Chapter 13 Dynamics - scunderwood Specialist Mathematics Complete Worked Solutions 1 Chapter 13 –...
Transcript of Chapter 13 Dynamics - scunderwood Specialist Mathematics Complete Worked Solutions 1 Chapter 13 –...
Essential Specialist Mathematics Complete Worked Solutions 1
Chapter 13 – Dynamics
Solutions to Exercise 13A 1
a
i. r = 7i – 2i + 5j
= 5i + 5j
ii. | r | = 52 + 52
= 50
= 5 2
≈ 7.07 N
tan ° = 55
= 45
magnitude is 7.07 N, direction is 45°
b
i. r = 2j + 7i – 11i – 6j
= – 4i – 4j
ii. | r | = 16 + 16
= 32
= 4 2 N
≈ 5.66 N
tan ° = 44
= 45
magnitude is 5.66 N, direction is 225°
c
i. r = 3i – 4i – 5j
= – i – 5j
ii. | r | = (–1)2 + (– 5)2
= 26
≈ 5.10 N
tan ° = 5
= 78.7
magnitude is 5.10 N, direction is 258.7°
d
i. r = 8i + 8j – 5i + 2j
= 3i + 10j
ii. | r | = 32 + 102
= 109
≈ 10.44 N
tan ° = 103
= 73.3°
magnitude is 10.44 N, direction is 73.3°
y
x
5
5
q
y
xq
y
xq
–5
–1
y
x
10
q
3
Essential Specialist Mathematics Complete Worked Solutions 2
e
i. r = 6i – 6i + 3j – 7j
= – 4j
ii. | r | = 4 N
= 270 °
magnitude is 4 N, direction is 270 °
f
i. r = 5i + 15j + 5i – 15j
= 10i
ii. magnitude = 10 N
direction = 0°
2
R = F1 + F2 + F3
= (3i + 2j) + (6i – 4j) + (2i – j)
= 11i – 3j N
3
Using the cosine rule,
F2 = 162 + 122 – 2 16 12 cos (130°)
= 256 + 144 – 384 cos (130°)
= 646.83044…
F = 25.43 N
The magnitude of F is 25.43 N, correct to two
decimal places.
4
Using the cosine rule,
162 = 92 + F22 – 2 9 F2 cos (120°)
256 = 81 + F22 – 18F2 – 1
2
175 = F22 + 9F2
F22 + 9F2 – 175 = 0
Using the quadratic formula,
F 2 =
–9 ± 92
– 4 1 –175
2
=781 – 9
2
= 9.5 (F2 > 0)
| F2 | = 9.5 N, correct to two decimal
places.
5
F1 + F2 + F3 = F
F3 = F – F1 – F2
= (3i – 2j + k) – (2i – j + k) – (3i – j – k)
= – 2i + k
6
Component of F in the direction of motion
= 400 cos 15°
= 386.37033…
≈ 386 N
7
a
i.
a = 4 cos 40° i + 4 sin 40° j
b = 3i
r = 6.064i + 2.57j
y
x
4
9 16
F2
120°
400
15°
y
x
4 N
a
b40°
3 N
Essential Specialist Mathematics Complete Worked Solutions 3
ii. | r | = 2257.206.6
≈ 6.59 N
tan ° = 2×576×06
= 22° 59'
magnitude ≈ 6.59 N
direction ≈ 22° 59'
b
i.
a = 6 cos 10° i + 6 sin 10° j
b = 7 cos 20° i + 7 sin 20° j
c = 8 cos 30° i + 8 sin 30° j
r = (6cos 10° + 7cos 20° + 8cos 30°) i
+ (6sin 10° + 7sin 20° + 8sin 30°) j
≈ 19.41i + 7.44j
ii. | r | = 2244.741.19
≈ 20.79 N
tan = 7×44
19×41
≈ 20° 57'
magnitude ≈ 20.79 N
direction ≈ 20° 57'
c
i.
a = 7 cos 20° i + 7 sin 20° j
b = – 6 cos 30° i + 6 sin 30° j
r = (7 cos 20° – 6 cos 30°) i
+ (7 sin 20° + 6 sin 30°) j
= 1.382i + 5.394j
ii. | r | = 49 + 36 – 84 cos 50°
≈ 5.57 N
tan ° =
7 sin 20° + 6 sin 30°7 cos 20° - 6 cos 30°
= 75° 38'
magnitude ≈ 5.57 N
direction ≈ 75° 38'
d
i.
a = 5 cos 27° i + 5 sin 27° j
b = – 5 sin 27° i – 5 cos 27° j
r = 2.19i – 2.19j
ii. | r | = 22)19.2(19.2
≈ 3.09 N
tan ° = 1
= 45°
magnitude ≈ 3.09 N
direction ≈ 315°
e
i.
a = 10 cos 25° i + 10 sin 25° j
b = 10 cos 25° i – 10 sin 25° j
r = 18.13i
ii. | r | ≈ 18.13 N
tan ° = 0
magnitude ≈ 18.13 N
direction ≈ 0°
y
x
8 N (c)
10°6 N (a)
7 N (b)
10°10°
x
y
ab20°30°
7 N6 N
x
ya
b
27°
27°
5 N
5 N
x
ya
b
25°
10 N
25°
10 N
Essential Specialist Mathematics Complete Worked Solutions 4
f
i.
a = 8 cos 30° i + 8 sin 30° j
b = – 2 cos 45° i + 2 sin 45° j
c = –10 cos 40° i – 10 sin 40° j
r = (8 cos 30° – 2 cos 45° – 10 cos 40°) i
+ (8 sin 30° + 2 sin 45° – 10 sin 40°) j
r ≈ – 2.15i – 1.01j
ii. | r | = 2201.115.2
≈ 2.37 N
tan = 1×012×15
= 25.27° (using exact values for the i
and j components)
magnitude ≈ 2.37 N
direction ≈ 180° + 25.27°
≈ 205.27°
≈ 205° 17’
8
a
Using the cosine rule,
R2 = 42 + 32 – 2 4 3 cos 140°
= 16 + 9 – 24 cos 140°
= 43.38506…
R 6.59 N
Using the sine rule,
sin(40 - q)°
3=
R
140sin
40 – = 17.02354…
= 22.97645…
The resultant force has magnitude 6.59 N and
direction 22° 59'.
c
Using the cosine rule,
R2 = 62 + 72 – 2 6 7 cos 50°
= 36 + 49 – 84 cos 50°
= 31.00584…
R 5.57 N
Using the sine rule,
sin(180 - (q + 30))°
7 =
R
50sin
180 – ( + 30)= 74.36726…
= 75.63273…
The resultant force has magnitude 5.57 N and
direction 75° 38'.
e
Using the cosine rule,
R2 = 102 + 102 – 2 10 10 cos 130°
= 200 – 200 cos 130°
= 328.55752…
R 18.13 N
The resultant force has magnitude 18.13 N and
is in the positive direction of the x axis.
