AQA A level Mechanics sample chapters

104 Pages·2017·15.71 MB·English

Contents
Getting the most from this book v 7 Circular motion 155
1 Kinematics 1 7.1 Introduction to circular motion 156
7.2 Circular motion with constant
1.1 The language of motion in one
speed 160
dimension 2
7.3 The conical pendulum 162
1.2 The constant acceleration formulae 11
7.4 Banked tracks 167
1.3 Variable acceleration 16
8 Hooke’s law 175
2 Motion in more than one dimension 24
8.1 Strings and springs 176
2.1 Motion in more than one dimension 25
8.2 Using Hooke’s lllaaawww wwwYYYYYYYYYYYYYYYYYYYYYiiittthhh more
2.2 The equation of a path 32
than one sssppprrriiinnnggg ooorrr ssstttrrriiinnnggg 182
2.3 Path of a projectile 37
8.3 Work and eeennneeePPPPPPPPPPPPPPPPPPPPPrrrgggyyy 188
3 Forces and motion 44 8.4 VVVeeerrrtttiiicccaaalll mmmoootttiiiooonnn iiinnnvvvooolllvvviiinnnggg elastic
fffooorrrccceeesss OOOOOOOOOOOOOOOOOOOOO 193
3.1 Forces and Newton’s laws of motion 45
3.2 Working in vectors 52 9 DDDiiimmmeeennnsssiiiooonnnaaalll aaannnaaalllyyysssiiisss 201
3.3 Forces in equilibrium 61 CCCCCCCCCCCCCCCCCCCCC
999...111 IIInnntttrrroooddduuuccctttiiiooonnn to dimensional
3.4 Finding resultant forces 69 aaannnaaalllyyysssiiisss 202
999...222 TTThhheee dddiiimmmeeennnsssiiiooonnnsss of further quantities 203
4 A model for friction 777888
TTTTTTTTTTTTTTTTTTTTT
999...333 OOOttthhheeerrr systems of units 204
4.1 A model for friction 777999
9.4 Dimensional consistency 206
4.2 Modelling with friction FFFFFFFFFFFFFFFFFFFFF 888000
9.5 Finding the form of the
Practice Questions: Set 1 888999 relationship 207
AAAAAAAAAAAAAAAAAAAAA
9.6 The method of dimensions 209
5 WWWooorrrkkk,,, eeennneeerrrgggyyy aaannnddd pppooowwweeerrr 91
RRRRRRRRRRRRRRRRRRRRR Practice Questions: Set 2 214
5.1 Energy and mmmooommmeeennntttuuummm 92
5.2 Work and eeennneeerrrgggyyy 92 10 Moments of forces 217
DDDDDDDDDDDDDDDDDDDDD
5.3 GGGrrraaavvviiitttaaatttiiiooonnnaaalll pppooottteeennntttiiiaaalll eeennneeerrrgggyyy 100
10.1 Introduction to moments 218
5.4 Work aaannnddd kkkiiinnneeetttiiiccc eeennneeerrrgggyyy for
10.2 The moment of a force which
tttwwwooo---dddiiimmmeeennnsssiiiooonnnaaalll mmmoootttiiiooonnn 104
acts at an angle 230
5.5 Power 108
10.3 Sliding and toppling .......................238
5.6 Work and energy with
variable forces 111 11 Centre of mass 245
11.1 The centre of mass 246
6 Impulse and momentum 120
11.2 Centre of mass of two- and
6.1 Impulse 121
three-dimensional bodies 250
6.2 Conservation of momentum 125
11.3 Calculating volumes 261
6.3 Newton’s law of impact 131
11.4 Centres of mass 267
6.4 Impulse and momentum in
11.5 Centres of mass of plane regions 279
more than one dimension 139
6.5 Oblique impact of smooth
elastic spheres 147
iii
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12 Circular motion 2 289
12.1 Circular motion with
variable speed ................................290
Practice Questions: Set 3 308
Appendix: An introduction
to radians 312
