49 s ii 0.3 m
2 a 1 m b 1.36 m s−1 (3 s.f.)
7 a 0.351s (3 s.f.) b 2.56 J
3 a 5 m b π s 8 a 2.45 m s−1 (3 s.f.) b 1.25 (3 s.f.)
4 _34 m s−1 9 a F = ma ⇒ T B − T A = 0.4x ¨
5 17.9 m s−1 (3 s.f.) Hooke’s law: T = _λ _lx _
AP: extension = 0.8 + x
6 a 1.26 m (3 s.f.) b x = 1.26 sin 4t
∴ T A = _ 1_2_(_01_._.8 2__ +__x_) = 10(0.8 + x)
7 a 0.133 m (3 s.f.) b 0.0141 m (3 s.f.)
8 a 1.37 m (3 s.f.) b 0.684 s (3 s.f.) BP: extension = 0.8 − x
9 a 1.00 m s−1 (3 s.f.) b 0.922 m s−1 (3 s.f.)
10 9.25 J (3 s.f.) b _32 s to fall 0.6 m ∴ T B = _ 1_2_(_01_._.8 2__ −__x_) = 10(0.8 − x)
11 a 1.26 m s−1 (3 s.f.) ∴ 10(0.8 − x) − 10(0.8 + x) = 0.4x¨
12 0.0738 s (3 s.f.)
−20x = 0.4x¨ ⇒ x¨ = −50x ∴ S.H.M.
13 a x = 4 sin 2t ⇒ x˙ = 8 cos 2t ⇒ x¨ = −16 sin 2t ⇒ x¨ = 4x b 3.6 J
∴ S.H.M.
239
Answers
10 a F = ma ⇒ T B − T A = mx¨ Hooke’s law: T = _λ _lx _ = _ 3_1_._3_6_(0_x_. 4_+_ 0__.0__5_)
Hooke's law: T = _λ _lx _ 0.4g − _ 3 _1_._3_6_(0_x_. 4_+_ 0__.0__5_) = 0.4x¨
AP: extension = 1.5l + x ⇒ TA = _ 3_m__g_(1__.l 5 _l__+__x_) x ¨ = − _30_1_. .4_3_ 26_ x ∴ S.H.M.
PB: extension = 1.5l − x ⇒ T B = _ 3_m__g_(1__.l 5 _l__−__x_) c 0.449 s, 0.07 m d 0.98 m s−1
⇒ _ 3_m__g_(1__.l 5 _l__+__x_) − _ 3_m__g_(1__.l 5 _l__−__x_) = mx¨ e 0.156 s to rise 11 cm (3 s.f.)
6 a 0.416 s (3 s.f.) b 0.0574 s (3 s.f.)
− _ _6__m___l_ g_x_ = mx¨ ⇒ x¨ = − _6_lg _ x ∴ S.H.M. 7 a 0.221 m b 0.517 s (3 s.f.)
√ b 2π _ 6_lg _ c 1.5l 8 a 1.69 m (3 s.f.)
b T A = _ 1_5_(0_e_.6_+_ _x_) , T B = _ 1_5_(_1_.6__0−_. 6_( _e_+__x_)_)
_____ ____
F = ma
1.5g + _ 1 _5_(_1_.6__0−_. 6_( _e_+__x_)_) − _1_5_(0_e_.6_+_ _x_) = 1.5x¨
d √ 12gl (or 2√ 3gl )
11 a F = ma ⇒ TB − T A = 0.5x ¨ 1.5g + 40 − 25(e + x) − 25(e + x) = 1.5x¨
Hooke's law: T = _λ _lx _ 1.5g + 40 − 50e − 50x = 1.5x¨
AP: extension = 1 + x ∴ T A = _ 1_5_(_11__ +__x_) = 15(1 + x) From part a, 50e = 1.5g + 40
BP: extension = 1.5 − x ∴ 1.5x¨ = − 50x ⇒ x¨ = − _1_30 0_ x ∴ S.H.M.
c 0.398 (3 s.f.)
∴ TB = _ 1_5_(_11_._.5 5__ −__x_) = 10(1.5 − x)
9 a 12.5 m s−1 (3 s.f.) b 10.4 m (3 s.f.)
c 1.56 s (3 s.f.)