30 6
20 30
7
R
10
25
130 10
R
25
25
Essential Specialist Mathematics Complete Worked Solutions 5
9
a Let a = 3i + 4j
a = 15
(3i + 4j)
b = – 4i + 3j
^b = 1
5(– 4i + 3j)
c = – 2j
c = – j
r = 165
(3i + 4j) + 125
(– 4i + 3j) – 15j
= (485
– 485
) i + (645
+ 365
– 15) j
= 5j
b | r | = 5 N
direction = 90°
10
a resolved part = 12 cos 20° = 11.28 N
b resolved part = 15 cos 65° = 6.34 N
c resolved part = 8 cos 90° = 0 N
d resolved part = 11 cos 145° = – 9.01 N
11
a 8 cos 40° + 12 cos 15° ≈ 17.72 N
b 8 + 12 cos 55° ≈ 14.88 N
12
a Let a = 2i – j and b = 7i + 3j
then a = 1
5(2i – j)
b . a = (7i + 3j) . 1
5(2i – j)
= 1
5(14 – 3)
= 11 5
5
b a
a =
11 55
1
5(2i – j)
= 115
(2i – j)
The component of (7i + 3j) N in the direction
of 2i – j is 115
(2i – j) N.
b Let a = 3i + 4j and b = 2i – 3j
then a = 15
(3i + 4j)
b . a = (2i – 3j) . 15
(3i + 4j)
= 15
(6 – 12)
=
-65
b a
a =
-65
15
(3i + 4j)
=
-625
(3i + 4j)
The component of (2i – 3j) N in the direction of
3i + 4j is
-625
(3i + 4j) N.
13
a 8 + 11 cos 155° ≈ –1.97 N
b 10 + 11 cos 115° ≈ 5.35 N
c 11 + 10 cos 115° + 8 cos 155° ≈ – 0.48 N
14
15 cos 25° + 25 cos 80° + 50 cos 115°
≈ –3.20 N
15
tan = 2×56
= 22.62°
10
155° 115°
11
8
B
10 NC A
6·56
2·5
q
24 N
Essential Specialist Mathematics Complete Worked Solutions 6
a 10 + 24 cos 22.62° = 10 + 24 6
6×5
≈ 32.15 N
b 24 + 10 cos 22.62° = 24 + 10 6
6×5
≈ 33.23 N
16
a
| | OX®
: 3 + 2 cos 50° ≈ 4.29 N
OX®
: 2 sin 50° ≈ 1.53 N
| r | = 2253.129.4
≈ 4.55 N
tan = 1×534×29
= 19.67° = 19° 40’
4.55 N at angle of 19° 40’ to OX®
b
| | OX®
: 10 cos 30° + 5 cos 110° + 2 cos 135°
≈ 5.54 N
OX®
: 10 sin 30° + 5 sin 110° + 2 sin 135°
≈ 11.11 N
| r | = 22)11.11()54.5(
= 14.154
≈ 12.42 N
tan ° = 11×115×54
≈ 63.52° = 63° 31’
12.42 N at angle of 63° 31’ to OX®
17
| | 10 N: 10 + 7 cos 50° ≈ 14.50 N
10 N: 5 sin 50° ≈ 5.36 N
| r | = 2236.55.14
≈ 15.46 N
18
a
| | 10 N: 10 + 8 cos 60° = 14 N
10 N: 8 sin 60° – P
8 sin 60° – P = 0
P = 8 sin 60°
= 4 3 N
≈ 6.93 N
b | r | = 142 + 02
= 14 N
19
OX®
: P + 5 sin 35° – 7 sin 35° = 0
P – 2 sin 35° = 0
P = 2 sin 35°
P ≈ 1.15 N
1·53
4·29O X
50°
7 N
10 N
60°
8 N
P N
10 N
Essential Specialist Mathematics Complete Worked Solutions 7
Solutions to Exercise 13B
1
a P = 2 kg 5 m/s = 10 kg m/s
b P = 0.3 kg 0.03 m/s = 0.009 kg m/s
c P = 1000 kg 30 ´ 1000
3600 m/s
= 8333 13
kg m/s
d P = 6 kg 10 m/s = 60 kg m/s
e P = 3000 kg 50 5
18 m/s
= 41 666 23
kg m/s
2
a P = 10(i + j) kg m/s
b
i. P = 10(5i + 12j) kg m/s
ii. | P | = 102(52 + 122)
= 10 25 + 144
= 130 kg m/s
3
change in momentum = final momentum –
initial momentum
a change in momentum
= 10 3 – 10 6
= 30 – 60
= – 30 kg m/s
b change in momentum
= 10 10 – 10 6
= 100 – 60
= 40 kg m/s
c change in momentum
= 10 3 – 10 – 6
= 30 + 60
= 90 kg m/s
4
a 5 kg = 5 g N
= 5 9.8 N
= 49 N
b 3 tonnes = 3 1000 kg
= 3 1000 9.8 N
= 29 400 N
c 60 g = 0.06 kg
= 0.06 9.8 N
= 0.588 N
5
a Resultant force, F = 8 4 = 32 N
b F = ma
a = 1mF =
510
= 12
m/s2
6
a F = ma m = | F || a |
= 102×5
= 4
kg
b | F | = m | a |
= 2 3.5
= 7 N
7
R – mg = ma
96 = ma + mg
96 = m(a + g)
96
a + g = m
m = 96
1.2 + g
m 8.73 kg
Essential Specialist Mathematics Complete Worked Solutions 8
8
(F – 75g)j = ma
a =
F - 75g
75
–1 =
F - 75g
75
– 75 = F – 75g
F = 75g – 75
F = 660 N
9
2.5g – mg = m
. .x
. .x = 2
2.5g = mg + 2m
2×5g
g + 2 = m
m = 2.076 kg
The reading would be 2.076 kg wt.
10
F = ma
= 9 10–31 6 1016
= 5.4 10–14 N
11
F = ma
2i + 10j = 2a
a = i + 5j
12
F = ma
(8i + 2j + (2i – 6j) = 10a
10a = 10i – 4j
a = i – 25
j
13
a
mg – 2·5g = m
m (g – 1) = 2.5g
m =
2 × 5g
g - 1
When at rest the reading is
2 × 5g
g - 1 kg wt
≈ 2.78 kg wt.
b
T – mg = 2m
T = mg + 2m
=
(g + 2) ´ 2 × 5g
g - 1
≈ 3.35 kg wt.