Extension material: Velocity and
acceleration for motion in a Y
circle 315
P
Answers ..............................................317
Index ....................................................337 O
C

T
F
A
R
D
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3
Forces and motion

Y
P
O
C

TTTTTTTTTTT
FFFFFFFFFFF
AAAAAAAAAAA
RRRRRRRRRRR
In the beginning (if
(cid:31)(cid:31)(cid:31) TTThhhiiisss tttiiiggghhhtttrrrooopppeee wwwaaalllkkkeeerrr iiisss ssstttaaatttiiiooonnnaaarrryyy... TTThhheeerrreee aaarrreee aaa nnnuuummmbbbeeerrr ooofff fffooorrrccceeesss aaaccctttiiinnnggg ooonnn
there was such a thing) ttthhheee wwwaaalllkkkeeerrr wwwhhhiiiccchhh cccaaannnccceeelll eeeaaaccchhh ooottthhheeerrr ooouuuttt,,, rrreeesssuuullltttiiinnnggg iiinnn nnnooo mmmoootttiiiooonnn... IIInnn ooorrrdddeeerrr
GGGoooddd cccrrreeeaaattteeeddd NNNeeewwwtttooonnn’’’sss DDDDDDDDDDD fffooorrr ttthhhaaattt tttooo bbbeee pppooossssssiiibbbllleee,,, ttthhheee cccaaabbbllleee mmmuuusssttt mmmaaakkkeee sssmmmaaallllll aaannngggllleeesss tttooo ttthhheee hhhooorrriiizzzooonnntttaaalll
lllaaawwwsss ooofff mmmoootttiiiooonnn tttooogggeeettthhheeerrr sssooo ttthhhaaattt ttthhheee vvveeerrrtttiiicccaaalll cccooommmpppooonnneeennntttsss ooofff ttthhheee ttteeennnsssiiiooonnn cccaaannn cccaaannnccceeelll ooouuuttt ttthhheee wwweeeiiiggghhhttt ooofff
the walker. In that case, the tensions in the cable will be greater than the
wwwiiittthhh ttthhheee nnneeeccceeessssssaaarrryyy
walker’s weight.
masses and forces.
This is all; everything
beyond this follows
from the development
of appropriate
mathematical methods
by means of deduction.
Albert Einstein
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1 Forces and Newton’s laws of
3
motion
Modelling vocabulary
C
h
Mechanics is about modelling the real world. In order to do this, suitable a
p
simplifying assumptions are often made so that mathematics can be applied to t
e
situations and problems. This process involves identifying factors that can be r 3
neglected without losing too much accuracy. Here are some commonly used F
modelling terms which are used to describe such assumptions: o
r
c
e
(cid:29) negligible: small enough to ignore s
(cid:29) inextensible: for a string with negligible stretch YYYYYY an
(cid:29) light: for an object with negligible mass d m
(cid:29) particle: an object with negligible dimensions PPPPPP o
(cid:29) sssmmmoooooottthhh::: fffooorrr aaa sssuuurrrfffaaaccceee wwwiiittthhh nnneeegggllliiigggiiibbbllleee fffrrriiiccctttiiiooonnn ti
o
(cid:29) uniform: the same throughout. n
OOOOOO
Forces
CCCCCC
AAA fffooorrrccceee iiisss dddeeefififi nnneeeddd aaasss ttthhheee ppphhhyyysssiiicccaaalll qqquuuaaannntttiiitttyyy ttthhhaaattt cccaaauuussseeesss aaa ccchhhaaannngggeee iiinnn mmmoootttiiiooonnn... AAAsss iiittt
dddeeepppeeennndddsss ooonnn mmmaaagggnnniiitttuuudddeee aaannnddd dddiiirrreeeccctttiiiooonnn,,, iiittt iiisss aaa vvveeeccctttooorrr qqquuuaaannntttiiitttyyy...