∴ 10(1.5 − x) − 15(1 + x) = 0.5x¨ √ √ √ Challenge ___ _____ ___
− 25x = 0.5x¨ ⇒ x¨ = − 50x ∴ S.H.M.
_2 ω_π_ = _ 2___π___ = _ 5π_ √ _2_ Particle P: T = 2π _m_λ_l = 2π _5_mm__lg _ = 2π _5_lg _
period = √ 50
__________ ________
b 1 m √ √ Particle Q: 3T = 2π _l(_m_5_+_m_kg_m __) = 2π _l(_1_5_+g_ _k_)
Exercise 5E ___ ________
√ √ 6π _l(_1_5_+g_ _k_) so _ _ _____
1 a 1.64 m (3 s.f.) _5_lg _ = 2π
b F = ma ⇒ 0.75g − T = 0.75x¨ 3 √ l = 2 √ l × √ 1 + k
Hooke’s law: T = _λ _lx _ = _8_0_(1_x_.5_+_ _e_) squaring both sides 9 = 4 + 4k so k = _ 45_
∴ 0.75g − _8 _0_(1_x_.5_+_ _e_) = 0.75x¨
from part a, 0.75g = _8 1_0._5e_ Mixed exercise 5
∴ 0.75x ¨ = − _18_._05_ x ⇒ x ¨ = − _1_.5__8×_0_0x_ ._7_5_ ∴ S.H.M.
c 0.745 s (3 s.f.) 1 a 108
d 0.296 m (3 s.f.)
2 a 0.049 m (3 s.f.) b 11.8 m (3 s.f.)
b 0.444 s (3 s.f.) 2 a x˙ = _ −_32_ 0_ (3t + 4) _12 + _ 73_6_
c 2.83 m s−1 (3 s.f.)
3 a 1.5 m s−1 b 18.7 m from O (3 s.f.)
b In equilibrium:
3 _ 3_k1_U _
( ) 4 _kg_ _12 ( 1 − e−2kD ) _12
5 56.1 m (3 s.f.)
R(↑): T = 2mg 6 d = _ 1_10_ b_ ln (1 + bU2) ma ⇒ − g( _c_ 2_c+_ 2_ v_ _2 ) = v _dd_xv_
Hooke’s Law: T = _λ _lx _ = _1λ_.e_5_ ∴ _ 1λ_.e_5_ = 2g 7 a R(↑): − mg − _m _c_g _ 2v_ _2 =
When oscillating: F = ma ⇒ 2g − T = 2x¨ ∫⇒∫ g dx = − c 2 _c_ 2_+v__ v_ _2 dv ⇒ gx = A − _c 2_ 2_ ln( c 2+ v 2)
Hooke’s Law: T = _λ _(e1__.+5__ x_) At x = 0, v = V: 0 = A − _c 2_ 2_ ln( c 2+ V 2)
∴ 2g − _λ _(e1__.+5__ x_) = 2x ¨ ⇒ A = _c 2_ 2_ ln( c 2+ V 2)
Hence gx = _c 2_ 2_ ln( c 2+ V 2) − _ c2_ 2_ ln( c 2+ v 2)
⇒ _ 1λ_.e_5_ − _ λ_(e1__.+5__ x_) = 2x ¨ ⇒ − _1λ_.x_5_ = 2x ¨ = _ c2_ 2_ ln( _cc_ 22_++__Vv_ _22 )
∴ x¨ = − _3λ_ x as λ . 0, this is S.H.M. ⇒ x = _2c _ g2_ ln( _ cc _ 22_++__Vv_ _22 )
c 48 d 0.375 m
4 a 0.138 s (3 s.f.) b 1.61 m s−1 At the greatest height v = 0: x = _2c _ g2_ ln( _ c_ 2_c+_ 2_ V_ _2 )
5 a 31.4 N (3 s.f.)