14
u = 50 5
18 =
1259
m/s
v = 0
t = 10
Use v = u + at
a =
v - ut
=
0 - 1259
10
=
-2518
Now m = 25 1000
= 25 000 kg
F = ma
= 25 000
-2518
= – 34 722 29
N
F
75g
2·5g
mg
+ve
2·5g
mg
T
mg
Essential Specialist Mathematics Complete Worked Solutions 9
15
R – mg = ma
R = mg + ma
= m(g + a)
= 10(9.8 + 1.5)
= 113 N
16
F = ma
= 16 0.6i
= 9.6i
and F = F1 + F2 + F3
F3 = F – F1 – F2
= 9.6i – (–10i – 15j) – (16j)
= 19.6i – j
17
u = 5, t = 3, v = 8
Using v = u + at,
a =
v - ut
=
8 - 53
= 1
F = ma
= 5 1
= 5 N
18
F = ma
a = 1m (F)
= 14
((8i + 12j) + (6i – 4j))
= 14
(14i + 8j)
= 72
i + 2j
19
F = 600 – 550 = 50
F = ma
a = Fm = 50
300 = 1
6
u = 0, t = 3
Using v = u + at
v = 0 + 16
3
= 12
The velocity after three seconds is 12
m/s.
20
R – 85g = ma
R = 85 9.8 + 85 – 2
= 85 7.8
= 663
The reaction force is 663 N.
21
a
F = (4g – Fr)i + (R – 10g) j
F = ma
(4g – Fr) i + (R – 10g)j = 10ai
i component: 4g – Fr = 10a 1
j component: R – 10g = 0
R = 10g 2
Fr = R
= 0.2 10g
= 2g 3
Substitute 3 into 1
4g – 2g = 10a
2g = 10a
R
10 g
+ve
R
85 g
+ve
4g
R
0·2R
10g
Essential Specialist Mathematics Complete Worked Solutions 10
a = g
5
a ≈ 1.96 m/s2
b
v = at
= 1.96(10)
= 19.6 m/s
22
Resultant force in direction of motion
= 8000g – 4000g
= 4000g
4000g = 200 103 . .x
. .x =
4000g
200 ´ 103
= g
50
.x =
g
50t + c
but .x = 0 when t = 0
c = 0
Hence .x =
g
50t
30 km/h = 30 000 m/h = 3
25 m/s
hence when x = 3
25:
3
25 =
g
50t
t = 50 ´ 25
3g = 42.517
It takes 42.517 seconds to go from rest to 30 km/h.
23
F = ma
One man: P – FR = 250 0.15
P – FR = 37.5 1
Two men: 2P – FR = 250 0.4
2P – FR = 100 2
2 – 1 gives P = 62.5
Substitute P into 1
62.5 – FR = 37.5
FR = 25 N
Pushing force = 62.5 N, and
Resistance = 25 N
24
(F – 20 000)i = ma
a =
F - 20000200 000
0.2 =
F - 20000200 000
F = 0.2 200 000 + 20 000
F = 60 000 N
Now,
– 20 000 i = ma
a =
-20 000200 000
a = – 110
m/s2
8000g N
R
20 ´ 200g
200 ´ 103g N
P
N
FR
250g
F
200 000g
20 000 N
N
20 000
Essential Specialist Mathematics Complete Worked Solutions 11
25
a = 0 since velocity is constant
10 – R = 10 0
= 10R
Now R – 10g = 0
R = 10g
= 10
10g = 1
g
= 5
49
26
a
FR = R
= 0.025 0.1g
= 0.0245 N
b
– 0.0245i = ma
a =
-0 × 02450 ×1
= – 0.245 m/s2
u = 10, t = 20
Using v = u + at
= 10 – 0.245 20
= 5.1 m/s
27
(24 – R) i + (R – 4g)j = ma i
R – 4g = 0
R = 4g
24 – 4g = 0
4g = 24
= 6g ≈ 0.612
28
a F = ma
(T – 200g) j = ma j
a = 0 m/s
T – 200g = 0
T = 200g N
T = 1960 N
b (T – 200g) j = ma j
ma = T – 200g
200 0.5 = T – 200g
T = 200g + 100 = 2060 N
29
a
For a smooth surface, = 0
(10 – 0) i + (R – 5g)j = 5a i
10 = 5a
a = 2 m/s2
b
R = 5g (20 – FR) i = ma i
20 – 0.3 5g = 5a
a =
20 - 1 × 5g
5
= 1.06 m/s2
10
R
10g
mR
R
0·1g
0·025R
T
200g
20 N
R
FR
5g
4g
24 N
R
R
Essential Specialist Mathematics Complete Worked Solutions 12
Solutions to Exercise 13C 1
i : mg sin 45° i = ma i
ma = mg sin 45°
a = g sin 45°
= 9×8
2
≈ 6.93 m/s2
2
i : (– R + mg sin 45°) i = ma i
– R + mg
2 = ma 1
j : R = mg cos 45°
R = mg
2 2
Substitute 2 into 1
– mmg
2 +
mg
2 = ma
a = 2
gg
a = 2
)1( g m/s2
3
j : R = 60g cos 60°
= 30g
i : (FR – 60g sin 60°) i = ma i
ma = FR – 60g sin 60°
60 –8 = FR – 60g sin 60°
FR = 60g sin 60° – 60° 8
= 29.223 N
4
a
i component: 10 cos 30° = 5a
5 3 = 5a
a = 3 m/s2
b
vertical: R + 20 sin 30° = 5g
R = 5g – 10
horizontal: (20 cos 30° – FR) i = ma i
ma = 20 cos 30° – FR
5a = 20 3
2 – 0·3(5g – 10)
a =
10 3 - 1 × 5g + 3
5
= 1.124 m/s2
5
vertical: R + F sin 45° = 2g
R = 2g – F
2
=
4g - 2F
2
jR
mg45°
i
jmR
R
mg45°
i
j
FR
R
60g
60°
i
Essential Specialist Mathematics Complete Worked Solutions 13
horizontal: (F cos 45° – R) i = 2a
( 2F2
– 12
(
4g - 2F
2)) = 2a
2a =
2 2F - 4g + 2F
4
8a = 3 2 F – 4g
F = 8a + 4g
3 2
= 8(g4 ) + 4g
3 2
= 6g
3 2
= 2g
2
= 2 g N
6
vertical:
R + 30 sin 30° – 20g = 0
R = 20 9.8 – 30 12
= 196 – 15
= 181 N
7
Resolving in the j direction:
R = mg cos
Resolving in the i direction:
P – R – mg sin = ma
a = 1m (P – mg cos – mg sin )
= Pm – g cos – g sin
8
Resolving in the i direction:
– mg sin 30° = ma = ma
–mg
2 = ma
a = –g
2
a = –g
2 i
9
Resolving in the j direction:
R – mg cos 60° = 0
R – mg
2 = 0 = 0
R = mg
2
Resolving in the i direction:
mg sin 60° – R = ma
mg 3
2 – 0.8
mg
2 = ma
g 3
2 –
4g
10 = a
a = 10
4g35g
= 10
)435 g( m/s2
= 4.57 m/s2
s = 5, u = 0, v = ?