Note FFFFFooooorrrrrccccceeeeesssss cccccaaaaannnnn ssssstttttaaaaarrrrrttttt mmmmmoooootttttiiiiiooooonnnnn,,,,, ssssstttttoooooppppp mmmmmoooootttttiiiiiooooonnnnn,,,,, ssssspppppeeeeeeeeeeddddd uuuuuppppp ooooorrrrr ssssslllllooooowwwww dddddooooowwwwwnnnnn ooooobbbbbjjjjjeeeeeccccctttttsssss,,,,, ooooorrrrr ccccchhhhhaaaaannnnngggggeeeee
ttttthhhhheeeee dddddiiiiirrrrreeeeeccccctttttiiiiiooooonnnnn ooooofffff ttttthhhhheeeeeiiiiirrrrr mmmmmoooootttttTTTTTTTTTTTTTTTTTTTTiiiiiooooonnnnn..... IIIIInnnnn rrrrreeeeeaaaaalllll sssssiiiiitttttuuuuuaaaaatttttiiiiiooooonnnnnsssss,,,,, ssssseeeeevvvvveeeeerrrrraaaaalllll fffffooooorrrrrccccceeeeesssss uuuuusssssuuuuuaaaaallllllllllyyyyy aaaaacccccttttt ooooonnnnn
g varies around the
aaannn ooobbbjjjeeecccttt... TTThhheee sssuuummm ooofff ttthhheeessseee fffooorrrccceeesss,,, kkknnnooowwwnnn aaasss ttthhheee rrreeesssuuullltttaaannnttt fffooorrrccceee,,, dddeeettteeerrrmmmiiinnneeesss
world, with 9.80 m s−2
wwwhhheeettthhheeerrr ooorrr nnnooottt ttthhheeeFFFFFFFFFFFFFFFFFFFFrrreee iiisss aaa ccchhhaaannngggeee ooofff mmmoootttiiiooonnn...
being a typical value.
Singapore, at 9.766, TTThhheeerrreee aaarrreee ssseeevvveeerrraaalll tttyyypppeeesss ooofff fffooorrrccceee ttthhhaaattt yyyooouuu ooofffttteeennn uuussseee...
has one of the lowest AAAAAAAAAAAAAAAAAAAA
values, and Helsinki,
with 9.825, has one of TTThhheee fffooorrrccceee ooofff gggrrraaavvviiitttyyy
RRRRRRRRRRRRRRRRRRRR
the highest. You need
EEEvvveeerrryyy ooobbbjjjeeecccttt ooonnn ooorrr nnneeeaaarrr ttthhheee EEEaaarrrttthhh’’’sss sssuuurrrfffaaaccceee iiisss pppuuulllllleeeddd vvveeerrrtttiiicccaaallllllyyy dddooowwwnnnwwwaaarrrdddsss bbbyyy ttthhheee
to look carefully at the
level of accuracy to fffooorrrccceee ooofff gggrrraaavvviiitttyyy... TTThhheee sssiiizzzeee ooofff ttthhheee fffooorrrccceee ooonnn aaannn ooobbbjjjeeecccttt ooofff mmmaaassssss M kg is Mg newtons
DDDDDDDDDDDDDDDDDDDD
which g iiisss gggiiivvveeennn iiinnn aaannnyyy wwwhhheeerrreee ggggggg iiiiisssss aaaaa cccccooooonnnnnssssstttttaaaaannnnnttttt wwwwwhhhhhooooossssseeeee vvvvvaaaaallllluuuuueeeee iiiiisssss aaaaabbbbbooooouuuuuttttt 99999.....88888 mmmmm sssss−2. The force of gravity is also
ppprrrooobbbllleeemmm... YYYooouuurrr fififi nnnaaalll kkknnnooowwwnnn aaasss ttthhheee weight of the object.
aaannnssswwweeerrr ssshhhooouuulllddd aaalllwwwaaayyysss
rrreeeflflfl eeecccttt ttthhheee aaaccccccuuurrraaacccyyy ooofff
the given information. Tension and thrust
When a string is pulled, as in Figure 3.1, it exerts a tension force opposite
to the pull. The tension acts along the string and is the same throughout the
string. A rigid rod can exert a tension force in a similar way to a string when it
is used to support or pull an object. It can also exert a thrust force when it is
in compression, as in Figure 3.2. The thrust acts along the rod and is the same
throughout the rod.
The tension on either side of a smooth pulley is the same, as shown in
Figure 3.3.