b F = ma ⇒ 0.4g − T = 0.4x¨ = _2c_ g2_ ln( 1 + _Vc _ 22_ )
240 On line Full worked solutions are available in SolutionBank.
Answers
b − g _ c_2 _c+_2 _v_2 = _ dd_vt_ 10 a P = Fv ⇒ F = _Pv _
∫⇒ ∫ gd t = −c2 ___1_ __ dv R(→): F − _m __k3_ v _ _2 = ma ⇒ _ Pv_ − _ m__k3_ v _ _2 = mv _dd_xv_
c2 + v2 3P − mkv 3= 3mv 2 _dd_xv_ ⇒ 3mv 2 _dd_xv_ = 3P − mkv 3
⇒ gt = −c2 arctan ( _vc_ ) + C ( ) b _ 1k_ ln _ 8 _(3__P_2−_1_ Pm_ _k_v_3_)
⇒ t = − _gc_ arctan ( _vc_ ) + D
When t = 0, v = V, so D = _ gc_ arctan ( _ Vc_ ) 11 a F = _x k_ 2_ when x = R, F = mg
So t = _ gc_ arctan ( _Vc_ ) − _ gc_ arctan ( _ vc_ ) ∴__ _xk__ 2_ _ = mg, k = mg R 2
____
√ √ b _ 8_R5_ g_ or 2 _ 2_R5_ g_
v = 0 ⇒ t = ( _gc_ ) arctan ( _ Vc_ )
12 a _54_0π_ (or 0.251 (3 s.f.))
8 Let the mass of the particle be m. b 0.203 m s−1 (3 s.f.)
Let the resistance be kv 2, where k is a constant of
proportionality. c 0.318 m (3 s.f.)
If U is the speed for which the resistance is equal to the
weight of the particle then 13 a x = 2.5
k U 2 = mg ⇒ k = _mU _ g2_ . Hence the resistance is _m _U_g_ 2v_ _2
R(↑): F = ma ⇒− mg − _m _U_g_ 2v_ _2 = ma⇒− _g_(U__ 2U__+ 2_ _v_ 2_) = _dd_vt_ b v2 = 25x − 5x2 + 32.75
14 a x = 3 sin( _4π_ t) ⇒ x˙ = _ 34_π_ cos( _4π_ t)
⇒ x¨ = − 3(_ 4π_ ) 2 sin( _4π_ t)
∫⇒∫ g dt = − U 2 _U _ 2__1+_ _v_ _2 dv ⇒ gt = A − U 2 × _ U1_ arctan( _ Uv_ ) ⇒ x ¨ = − (_ 4π_ ) 2 x ∴ S.H.M.
3 m
b Amplitude =
When t = 0, v = U: 0 = A − U arctan 1 Period = 8 s
⇒ A = U arctan 1 = _π 4_U_ c _34_π_ m s−1 (or 2.36 m s−1 (3 s.f.))
Hence gt = _π 4_U_ − U arctan( _ Uv_ ) ⇒ t = _π4 _Ug_ − _Ug_ arctan( _Uv_ ) d 0.405 s (3 s.f.)
15 a 1.54 s
Let the time of ascent be T. b 0.116 m
When t = T, v = 0: T = _π4 _Ug_ − _Ug_ arctan 0 = _π4_Ug_ c 1.03 s
b Equation can be written as − _g _(U__ 2U__+ 2_ _v_ 2_) = v _dd_xv_ 16 a F = ma ⇒ −T = 0.6x¨
Hooke’s law: T = _λ _lx _ = _22_5._5x_ = 10x
xs¨. f=.)− _01_._06_ x
∫⇒ ∫ g dx = −U 2 _ U_ 2__v+_ _v_ _2 dx ⇒gx = B − _U 2_ _2 ln(U 2+ v 2) ⇒ −10x = 0.6x¨ ⇒ ∴ S.H.M.
b Period is 1.54 s (3
When x_Ug 2_x= _2 l=0n,(_U 22_v U_2 =l 2n)U (2: U0 2=) −B_ U−2_ _2 _ U l2_n _2 ( Uln 2(2+U v 2 )2 ) Amplitude is (4 − 2.5) m = 1.5 m
⇒B=
Hence c P takes 0.468 s to move 2 m from B (3 s.f.)