Use v2 = u2 + 2as
v = 557.420
= 6.76
speed = v = 6.76 m/s
Essential Specialist Mathematics Complete Worked Solutions 14
10
j : R = mg cos 20°
i : – 0.25R – mg sin 20° = ma
– 0.25(mg cos 20°) – mg sin 20° = ma
a = g(– sin 20° – 0.25 cos 20°)
= – 0.577g
≈ – 5.65 m/s2
Using v2 = u2 + 2as,
s =
2(10)
2( 5.65)
= 8.84 m
When retuning:
i : 0.25R – mg sin 20° = ma
0.25(mg cos 20°) – mg sin 20° = ma
a = g(0.25 cos 20° – sin 20°)
a = –1.05 m/s2
Using v2 = u2 + 2as
= 2 –1.05 – 8.84
v = 58.18
= 4.31 m/s
11
If tan =4
3 , then sin =
5
4
mg sin i + (R – mg cos ) j = ma
a
ma = mg sin
a = g sin
= 4g
5
≈ 7.84 m/s2
u = 0, s = x,
Using v2 = u2 + 2as
= 2 4g
5 x
= 8gx
5
v = 8gx
5
= 8gx
5
5
5
= 40 gx
25
= 40 gx
5
= 25
10gx m/s
b
Resolving vertically:
R – mg = 0
R = mg
Resolving horizontally:
0.3R = ma
0.3mg = ma
a = 0.3g
= 2.94 m/s2
So,
u = 2 10gx
5
a = –2.94 m/s2
v = 0
s = ?
using v2 = u2 + 2as
s =
v2 - u2
2a
jR
0·25R
20°
i
mg
j
R 0·25R
20°
i
mgR
mg
0·3R
Essential Specialist Mathematics Complete Worked Solutions 15
s =
–4
25 10 gx
–6g
10
s = 8gx
5
10
6g
s =8x
3 m
12
a Resolve perpendicular to the plane:
F sin + R = Mg
Resolve parallel to the plane:
F cos – R = Ma
a = 1M
(F cos – (Mg – F sin ))
= FM
(cos + sin ) – g
b
Resolve perpendicular to the plane:
F sin + Mg = R
Resolve parallel to the plane:
F cos – R = Ma
a = 1M
(F cos – (F sin + Mg))
= FM
(cos – sin ) – g
13
sin = 1
20
a
Resolve in the i direction
(FR – 1000g sin ) = 0
FR – 1000g
20 = 0
FR – 50g = 0
FR = 50g
= 490 N
b
(F – mg sin – FR) i + (R – mg cos ) j = ma i
F – FR – mg sin = ma
F – 1000g
20 –
1000g
20 = 1000 (a = 1)
F = 1000 + 1000g
20 2
= 1980 N
14
sin = 35
cos = 45
a
Resolve perpendicular to the plane:
N = 0.5g cos
= 0.4g
Resolve parallel to the plane:
–3N8
– 0.5g sin = 0.5 . .x
–8
3
5
2g –
g
2
5
3 = 1
2. .x
–3g
20 –
3g
10 = 1
2. .x
–9g
10 =
. .x
d ( 1
2 v2 )
dx = –
9g
10
R
Mg
F
mR q
R
Mg F
mR
q
R
1000gq
F
N
0·5gq
3N8
Essential Specialist Mathematics Complete Worked Solutions 16
12
v2 = 6 – 3g
10x
12
v2 = –9gx
10 + c
When x = 0, v = 6
18 = c
12
v2 = –9gx
10 + 18
When v = 4, –10 = –9gx
10
i.e. 10 ´ 109g
= x
1009g
= x
x 1.13 m
b
When v = 0, –9gx
10 = –18
x = 20g
it goes 20g metres up the plane
i.e. when t = 0, x = 20g , v = 0
Resolve parallel to the plane: 3N8
– 0·5g sin = 0·5 . .x
8
3
5
2g –
g
2
5
3 = 1
2. .x
3g
20 –
3g
10 = 1
2. .x
–3g
10 =
. .x
. .x = –
3g
10
d ( 1
2 v2 )
dx = –
3g
10
12
v2 = –3g
10x + c
When x = 20g , v = 0
0 = –3g
10 20
g + c
c = 6
When x = 0, v2 = 12
i.e v = 12
= 2 3 m/s
3.46 m/s
15
Resolve in the j direction:
10 – 8 cos 60° = 6
Resolve in the i direction:
8 cos 30° – P
F = (8 cos 30 – P)i + 6j
| F | 2 = 48 – 8 3 P + P2 + 36
= 84 – 8 3 P + P2
For | F | = m | a |
84 – 8 3 P + P2 = 100
P2 – 8 3 P – 16 = 0
P = 8 3 ± 192 + 64
2
= 8 3 ± 16
2
= 4 3 + 8 or 4 3 – 8
= (4 3 + 8) N
P = 4 3 + 8 is the required force.
16
(F – R5
– 5g sin 30°) i + (R – 5g cos 30°) j = mai
a
R – 5g cos 30° = 0
R = 5g 3
2
and
F – R5
– 5g sin 30 = ma
N
0·5gq
3N8
O
+ve
Essential Specialist Mathematics Complete Worked Solutions 17
F – g 3
2 –
5g
2 = 5 1.5
F = 5 1.5 + g 3 + 5g
2
= 40.49 N
b
Resolve perpendicular to the plane:
F cos 60° + R = 5g cos 30°
R = 5g 3
2 – F
2
= 5g 3
2 – 1
2(15
2 +
g 3 + 5g
2)
=
9g 3 - 5g
4 –
154
= g
4(9 3 – 5) –
154
Resolve parallel to the plane:
F cos 30° – 15
R – 5g cos 60° = 5 . .x
. .x = 1.22 m/s2
30°5g
F
30°
R
(152
+ g 3 + 5g
2)
32
– g
20(9 3 – 5) –
1520
– 5g
2 = 5
. .x
Essential Specialist Mathematics Complete Worked Solutions 18
Solutions to Exercise 13D
1
a 10g – T = 10a 1
T – 8g = 8a 2
1 + 2 gives 2g = 18a
a = 2g
18
= g
9
Substitute into 2
T – 8g = 8g
9
T = 8g + 8g
9
= 80g
9
87.1 N
b a = g
9 m/s2 1.09 m/s2
2
a
F = 10 N
F = ma
10 = 11a
a = 1011
0.91 m/s2
b
F – S = ma
10 – S = 6 1011
S = 10 – 6011
T = S = 4.55 N
3
a
T – 1.5g = 1.5a 1
2g – T = 2a 2
1 + 2 gives 0.5g = 3.5a
a = 0×5g
3×5
= g
7
From 2 2g – T = 2(g
7)
T = 2g – 2g
7
= 12g
7
= 16.8 N
b
a = g
7 =
9×87
= 1.4 m/s2
4
a
Mg – T = M
Mg – M = T
M (g – 1) = T
M =
Tg - 1
1
10g8g
T T
R
11g
F
S
R
6g
F
1·5 kg
1·5g2g
T T
2 kg
Mg25°5g
T
T
R
Essential Specialist Mathematics Complete Worked Solutions 19
T – 5g sin 25° = 5
T = 5g sin 25° + 5
= 5(g sin 25° + 1) 2
From 1 M =
5(g sin 25° + 1)
g - 1
= 2.92 kg
b
From 2 T = 5(g sin 25° + 1)
= 25.71 N
5
a
8g – T = 8a
a = g – T8
and T = 4a