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Forces and Newton’s laws of motion
T (tension pulling
W (weight of block)
down on ceiling)
T (thrust pushing
up on block)
T
T (tension pulling up
on object) T (thrust pushing
down on floor)
T
mg (weight of object) R (reaction from floor)
Figure 3.1 Figure 3.2 Figure 3.3
Normal reaction
AAA bbbooooookkk rrreeessstttiiinnnggg ooonnn aaa tttaaabbbllleee iiisss sssuuubbbjjjeeecccttteeeddd tttooo tttwwwooo fffooorrrccceeesss,,, iiitttsss wwwYYYYYYYYYYYYYYeeeiiiggghhhttt aaannnddd ttthhheee normal
reaction ooofff ttthhheee tttaaabbbllleee... IIIttt iiisss cccaaalllllleeeddd nnnooorrrmmmaaalll bbbeeecccaaauuussseee iiitttsss llliiinnneee ooofff aaaccctttiiiooonnn iiisss nnnooorrrmmmaaalll
(((aaattt rrriiiggghhhttt aaannngggllleeesss))) tttooo ttthhheee sssuuurrrfffaaaccceee ooofff ttthhheee tttaaabbbllleee... SSSiiinnnccceee ttthhheee bbbooooookkk iiisss iiinnn eeeqqquuuiiillliiibbbrrriiiuuummm,,,
PPPPPPPPPPPPPP
ttthhheee nnnooorrrmmmaaalll rrreeeaaaccctttiiiooonnn iiisss eeeqqquuuaaalll aaannnddd oooppppppooosssiiittteee tttooo ttthhheee wwweeeiiiggghhhttt ooofff ttthhheee bbbooooookkk;;; iiittt iiisss aaa
Note
positive force.
OOOOOOOOOOOOOO
In Figure 3.4, the R (((nnnooorrrmmmaaalll rrreeeaaaccctttiiiooonnn)))
normal reaction is TTThhheee nnnooorrrmmmaaalll rrreeeaaaccctttiiiooonnn
vertical but this is not fffooorrrccceee iiisss aaalllsssooo ooofffttteeennn
CCCCCCCCCCCCCC
always the case. For cccaaalllllleeeddd ttthhheee nnnooorrrmmmaaalll
example, the normal contact force.
reaction on an object on

a slope is perpendicular mmmggg (((wwweeeiiiggghhhttt)))
TTTTTTTTTTT
to the slope.
Figure 3.4
IIIfff ttthhheee bbbooooookkk iiisss aaabbbFFFFFFFFFFFooouuuttt tttooo lllooossseee cccooonnntttaaacccttt wwwiiittthhh ttthhheee tttaaabbbllleee (((wwwhhhiiiccchhh mmmiiiggghhhttt hhhaaappppppeeennn,,, fffooorrr
iiinnnssstttaaannnccceee,,, iiifff ttthhheee tttaaabbbllleee iiisss aaacccccceeellleeerrraaatttiiinnnggg rrraaapppiiidddlllyyy dddooowwwnnnwwwaaarrrdddsss))),,, ttthhheee nnnooorrrmmmaaalll fffooorrrccceee
bbbeeecccooommmeeesss zzzeeerrrooo...
AAAAAAAAAAA
FFFrrriiiccctttiiiooonnnaaalll fffooorrrccceee
RRRRRRRRRRR
IIInnn ttthhhiiisss dddiiiaaagggrrraaammm,,, ttthhheee bbbooooookkk ooonnn ttthhheee tttaaabbbllleee iiisss bbbeeeiiinnnggg pppuuussshhheeeddd bbbyyy aaa fffooorrrccceee P parallel to the
sssuuurrrfffaaaccceee... TTThhheee bbbooooookkk rrreeemmmaaaiiinnnsss aaattt rrreeesssttt bbbeeecccaaauuussseee P is balanced by a frictional force, F,
DDDDDDDDDDD
iiinnn ttthhheee oooppppppooosssiiittteee dddiiirrreeeccctttiiiooonnn tttooo P. The magnitude of the frictional force is equal to
ttthhheee pppuuussshhhiiinnnggg fffooorrrccceee P = F
R (normal reaction)
The frictional force
P is also sometimes
F (friction) called frictional
contact force.
mg (weight)
Figure 3.5
If P is increased and the book starts to move, F is still present but now P > F.
Friction always acts in the opposite direction to the motion. Friction may
prevent the motion of an object or slow it down if it is moving.
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Driving force
3
In problems about moving objects such as cars, all the forces acting along the
line of motion can usually be reduced to two or three: the driving force, the
resistance to motion and, possibly, a braking force.