17 a F = ma ⇒ T B − T A = 0.4x ¨
HT Ao o=k_ 2e_’.s5__(l0a_0.w_. 46_: _+T__x=_) ,_λ T_lx _ B = _ 2_.5__(0_0._. 46_ _−__x_)
⇒x = _U2 _g _2 ln( _ U_ 22__ U+__ 2v_ _2 )
Let the total distance ascended be H. ∴ _ 2_.5__(0_0._. 46_ _−__x_) − _ 2_.5__(0_0._. 46_ _+__x_) = 0.4x¨
When h = H, v = 0: H = _U2 _g _2 ln( _2_U U_ _2 2 _ ) = _ U2_g _2 ln 2 ⇒ −5x = 0.4x¨ ⇒ x ¨ = − _0_.6__×5__ 0__.4_ x ∴ S.H.M.
9 a R(→): − mk(V0 2 + 2v 2) = ma ⇒ −k(V 0 2 + 2v 2) = v _dd_xv_ b 1.38 s (3 s.f.)
∫⇒∫ kdx = − _V_ 02_ _+v__ 2_v_ 2_ dv ⇒ kx = A − _14 _ ln( V 02 + 2v 2) c 0.229 s to reach D (3 s.f.)
At x = 0, v = V0: 0 = A − _14 _ ln( V 02 + 2V 02 ) 18 a Spring AP extension: 2.4 m
⇒ A = _14 _ ln( 3V 02 )
Hence kx = _41 _ ln( 3V 02 ) − _ 14_ ln( V 02 + 2v 2) Spring PB extension: 1.6 m
⇒ x = _4 1_k_ ln( _V _ 02_ 3_+V__ 202_ v_ 2_ )
b F = ma
When v = _12 _ V 0: x = _4 1_k_ ln( _ V_ 02_ 3_+V___ 1202__ V_ 0_2 ) T B − T A = 0.4x ¨
T A = _ 2_0_(_2_.5_4 _ _+__x_)
T B = _ 1_8_(_1_.3_6 _ _−__x_)
6(1.6 − x) − 4(2.4 + x) = 0.4x..
b 0.24 (2 d.p.) = _ 41_k_ ln( _ _323__ VV_ 00_22 ) = _ l4n__k 2_ , t = _0 k_.V_2_ 04_ −10x = 0.4x¨
x ¨ = −25x
∴ S.H.M.
c 4 m s−1
241
Answers
19 a λ = 3 g T = 0.5x¨ , T = _λ _lx _ = _ 3_g_(_01_._2. 2__+ __x_) b 162 m
b 0.5g − 6 a v (m s21)
⇒ 5g − _ 3 _g_(_01_._2. 2__+ __x_) = 0.5x¨ ∴ x¨ = _0_.5_−_3×_g_x1_ _.2_ = −5gx 15
∴ S.H.M. 7.5
c 0.898 s (3 s.f.) 2
O 45
d 2.01 m s−1 (3 s.f.)
e 0.406 m (3 s.f.) 10 t (s)
20 a F = ma ⇒ T B − T A = mx¨ b 2 , t , 5
Hooke’s law: T = _λ _lx _
T A = _ 5_m__g_2(l_l _−__x_) , T B = _ 5_m__g_2(l_l _+__x_) ∫ 4
∴ _ 5_m__g_2(l_l _−__x_) − _5_m__g_2(l_l _+__x_) = mx¨
√ ⇒ −_ _5___m_l_ g_x_ = − _5_l_g _ x c 3 t(t – 4) dt = [t3 – 12t]40 = –32
mx¨ ⇒ x¨ = c ∴ S.H.M. 0
b 2π _5_lg _ _ 41 √_ 5_g__l ∫5
3 t(t – 4) dt = [t3 – 12t]54 = 7
4
So distance travelled in the interval is 32 + 7 = 39 m
d 6.98 (3 s.f.)