a = g – 4a8
= g – a2
3a2
= g
a = 2g
3 =
9815
m/s2
b
T = 4a = 4 9815
= 26 215
N
6
a
4g – T = 4a
a = g – T4
and T – 2g sin 30° = 2a
T = g + 2a
= g + 2(g – T4
)
= g + 2g – T2
3T2
= 3g
T = 23
3g
= 2g
= 19.6 N
b
a = g – T4
= g – 19×6
4
= 4.9 m/s2
7
a
10g sin 30° – T = 10a 1
T – 5g sin 45° = 5a 2
1 + 2 gives
10g sin 30° – 5g sin 45° = 15a
5g – 5g 2
2 = 15a
10g - 5g 2
2 = 15a
a =
10g - 5g 2
30
=
2g - g 2
6
≈ 0.96 m/s2
b
T – 5g sin 45° = 5
2g - g 2
6
T = 5g sin 45° + 5
2g - g 2
6
≈ 39.4 N
T
8g
R
T4g
4g30°2g
T
T
R
T
5g30°
10g45°
R2
R1T
Essential Specialist Mathematics Complete Worked Solutions 20
8
i : (T – 20)i = ma i
T – 20 = 5 0.8
T = 24 N
j : mg – T = ma
T = m(g – 0.8)
24 N = 9m
m = 2.67 kg
9
a
T – 750 = ma
T – 750 = 5000 2
T = 10 750 N
b
40 000 – T – FR = 10 000a
40 000 – 10 750 – FR = 20 000
FR = 40 000 – 10 750 – 20 000
= 9250 N
10
T = 37.5 N
xg – T = xa 1
T – 3g = 3a 2
From 1 xg – 37.5 = xa
x(g – a) = 37.5
x =
37 × 5g - a 3
From 2 a = 37×5
3 – g
a = 2.7 m/s2 4
Substitute 4 into 3
x =
37 × 5g - 2 × 7
x = 5.28 kg
11
Engine:
Truck:
a
engine:
(60 000 – 40 000g sin – T) i = ma
60 000 – 18
(40 000g) – T = 40 000a
a = 140000
(60 000 – 5000g – T)
truck:
T – 18
(8000g) = 8000a
T = 8000a + 1000g
Substitute T into a:
a = 140000
(60 000 – 6000g – 8000a)
a = 64
– 6
40g – a
5
6a5
= 3020
– 3
20g
a =
30 - 3g
24
a = 0.025 m/s2
b
T = 8000(0.025) + 1000g
T = 10 000 N
750 N
R2
5000g
T TR1
10000g
40000 N
FR
3 kgx kg
T T
q
R60000 N
T40000g
ji
engine
q
RT
8000g
Essential Specialist Mathematics Complete Worked Solutions 21
12
a
5g – T = 5a 1
T = 20a 2
1 + 2 gives
5g = 25a
a = g
5 m/s2
T2 = 8a
= 8 g
5
= 5
8 g
= 15.68 N
b
T = 20a
= 20 g
5
= 4g N
= 39.2 N
c
a = g
5 m/s2
= 1.96 m/s2
13
u = 0, s = 3, t = 3
s = ut + 12
at2
s – ut = 12
at2
a =
2(s - ut )
t2
=
2(3 - 0)
9
= 69
m/s2
= 23
m/s2
0.2g – T = ma
1.96 – T = 0.2 23
T = 13775
N
and T – R = ma
13775
– 0.5g = 0.5 23
= (13775
– 0.5 23
) ÷ 0.5g
= 0.305
14
a
For A:
Resolve parallel to the plane:
T – R – 4g sin 30° = ma 1
Resolve perpendicular to the plane:
R – 4g cos 30° = 0
R = 2 3 g 2
Substituting 2 into 1 gives
T – 2 3 g – 2g = 4 3
For B:
6g – T = 6
T = 6(g – 1) 4
Substitute 4 into 3
6(g – 1) – 2 3 g – 2g = 4
4g – 10 = 2 3 g
=
4g - 10
2 3g
= 0.86
b
T = 6(g – 1) = 52.8 N
T
5g
R2
T12g
R1
8g
T2
a
mRT
0·2g
R
T0·5g
A
6g30°4g
T
TR
mR B
Essential Specialist Mathematics Complete Worked Solutions 22
15
a
4.2g sin – T = ma
4.2g 0.6 – T = 4.2 2
T = 2.52g – 8.4
= 16.296 N
b
T – R2 = ma
16.296 – 3g = 3 2
3g = 10.296
= 0.35
T
3g
4·2g
R2
R
mR2
q
T
Essential Specialist Mathematics Complete Worked Solutions 23
Solutions to Exercise 13E
1 F = (10 – t)2
using F = ma
10 a = (10 – t)2
a =(10 – t)
2
10
dv
dt=
(10 – t)2
10
v =1
10
(10 – t)2 dt
v = –1
30 (10 – t)
3+ c
When t = 0 , v = 0 :
c =100
3
v = –1
30 (10 – t)
3+
100
3
When t = 10 , v =100
3 m/s
i.e. the velocity after 6 seconds is100
3 m/s
Now,
x =
–1
30(10 – t)
3+
100
3
dt
x = –1
30
(10 – t)4
4 –1 +
100
3 t + d
x =1
120 (10 – t)
4+
100
3 t + d
When t = 0 , x = 0 :
d = –250
3
x =1
120 (10 – t)
4+
100
3 t –
250
3
When t = 10 , x = 250 m
Hence the distance travelled is 250 m
2
a F = 10 sin ( t)
using F = ma
5a = 10 sin ( t)
a = 2sin ( t)
v =
2sin ( t) dt
v = –2 cos ( t) + c
When t = 0 , v = 4 :
c = 6
v = –2 cos ( t) + 6
Now,
x =
(–2 cos ( t) + 6) dt
x = –2 sin ( t) + 6 t + d
When t = 0 , x = 0 :
d = 0
x = –2 sin ( t) + 6 t
b F = 10 + 5x
5a = 10 + 5x
a = 2 + x
d
dx 1
2 v
2
= 2 + x
1
2 v
2=
(2 + x) dx
v2
=
(4 + 2x) dx
v2
= 4x + x2
+ c
When v = 4 , x = 0 :
c = 16
v2
= 4x + x2
+ 16
v = 4x + x2
+ 16
When x = 4 , v = 48 = 4 3 m/s
Essential Specialist Mathematics Complete Worked Solutions 24
c F = 10 cos2(t)
2
2
2
2
2
2
5 10 cos ( )
2 cos ( )
2 cos
2 cos ( )
a t
a t
d xt
dt
dxt dt
dt
Using the identity 2 21cos ( ) 1 cos
2t t
2(1 cos ( )
1sin(2 ) c
2
dxt dt
dt
dxt t
dt
When 0, 0 :dx
t vdt
0
1sin(2 )
2
c
dxt t
d t
So,
2
1sin(2 )
2
1 1cos(2 )
4 2
x t t dt
x t t d
When 0, 0 :t x
2
2
1
4
1 1 1cos(2 )
4 2 4
12 cos(2 ) 1
4
d
x t t
x t t
3 F =100
( t + 5)2
6a =100
( t + 5)2
a =50
3( t + 5)2
dv
dt=
50
3( t + 5)2
v =50
3
1
( t + 5)2 dt
v =50
3
( t + 5)–1
–1 + c
v = –50
3( t + 5)+ c
When t = 0 , v = 10 :
c =40
3
v = –50
3( t + 5)+
40
3
When t = 10 , v =110
9 m/s
i.e. the velocity after 10 seconds is
100
9 m/s
Now,
x =
–50
3( t + 5)+
40
3
dt
x = –50