C
h
Resistance Driving force a
p
t
e
Braking force r
3
Figure 3.6 F
o
r
Example 3.1 c
Figure 3.7 shows a block A of mass 10 kg connected to a light scale pan by a e
s
llliiiggghhhttt iiinnneeexxxttteeennnsssiiibbbllleee ssstttrrriiinnnggg ttthhhaaattt pppaaasssssseeesss ooovvveeerrr aaa llliiiggghhhttt sssmmmoooooottthhh pppuuulllllleeeyyy... TTThhheee ssscccaaallleee pppaaannn a
YYYYYYYYYYY n
hhhooollldddsss bbbllloooccckkk BBB,,, aaalllsssooo ooofff mmmaaassssss 111000 kkkggg... TTThhheee sssyyysssttteeemmm iiisss iiinnn eeeqqquuuiiillliiibbbrrriiiuuummm... d
m
(((iii))) OOOnnn ssseeepppaaarrraaattteee dddiiiaaagggrrraaammmsss,,, ssshhhooowww aaallllll ttthhheee fffooorrrccceeesss aaaccctttiiinnnggg ooonnn eeeaaaccchhh ooofff ttthhheee mmmaaasssssseeesss,,,
PPPPPPPPPPP o
t
the scale pan and the pulley. i
o
n
(((iiiiii))) FFFiiinnnddd ttthhheee vvvaaallluuueee ooofff ttthhheee ttteeennnsssiiiooonnn iiinnn ttthhheee ssstttrrriiinnnggg...
OOOOOOOOOOO
A
(((iiiiiiiii))) FFFiiinnnddd ttthhheee ttteeennnsssiiiooonnn iiinnn ttthhheee rrroooddd hhhooollldddiiinnnggg ttthhheee pppuuulllllleeeyyy...
B
(((iiivvv))) FFFiiinnnddd ttthhheee nnnooorrrmmmaaalll rrreeeaaaccctttiiiooonnn ooofff CCCCCCCCCCCBBB ooonnn ttthhheee ssscccaaallleee pppaaannn...
Figure 3.7
Solution
TTTTTTTTTTTTTTTTTTTT
(i) NNNNNooooorrrrrmmmmmaaaaalllll rrrrreeeeeaaaaaccccctttttiiiiiooooonnnnn Tension in rod
TTTeeennnsssiiiooonnn iiinnn ttthhheee ssstttrrriiinnnggg Tension in the string
ooofff ssscccaaallleee pppaaannn ooonnn BBB holding pulley
FFFFFFFFFFFFFFFFFFFF
TTT R T T1
AAAAAAAAAAAAAAAAAAAA
AAA B
RRRRRRRRRRRRRRRRRRRR
R
10g 10g T T
DDDDDDDDDDDDDDDDDDDD WWWeeeiiiggghhhttt ooofff AAA wwwrrriiitttttteeennn aaasss Normal reaction Tension in the
Weight of B
10g and acting of B on scale pan string
vertically downwards.
Figure 3.8
(ii) Block A is in equilibrium: ⇒ T = 10 g = 98 N.
(iii) The pulley is in equilibrium: ⇒ T = 2T = 196 N.
1
(iv) Block B is in equilibrium: ⇒ R = 10 g = 98 N.
Newton’s laws of motion
1 Every object continues in a state of rest or uniform motion in a straight line
unless it is acted on by a resultant external force.
2 The acceleration of an object is proportional to, and in the same direction as,
F is the resultant force.
the resultant of the forces acting on the object.
m is the mass of the
object. F = ma
a is the acceleration.
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Forces and Newton’s laws of motion
Notice that this is a vector equation, since both the magnitudes and directions
Historical of the resultant force and the acceleration are involved. If the motion is along a
note straight line it is often written in scalar form as F = ma.
Isaac Newton was
3 When one object exerts a force on another there is always a reaction, which
born in Lincolnshire
is equal and opposite in direction to the acting force.
in 1642. He was not an
outstanding scholar
either as a schoolboy
Equation of motion
or as a university
student, yet later in life
The equation resulting from Newton’s second law is often described as an
he made remarkable
contributions in equation of motion, as in the following examples.