Challenge 7 a v = (p + qt) −1
a = − q (p + qt) −2= − q v 2
b p = _2 1_0 , q = _1 _63_0 0_
ma = − _ (_mR__M+__Gx_) _ 2 so a = − _ (_R_M_+_G_x _)_ 2 ( ) c x = _1 _63_0 0_ ln 1 + _8 3_0 t
a v _dd_xv_ = − _(R__M+__Gx_ _) _2 maximum height, H, is reached 8 a = _d d_x_ ( _ 21_ v 2)= 4x
when v = 0 _ 12_ v 2 = ∫4x dx = 2x 2 + A
v 2 = 4x 2 + B
0 0 H _( R__M+__Gx_ _) _2 dx ⇒ 0 [ _ (R_M_+_G_x_ _) ] 0H
v dv =
u u
∫ ∫ = − [ _ v2_ 2_ ] At x = 2, v = 4
− _u2_ 2_ = _(R_M_+_G_H_ _) − _ M_R_G _ = MG(_ R__+1_ _H_ − _ R1_ ) = MG( _ RR__(−R__R+__−H__H)_ ) 16 = 16 + B ⇒ B = 0
v 2= 4x 2
− _u2_ 2_ = _(R_M_+_G_H_ _) − _ M_R_G _ = MG( _R__+1_ _H_ − _ R1_ ) = MG( _ RR__(−R__R+__−H__H)_ ) 9 a = _d d_x_ ( _ 21_ v 2)= 1 − 4x −2
_21_ v 2 = ∫(1 − 4x −2) dx
= x+ _4x _ + A
MG( _R_(_R_H+_ _H__) ) = _ u2_ 2_ ⇒ 2MGH = u 2(R 2 + RH)
v 2= 2x + _8x _ + B, where B = 2A
2MGH − RH u 2= u 2 R 2 H(2MG − R u 2) = u 2 R 2 __
At x = 1, v = 3 √ 2
H = _2 _M__GR_2_− u_2_R _ u__2 = _ _2__MR___R_ G__ u_ −_2 _u__2
18 = 2 + 8 + B ⇒ B = 8
v 2= 2x + _8x _ + 8
b H → ∞ as u 2 → _ 2_MR__ G_ ___________________________ At x = _32 _
_____ v 2= 2 × _23 _ + 8 × _23 _ + 8 = _4 39_
√ √ Escape velocity u = _2_MR__ G_ = _ 2_×__5__.9_8__× _6_.14_0_×_2 _41 _×0__ 66 _ ._7_×__ 1__0_ 1_1
10 8.76 m s−1 (3 s.f.)