3 log e( t + 5) +
40
3 t + d
When t = 0 , x = 0 :
d =50
3 log e(5)
x =50
3 log e
5
t + 5
+40
3 t
and when t = 10 :
x =50
3 log e
1
3
+400
3
x =
400
3–
50
3 log e(3) m
Essential Specialist Mathematics Complete Worked Solutions 25
4 F = 1 – sin t
4
m = 1
using F = m a
a = 1 – sin t
4
dv
dt= 1 – sin
t
4
v =
1 – sin t
4
dt
v = t + 4cos t
4
+ c
When t = 0 , v = 0 :
c = –4
v = t + 4cos t
4
– 4
So,
x =
t + 4cos t
4
– 4
dt
x =1
2 t
2+ 16 sin
t
4
– 4 t + d
When t = 0 , x = 0 :
d = 0
x =1
2 t
2+ 16 sin
t
4
– 4 t
5 F = 1 – cos 1
2 t
a using F = ma
m = 1
a = 1 – cos 1
2 t
v =
1 – cos 1
2 t
dt
v = t – 2sin 1
2 t
+ c
When t = 0 , v = 0 :
c = 0
v = t – 2sin 1
2 t
b If v = t – 2sin 1
2 t
then
x =
t – 2sin 1
2 t
dt
x =1
2 t
2+ 4cos
1
2 t
+ d
When t = 0 , x = 0 :
d = –4
x =1
2 t
2+ 4cos
1
2 t
– 4
6 F = 12 t – 3t2
using F = m a
4a = 12 t – 3 t2
a = 3 t –3
4 t
2
v =
3 t –3
4 t
2
dt
v =3
2 t
2–
1
4 t
3+ c
When t = 0 , v = 2 :
c = 2
v =3
2 t
2–
1
4 t
3+ 2
When t = 4 , v = 10 m/s
Hence the velocity after 4 seconds is
10 m/s
7 F =t
t + 1
Using long division t
t + 1= 1 –
1
t + 1
Using F = ma
m = 1
a = 1 –1
t + 1
v =
1 –1
t + 1
dt
v = t – log e( t + 1) + c
Essential Specialist Mathematics Complete Worked Solutions 26
When t = 0 , v = 0 :
c = 0
v = t – log e( t + 1)
When t = 10 , v = 10 – log e(11 ) 7.6 m/s
Hence the velocity after 10 seconds is
7.6 m/s
8 F = e
–t
2
a using F = ma
0.5 a = e
–t
2
a = 2e
–t
2
v =
2e
–t
2 dt
v = –4 e
–t
2+ c
When t = 0 , v = 0 :
c = 4
v = –4 e
–t
2+ 4
v = 4
1 – e
–t
2
m/s
b
c
distance travelled
=
0
30
(–4 e–0.5 t
+ 4) dt
=
8e–0.5 t
+ 4 t
30
0
= 8e–15
+ 120 – 8
112 m
Alternatively, using CAS to determine the
distance travelled in the first 30 seconds
we have
9 F ( t) =
14 – 2 t
100 t–2
0 t 5
t > 5
as F = ma
a =F
m=
F
10
a( t) =F ( t)
10=
1.4 – 0.2 t
10 t–2
0 t 5
t > 5
Note that the signed area under an a-t
graph gives change in velocity. This
concept can be used to determine the
speed of the body when t = 10.
a Sketch the a-t graph
Essential Specialist Mathematics Complete Worked Solutions 27
t1 2 3 4 5 6 7 8 9 10 11
a
0.2
0.4
0.6
0.8
1
1.2
1.4
Change in velocity
=1
2(0.4 + 1.4 ) 5 +
5
10
10 t–2
dt
= 4.5 +
–10 t–1
10
5
= 4.5 + 1
= 5.5
Alternatively, using CAS we have
When t = 0, v = 0.
When t = 10 , v = 0 + change in velocity
v = 5.5
Speed = |v | = 5.5 m/s
b
distance travelled can be found by
determining the area under the v-t graph.
For 0 t 5:
v =
(1.4 – 0.2 t) dt
v = 1.4 t – 0.1 t2
+ c
From part a, when t = 0, v = 0:
c = 0
v = 1.4 t – 0.1 t2 for 0 t 5
For t > 5:
v =
10 t–2
dt
v = –10
t+ d
From part a, when t = 5, v = 4.5:
d = 6.5
v = –10
t+ 6.5 for t > 5
Hence,
v( t) =
1.4 t – 0.1 t2
–10
t+ 6.5
0 t 5
t > 5
for t [0, 10 ], v(t) 0
Hence the distance travelled in the first 10
seconds
=
0
5
(1.4 t – 0.1 t2) dt +
5
10
–10
t+ 6.5
dt
= 275
6– 10 log e(2) m (using CAS)
Essential Specialist Mathematics Complete Worked Solutions 28
10 F = kv
using F = ma
m a = kv
a =kv
m
dv
dt=
kv
m
dt
dv=
m
kv
t =m
k
1
v dv
t =m
k log e(v) + c
When t = 0 , v = u :
c = –m
k log e(u)
t =m
k log e
v
u
kt
m= log e
v
u
e
kt
m=
v
u
v = ue
kt
m
So,
x = u
e
kt
m dt
x =um
k e
kt
m + d
When t = 0 , x = 0 :
d = –um
k
x =um
k e
kt
m–
um
k
1m
kt
ek
umx metres
11 F = – kv
using F = ma
m a = – kv
a = –kv
m
v dv
dx= –
kv
m
dv
dx= –
k
m
v = –k
m
1 dx
v = –k
m x + c
When x = 0 , v = V :
c = V
v = –k
m x + V
v = V –k
m x
12 F = b – cv
using F = ma
m a = b – cv
a =b – cv
m
dv
dt=
b – cv
m
dt
dv=
m
b – cv
t = m 1
b – cv dv
t = –m
c log e(b – cv) + d
When t = 0 , v = 0 :
d =m
c log e(b)
t =m
c log e
b
b – cv
ct
m= log e
b
b – cv
e
ct
m=
b
b – cv
Essential Specialist Mathematics Complete Worked Solutions 29
b – cv =b
e
ct
m
= b e
–ct
m
cv = b – b e
–ct
m
cv = b
1 – e
–ct
m
m
ct
ec
bv 1 m/s
Terminal velocity is the limiting velocity
as t
= b
c
1 – e–
=b
c (1 – 0)
=b
c
Hence the terminal velocity is c
b m/s
13
When the body is projected upwards:
using F = ma
2
2
22 2
m g kva
m
dv m g kvv
dx m
dv m g kv
dx m v
kv2
mg positive
2
dx mv
dv mg kv
2
2
2
2
log2
e
m kvx dv
k m g kv
mx m g kv c
k
When 0, :x v u
2
2
2
log ( )2
log2
e
e
mc m g kv
k
m m g kux
k m g kv
Maximum height is reached when 0v 2
2
log2
log 1 (1)2
e
e
m mg kux
k mg
m kux
k mg
Hence the maximum height reached = 2
log 12
e
m ku
k m g
Now take the highest point as the origin.