dynamics, optics,
astronomy, chemistry,
Example 3.2
music theory and YYYYYYYY
theology. He became AAAnnn eeemmmppptttyyy bbbooottttttllleee ooofff mmmaaassssss 000...555 kkkggg iiisss rrreeellleeeaaassseeeddd fffrrrooommm aaa sssuuubbbmmmaaarrriiinnneee aaannnddd rrriiissseeesss wwwiiittthhh aaannn
Member of Parliament acceleration of 0.75 m s−2... TTThhheee wwwaaattteeerrr cccaaauuussseeesss aaa rrreeesssiiissstttaaannnccceee ooofff 111...111 NNN...
for Cambridge PPPPPPPP
University and later (((iii))) DDDrrraaawww aaa dddiiiaaagggrrraaammm ssshhhooowwwiiinnnggg ttthhheee fffooorrrccceeesss aaaccctttiiinnnggg ooonnn ttthhheee bbbooottttttllleee aaannnddd ttthhheee
Warden of the Royal direction of its acceleration.
OOOOOOOO
Mint. His tomb in
(((iiiiii))) WWWrrriiittteee dddooowwwnnn ttthhheee eeeqqquuuaaatttiiiooonnn ooofff mmmoootttiiiooonnn ooofff ttthhheee bbbooottttttllleee...
Westminster Abbey
reads ‘Let mortals
(((iiiiiiiii))) FFFiiinnnddd ttthhheee sssiiizzzeee ooofff ttthhheee bbbuuuoooyyyaaannncccyyy fffooorrrccceee...
rejoice that there CCCCCCCC
existed such and so
great an ornament to
Solution
the Human Race’.
TTTTTTTTTTTTTT
(((((iiiii))))) TTTTThhhhheeeee fffffooooorrrrrccccceeeeesssss aaaaaccccctttttiiiiinnnnnggggg ooooonnnnn ttttthhhhheeeee bbbbbooooottttttttttllllleeeee aaaaannnnnddddd ttttthhhhheeeee aaaaacccccccccceeeeellllleeeeerrrrraaaaatttttiiiiiooooonnnnn aaaaarrrrreeeee ssssshhhhhooooowwwwwnnnnn iiiiinnnnn ttttthhhhhiiiiisssss
dddiiiaaagggrrraaammm...
FFFFFFFFFFFFFF
BN Buoyancy
AAAAAAAAAAAAAA
The acceleration is shown with a
RRRRRRRRRRRRRRFFFooorrrccceeesss aaaccctttiiinnnggg 0.75m s−2 different type of arrow. A double
ooonnn ttthhheee bbbooottttttllleee
arrow is often used.
1.1N
DDDDDDDDDDDDDD 0.5g N
The weight is always
Water resistance shown as mg for moving bodies.
Figure 3.9
(ii) The resultant force acting on the bottle is (B – 0.5 g – 1.1) upwards.
The resulting equation
B – 0.5 g – 1.1 = 0.5a
is called the equation of motion.
(iii) B – 4.9 – 1.1 = 0.375 0.5 × 0.75
B = 6.375
The buoyancy force on the bottle is 6.4 N.
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Example 3.3
A car of mass 900 kg travels at a constant speed of 20 m s−1 along a straight 3
horizontal road. Its engine is producing a driving force of 500 N.
(i) What is the resistance to its motion?
Later the driving force is removed and the car is brought to rest in a
C
time of 5 s with the same resistance to motion. h
a
(ii) Find the force created by the brakes, assuming it to be constant. p
t
e
r
3
Solution F
o
r
(i) The car is travelling at constant speed, so that the resultant force acting c
e
on the car is zero. s
a
a = 0 YYYYYYYY n
d
m
PPPPPPPP o
t
i
o
Resultant = 500 – R R N 555000000 NNN n
OOOOOOOO
Figure 3.10
Let the resistive force be RRR NNNCCCCCCCC...
500 – R = 0
R = 500
TTTTTTTTTTTTTTTTTTTTTTT
TTTTThhhhheeeee rrrrreeeeesssssiiiiissssstttttiiiiivvvvveeeee fffffooooorrrrrccccceeeee iiiiisssss 555550000000000 NNNNN.....
So you expect a to be
(((iiiiii))) TTThhheee cccaaarrr iiisss sssFFFFFFFFFFFFFFFFFFFFFFFlllooowwwiiinnnggg dddooowwwnnn...
negative.
a
AAAAAAAAAAAAAAAAAAAAAAA
B is constant, so that a
Resultant = 500 – B 555000000 NNN is also constant and you
RRRRRRRRRRRRRRRRRRRRRRR
can use the constant
BBB
acceleration formulae.