= 1.12 × 104 m s−1 ( )11 a v2 = 4k2 1 − _x _+2__ 1_
_________
Review exercise 2
√ b v = 2k 1 − _x _+2__ 1_
1 a = _dd _vt_ = e 2t _1 _1+__ x_ is _x _+1__ 1_ ,
As x is positive, positive and 1 − 1
So v , 2k
v = ∫ e 2t dt = _21 _ e 2t + A
12 a At the maximum value of v, a = 0
When t = 0, v = 0
0 = _12 _ + A ⇒ A = − _12 _ a = _1 1_2 (30 − x)= 0
2 v= _12 _ ( e2t − 1) b 11.2 m s−1 (3 s.f.) __c 13 b x = 30 − _1 x_22_ + 25
a v = 13 − 3e − _16 t b 4(π − √ 2 ) m v 2 = 5x
b _21 ( 1 − ln 2) m
3 a v = 8 − 4 cos _21 t 13 a 0.9 m
4 a v = 2e−2t − 1
b a = p – qx
5 a a = _dd _vt_ = 3(t + 4) − _12_ At x = 0, a = 20
v = − 3 ∫ (t + 4) −_ 21_ dt = A − 6(t + 4) _12_ 20 = p – 0 ⇒ p = 20
When t = 0, v = 18 a = 20 – qx
18 = A − 6 ×__2__⇒_ A = 30 At x = 2, a = 12
v = 30 − 6√ t + 4
12 = 10 − 2q ⇒ q = 4
c 1 m
242 On line Full worked solutions are available in SolutionBank.
Answers
14 a v 2 = 60 − _2 _x7_+_2_ 1_ 25 a F = ma
b 0.1 m
_4_8_0__0_0_ = 800a
15 a a = _d d_x_ ( _21_ v 2) = _ 2_kx_ _2 + _ 4_kc_ 2_ ( t + 2) 2
∫_ 12_ v 2 = (_ 2_kx_ _2 + _ 4_kc_ _2 ) dx = − _2 k_x_ + _ 4k_c_x _2 + A a = 60(t + 2) −2
v = ∫ 60 (t + 2) −2 dt
− _kx_ + _ 2k_c_x _2 + = A − _t _6+_0_2 _
When t = 0 v = 0
v 2 = B 3A0−−_6 2__0t _ 6+⇒_0_2 _A =
0 = 30
__ v =
√ At x = 2a, v = − _ck_
_ ck_ = − _2k_c_ + _ 22_kc_ c2_ + B ⇒ B = _2 k_c_ As t → ∞ , _t _6+_0_2 _ → 0, v → 30
At x = c As t increases, the car approached a limiting speed
of 30 m s−2
v 2 = − _ck_ + _ 2k_c_c _2 + _ 2k_c_ = 0
b (180 − 120 ln 2) m
The particle comes to instantaneous rest where 26 a 6 b 17 m
x = c. 27 a _g_(_e _3 3_et__ −3_t _1__ ) m s−1 b _9g___ (5 + e −6) m
b a = _d d_x_ ( _ 12_ v 2) = _ 2_kx_ _2 − _ 4_kc_ _2 28 gk 2 ln(_ g_k_g_+k_ _U_ ) + Uk
∫ _12_ v 2 = ( _2_kx_ _2 − _ 4_kc_ _2 ) d x = − _2 k_x_ − _ 4k_c_x _2 + C 29 a As F ∝ _ x1_ 2_ , F ∝ − _xk_ 2_
v 2 = − _kx_ − _ 2k_c_x _2 + D At x = R, F = − mg
− mg = − _R k_ 2_ ⇒ k = mg R 2
At x = c, v = 0
0 = − _ck _ − _ 2k_c_c _2 + D ⇒ D = _23 _ck_ F = − _m _xg__ 2R _ _2
Hence ___
√ b _ g6_R_
v 2 = − _kx_ − _ 2k_c_x _2 + _ 23_ck_ = _ −_ 2_k_c_ 2_ _−2 _k_c _x 2x_ 2_ +_ _3_k_c_x_
30 a F = ma
= − _2_c_k_ 2 _x_ (x − c)(x − 2c) − _cx_m 2_ = ma
a = − c x −2
When v = 0, (x – c)(x – 2c) = 0 _12_ v 2= − ∫ c x −2dx = _xc _ + A
v 2 = _ 2x_c_ + B
x = c, 2c
After leaving B (x = c), the particle next comes to When x = R, v = U
rest at A (x = 2c).
16 a T = _56 _ b OA = 10ln 2 m U 2 = _ 2R_c_ + B ⇒ B = U 2 − _ 2R_c_
17 a k ln( _k__+k_ _U_ ) + U m b ln(_ k__+k_ _U_ ) s v 2 = _ 2x_c_ + U 2 − _ 2R_c_ = U 2+ 2c ( _1x_ − _ R1_ )
b _ 12 RU 2
18 a _ 1_3_(_e_ _15__3 0_t_ −__1_ _) m s−1
(e + _15_30_t 1) 31 a F = ma
− _xk_ 2_ = ma
b As _ ((ee__ xx_+−__11__)) , 1, v , 13 m s−1 a = − _m _k_ a _ _2
so speed cannot exceed 13 m s−1
19 a _3_0_(_e_e _5t__ _ 5t_ − __1_ _) m s−1 b 30 m s−1 At x = R, a = − g
20 a T = _a1 _ ( arctan _2a_0_ − arctan _1a_2_ ) − g = − _m _k_ R_ _2
b _21_ ln( _aa _ 22__++__41_04_04_ ) m k = mg R 2
21 53.6 (3 s.f.)
− _m_xg__ 2R _ _2 = ma
a = v _dd_xv_ = − _gx_ R _ 2 _2
22 a v 2 = 20 − 16e−0.1x
b x = 10 ln 4 b X = _2_g_2R_g_R−_ 2_U _ _2
32 _23
c For all x, e−0.1x . 0 33 a 19.7 N (3 s.f.) b 5.44 m s−1 (3 s.f.)
So v2 = 20 – _1_6_e−0.1x , 20 b 12 m s−1 b 2.6 m s−1
Hence v , √ 20 b (4 + 20e−1) m 34 a 1.3 m d 0.79 s (2 d.p.)
23 a v 2 = 2x2 − x3 + 144 c 5.2 m s−2
24 a t = −2 ln( _25 )or 2 ln( _ 52 )
243
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