kv2
using F = ma
2
2
Fa
m
m g kva
m
dv m g kvv
dx m
mg
positive
Essential Specialist Mathematics Complete Worked Solutions 30
2
2
2
2
2
log ( )2
e
dx m v
dv m g kv
m kvx dv
k m g kv
mx m g kv c
k
At highest point 0, 0 :x v
2
log ( )2
log (2)2
e
e
mc m g
k
m m gx
k m g kv
The object returns to the point of
projection when (1) = (2).
m
2k log e
1 +ku
2
m g
=m
2k log e
m g
m g – kv2
1 +ku
2
m g=
m g
m g – kv2
m g + ku
2
m g=
m g
m g – kv2
m g
m g + ku2
=m g – kv
2
m g
(m g )
2
m g + ku2
= m g – kv2
kv2
= m g –(m g )
2
m g + ku2
kv2
=m g (m g + ku
2)
m g + ku2
–(m g )
2
m g + ku2
kv2
=m gku
2
m g + ku2
v2
=m gu
2
m g + ku2
v =m gu
2
m g + ku2
v = um g
m g + ku2
Speed = |v | = um g
m g + ku2
14 F =4
x
using F = m a
0.2 a =4
x
a =20
x
d
dx 1
2 v
2
=20
x
1
2 v
2=
20
x dx
1
2 v
2= 20 log e(x) + c
When v = 0 , x = 1 :
c = 0
1
2 v
2= 20 log e(x)
v2
= 40 log e(x)
v = ± 40 log e(x)
v = 40 log e(x)
(as the body is moving on the positive x
axis). As required.
15
vF
25
50 Usea
a =50
25 + v
dv
dt=
50
25 + v
dt
dv=
25 + v
50=
1
2+
v
50
t =
1
2+
v
50
dv
t =1
2 v +
v2
100+ c
When t = 0 , v = 0 :
c = 0
t =1
2 v +
v2
100
Essential Specialist Mathematics Complete Worked Solutions 31
When t = 50 , v = 25 + 25 = 50
Hence, when t = 50 , v = 50 as required
b a =50
25 + v
v dv
dx=
50
25 + v
dv
dx=
50
25 v + v2
dx
dv=
25 v + v2
50=
1
2 v +
1
50 v
2
x =
1
2 v +
1
50 v
2
dv
x =1
4 v
2+
1
150 v
3+ c
When x = 0 , v = 0 :
c = 0
x =1
4 v
2+
1
150 v
3
When v = 50 , x = 625 +2500
3=
4375
3 m
Hence when v = 50 the distance from
O to P is 4375
3 m
1000 use 50For
2v
Ft c
v dv
dx= –
v2
1000
dv
dx= –
v
1000
dx
dv= –
1000
v
x = –1000
1
v dv
x = –1000 log e(v) + c
When v = 50 , x =4375
3:
c =4375
3+ 1000 log e(50 )
x = 1000 log e
50
v
+4375
3
When v = 25 ,
x = 1000 log e(2) +4375
3 2151.48 m
Hence when v = 25 the distance from
O to P is 2151.48 m
Essential Specialist Mathematics Complete Worked Solutions 32
Solutions to Exercise 13F
1
a
PQR = 180° – 120° = 60°
RPQ = 180° – 80° = 100°
b
QPR = 180° – 100° = 80°
RQP = 180° – 120° = 60°
c
QPR = 180° – 115° = 65°
PRQ = 180° – (90° + 65°) = 25°
2
a
b
P2 = 22 + 52
P = 29 N
c
1
2tan(180 )
5
2180 tan
5
180 21.80
= 158.20 °
Hence the angle the P N force makes with the 5
N force is 158.20 °
3
a
b
P2 = 72 – 52
P = 24
P = 62 N
c
1
5cos(180 )
7
5180 cos
7
180 44.42
= 135.58 °
Hence the angle the 5 N force makes with the 7
N force is 135.58°
4
a
b
Using the cosine rule,
102 = P2 + P2 – 2P2 cos 140°
100 = 2P2 (1 – cos 140°)
P2 = 28.31
P = 5.32 N
R
QP
100°
60°
20°
R
Q
P80°
60°
40°
R
QP
65° 25°
5
P2
180°– q
180°– q
P
7
5
P
10Q
140°
Essential Specialist Mathematics Complete Worked Solutions 33
180 – = tan–1
2
5
180 – = 21.80 °
5
a
Firstly, ° = 360° – (115° + 80°)
= 165°
Qsin 165°=
Psin 115° =
5sin 80°
Q = 5
sin 80° sin 165°
Q = 1.31 N
P = 5
sin 80° sin 115°
= 4.60 N
Hence, N 31.1 N, 60.4 QP
b
Qsin 160°
= 5
sin 130° = P
sin 70°
Q = 5
sin 130° sin 160°
= 2.23 N
P = 5
sin 130° sin 70°
= 6.13 N
Hence, N 23.2 N, 13.6 QP
6
Using the cosine rule,
P2 = 102 + 102 – 2(10)(10) cos 130°
P2 = 328.56
P = 18.13 N
7
Using Lami’s theorem,
Psin 120°
= 5g
sin 140°
P = 66.02 N
8
Along the force 10 N, the sum of the resolved
parts in newtons is
10 + 5 cos 120° + 5 3 cos 210° = 0
Along the force 5 N, the sum of the resolved
parts in newtons is
5 + 10 cos 120° + 5 3 cos 90° = 0
Along the 5 3 N force, the sum of the
resolved parts in newtons is
5 3 + 10 cos 210° + 5 cos 90° = 0
the particle is in equilibrium.