FFFiiiggguuurrreee 333...111111
DDDDDDDDDDDDDDDDDDDDDDD
TTThhheee eeeqqquuuaaatttiiiooonnn ooofff mmmoootttiiiooonnn iiisss –––B – 500 = 900a ①
000 === 222000 +++ a × 5 Use v = u + at with
u = 20, v = 0 and t = 5.
a = –4 m s−2
Substituting in ① –B – 500 = 900 × –4
⇒ B = 3100 N
The braking force is 3100 N.
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Forces and Newton’s laws of motion
Example 3.4
Two boxes A and B are descending vertically supported by a parachute. Box A
has mass 100 kg. Box B has mass 75 kg and is suspended from box A by a light
vertical wire. Both boxes are descending with acceleration 3 m s−2.
(i) Draw a labelled diagram showing all the forces acting on box A and
another diagram showing all the forces acting on box B.
(ii) Write down separate equations of motion for box A and for box B.
(iii) Find the tensions in both wires.
Solution
A
(i) YYYYYYYYYYYYYYYYY
T
2
T
1
B PPPPPPPPPPPPPPPPP
Figure 3.12
A OOOOOOOOOOOOOOOOO B
3ms–2 333mmmsss–––222
CCCCCCCCCCCCCCCCC
T
2 75g
111000000ggg

FFFFFiiiiiggggguuuuurrrrreeeee 33333.....1111133333 AAAAAcccccccccceeeeellllleeeeerrrrraaaaatttttiiiiiooooonnnnn dddddooooowwwwwnnnnnwwwwwaaaaarrrrrdddddsssss.....
TTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT
(((((iiiiiiiiii))))) BBBBBoooooxxxxx AAAAA::::: TTTTThhhhheeeee rrrrreeeeesssssuuuuullllltttttaaaaannnnnttttt dddddooooowwwwwnnnnnwwwwwaaaaarrrrrdddddsssss fffffooooorrrrrccccceeeee iiiiisssss T + 100 g – T so the
2 1
eeeqqquuuaaatttiiiooonnn ooofff mmmoootttiiiooonnn   iiisss
Note
FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF
TTTTTTT +++++++ 111111100000000000000ggggggggg ––––– TTTTTTT === 333000000 ① Since 100a = 100 × 3 = 300
T is the tension in the 22222 11111
1
blue wire linking A to BBBAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAoooxxx BBB::: TTThhheee rrreeesssuuullltttaaannnttt dddooowwwnnnwwwaaarrrdddsss fffooorrrccceee iiisss 777555g – T so the equation of
the parachute. T is the 2
2 mmmoootttiiiooonnn iiisss
tension in the green
wire linking A to B. RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR 77777775555555ggggggggg ––––– T = 225 ② Since 75a = 75 × 3 = 225
2
T ≠ T
1 2 (((iiiiiiiii))) FFFrrrooommm ②: T = 75 × 9.8 – 225
2
DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD
= 510 N
DDDiiissscccuuussssssiiiooonnn pppoooiiinnnttt
Substituting in ①: T = 510 + 100 × 9.8 – 300
IIIsss ttthhheee llleeevvveeelll ooofff aaaccccccuuurrraaacccyyy 1
gggiiivvveeennn iiinnn ttthhheeessseee aaannnssswwweeerrrsss = 1190 N
justifi ed?
The tension in the blue wire linking A to the parachute is 1365 N.
The tension in the green wire linking A to B is 510 N.
Exercise 3.1
① Find the accelerations produced when a force of 100 N acts on an object
(i) of mass 15 kg
(ii) of mass 10 g
(ii) of mass 1 tonne.
② A bullet of mass 20 g is fi red into a wall with a velocity of 400 m s−1. The
bullet penetrates the wall to a depth of 10 cm. Find the resistance of the
wall, assuming it to be uniform.
50
9781510414426_CH03.indd 50 12/09/17 8:45 PM
③ A car of mass 1200 kg is travelling along a straight level road.
3
(i) Calculate the acceleration of the car when a resultant force of 2400 N
acts on it in the direction of its motion. How long does it take the car
to increase its speed from 4 m s−1 to 12 m s−1?
The car has an acceleration of 1.2 m s−2 when there is a driving force of
C
2400 N. h
a
(ii) Find the resistance to motion of the car. p
t
e
④ A load of mass 5 kg is held on the end of a string. Calculate the tension in r
3
the string when F
(i) the load is raised with an acceleration of 2.5 m s−2 or
c
(ii) the load is lowered with an acceleration of 2.5 m s−2 e
s
(iii) the load is raised with a constant speed of 2 m s−−−111 YYYYYYYYYYYYYYYYYY an
d
(iv) the load is raised with a deceleration of 2.5 mmm sss−−−222... m
⑤ A block A of mass 10 kg is connected to a PPPPPPPPPPPPPPPPPP o
t
block B of mass 5 kg byyy aaa llliiiggghhhttt iiinnneeexxxttteeennnsssiiibbbllleee io
n
ssstttrrriiinnnggg pppaaassssssiiinnnggg ooovvveeerrr aaa sssmmmoooooottthhh fififixxxeeeddd pppuuulllllleeeyyy... TTThhheee
OOOOOOOOOOOOOOOOOO
bbbllloooccckkksss aaarrreee rrreeellleeeaaassseeeddd fffrrrooommm rrreeesssttt wwwiiittthhh AAA 000...333 mmm
abovvveee gggrrrooouuunnnddd llleeevvveeelll,,, aaasss ssshhhooowwwnnn iiinnn FFFiiiggguuurrreee 333...111444...
(i) FFFiiinnnddd ttthhheee aaacccccceeellleeerrraaatttiiiooonnn ooofff ttthhheeeCCCCCCCCCCCCCCCCCC sssyyysssttteeemmm aaannnddd A B
ttthhheee ttteeennnsssiiiooonnn iiinnn ttthhheee ssstttrrriiinnnggg...
(ii) FFFiiinnnddd ttthhheee ssspppeeeeeeddd ooofff ttthhheee mmmaaasssssseeesss wwwhhheeennn AAA hhhiiitttsss 0.3m

the ground.
TTTTTTTTTTTTTTTT
(iii) HHHHHooooowwwww fffffaaaaarrrrr dddddoooooeeeeesssss BBBBB rrrrriiiiissssseeeee aaaaafffffttttteeeeerrrrr AAAAA hhhhhiiiiitttttsssss ttttthhhhheeeee flflflflfloooooooooorrrrr
Figure 3.14
aaannnddd ttthhheee ssstttrrriiinnnggg bbbeeecccooommmeeesss ssslllaaaccckkk???
FFFFFFFFFFFFFFFF
⑥ AAA bbbllloooccckkk AAA ooofff mmmaaassssss 111000 kkkggg iiisss lllyyyiiinnnggg
A
ooonnn aaa sssmmmoooooottthhh hhhooorrriiizzzooonnntttaaalll tttaaabbbllleee...
AAAAAAAAAAAAAAAA
LLLiiiggghhhttt iiinnneeexxxttteeennnsssiiibbbllleee ssstttrrriiinnngggsss cccooonnnnnneeecccttt
AAA tttooo pppaaarrrtttiiicccllleee BBB ooofff mmmaaassssss 666 kg and B
RRRRRRRRRRRRRRRR
pppaaarrrtttiiicccllleee CCC ooofff mmmaaassssss 444 kg, which C
hhhaaannnggg fffrrreeeeeelllyyy ooovvveeerrr sssmmmoooooottthhh pppuuulllllleeeyyysss
Figure 3.15
DDDDDDDDDDDDDDDD aaattt ttthhheee eeedddgggeee ooofff ttthhheee tttaaabbbllleee...
(((iii))) Draw force diagrams to show the forces acting on each mass.
(ii) Write down separate equations of motion for A, B and C.
(iii) Find the acceleration of the system and the tensions in the strings.
⑦ A truck of mass 1250 kg is towing a trailer of mass 350 kg along a horizontal
straight road. The engine of the truck produces a driving force of 2500 N.
The truck is subjected to a resistance of 250 N and the trailer to a resistance
of 300 N.
Figure 3.16
(i) Show, in separate diagrams, the horizontal forces acting on the truck
and the trailer.
(ii) Find the acceleration of the truck and trailer.
(iii) Find the tension in the coupling between the truck and the trailer.
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