80°
115°q°
P N
Q N
5 N
130°
70° q°
P N
Q N
5 N
50°
P N
10 N10 N
130°
10
P
10
Essential Specialist Mathematics Complete Worked Solutions 34
9
a
The resultant force, P, of the two 10 N forces is
shown above.
The direction of the resultant of the two forces
occurs along the bisector of the angle between
the forces. (this is the rhombus property)
b
P + 10 cos 155° + 10 cos 155° = 0
P – 18.13 = 0
P = 18.13 N
c
18.13 N is the magnitude of P at an angle of
155° with each 10 N force.
10
Along the force P N, the sum of the resolved
parts in newtons is
P + Q cos 80° + 23 cos 135° = 0 1
Perpendicular to the force P N, the sum of the
resolved parts in newtons is
23 cos 45° + Q cos 190° = 0 2
From 2 , Q =
-23 cos 45°cos 190°
3
Substitute 3 into 1
P = – Q cos 80° – 23 cos 135°
= 23 cos 45° ´ cos 80°
cos 190° – 23 cos 135°
P = 13.40 N
Q = 16.51 N
11
Resolve in the direction of Q
Q + 10 cos 60° + 15 cos 150° = 0
Q + 5 + 15 (–3
2) = 0
Q = (15 3
2 – 5) N
≈ 7.99 N
Resolve in the direction of P
P + 15 cos 120° + 10 cos 210° = 0
P + 15 (– 12
) + 10 –3
2 = 0
P = 15 + 10 3
2
≈ 16.16 N
12
a
b
Using the cosine rule,
102 = 162 + 82 – 2(8)(16) cos (180° – )
cos (180° – ) = 64
55
= 180° – 30.75° = 149.25°
50°
P N
10 N10 N
155° 155°10 N
120°15 N
P N
60°Q
10 N
16 N8 N
q
180°– q
8
10
16
Essential Specialist Mathematics Complete Worked Solutions 35
13
a
b
Using the cosine rule,
P2 = 32 + 52 – 2(3)(5) cos 80°
P2 = 28.79
P = 5.37 N
c
Using the cosine rule,
32 = 52 + 5.372 – 2(5)(5.37) cos (180° – )
cos (180° – ) = 0.835
= 180° – 33.41° = 146.59°
14
Note:
cos = 3
5
sin =4
5 Resolve horizontally
T cos (90° – ) = P cos (15°)
i.e. T sin = P cos (15°) 1
Resolve vertically
T cos = P cos 75° + 2g 2
From 1
T = P cos (15°)
sin q
= 5P cos (15°)
4
Substitute T into 2
5P cos (15°)
4
35
= P cos (75°) + 2g
P [ 3 cos 15°4
– cos (75°)] = 2g
P (
3 cos 15° - 4 cos 75°4
) = 2g
i.e. P =
8g
3 cos 15° - 4 cos 75°
≈ 42.09 N
T = 5P cos 15°
4
=
5 cos 15° ´ 8g
4(3 cos 15° - 4 cos 75°)
≈ 50.82 N
15
Method 1: Resolving forces
Resolve horizontally
5g = T1 cos + T2 cos
5g = T1 1213
+ T2 5
13 1
Resolve vertically
T1 cos = T2 cos
i.e. T1 5
13 = T2 12
13
i.e. T1 = 125
T2
Substitute in 1
5g = 125
1213
T2 + 5T213
65g = (144 + 25)T2
5
325169
= T2
T1 = 125
325169
g
T1 = 60g
13
≈ 45.23 N
P N
3 N5 N
q
100°
180°– q 5
P
3
80°
2g
P
Tq1·5
2
2·5
75°
105°90 – qq
Essential Specialist Mathematics Complete Worked Solutions 36
Method 2: Triangle of forces
T1 = 5g sin
= 5g 1213
= 60g
13
≈ 45.23 N
16
a
Perpendicular to P N
10 cos 140° + 12 cos (220° – ) = 0
cos (220° – ) =
-10 cos 140°12
= 220° – cos–1 (0.638)
= 169.67°
b
Along the force P N
P + 10 cos 50° + 12 cos (169.67° + 50°) = 0
P = –10 cos 50° – 12 cos 219.67°
= 2.81 N
5g
T1
q T2
Essential Specialist Mathematics Complete Worked Solutions 37
Solutions to Exercise 13G 1
a Limiting frictional force is Fmax = R
Fmax = 0.3 10 g = 3g N Fmax = 29.4 N
An applied force of 10 N generates a frictional
force of 10 N.
b An applied force of 30 N overcomes the
limiting friction force,
Fmax = 3g N = 29.4 N
c An applied force of 40 N overcomes the
limiting friction force,
Fmax = 3g N = 29.4 N
2
Resolving perpendicular to the plane:
R + Psin 40 ° – 1.2 g = 0 R = 1.2 g – Psin 40 °
N 2 If Pa then R = 1.2 g – 2sin 40 °
and limiting frictional force,
F max = 0.2 (1.2 g – 2sin 40 °) Fmax = 2.09 N
Resolving parallel to the plane:
2cos 40 ° – F = 0 F = 2cos 40 °
F = 1.53 N
Hence an applied force of 2 N generates a
frictional force of 1.53 N.
b If P = 3 N
then R = 1.2 g – 3sin 40 °
and limiting frictional force,
F max = 0.2 (1.2 g – 3sin 40 °) Fmax = 1.97 N
Resolving parallel to the plane:
3cos 40 ° – F = 0 F = 3cos 40 °
F = 2.30 N
Hence an applied force of 3 N overcomes the
limiting frictional force.
F = 1.97 N
3
a Fmax = R Fmax = 0.3 100 g Fmax = 30 g N
Hence the greatest horizontal force that can be
applied without moving the particle is 30g N.
R
P
1.2 g
F 40°
R
F
100g
Essential Specialist Mathematics Complete Worked Solutions 38
b
Resolving perpendicular to the plane:
R + Psin 60 ° – 100 g = 0 (1)
Resolving parallel to the plane:
Pcos 60 ° – F = 0 F = Pcos 60 °
R =1
2 P
0.3 R =1
2 P
R =5
3 P (2)
Substitute (2) into (1):
5
3 P + Psin 60 ° – 100 g = 0
P 5
3+
3
2
= 100 g
P = 386.94 N
Hence the magnitude of the greatest pulling
force that can be applied without moving the
particle is 386.94 N.
c
Resolving perpendicular to the plane:
R + Psin 60 ° – 100 g = 0 (1)
Resolving parallel to the plane:
– F – Pcos 60 ° = 0 F = – Pcos 60 °
0.3 R = –1
2 P
R = –5
3 P (2)
Substitute (2) into (1):
–5
3 P + Psin 60 ° – 100 g = 0
P
–5
3+
3
2
= 100 g
P = –1224.02 N
Hence the magnitude of the greatest pushing
force that can be applied without moving the
particle is 1224.02 N.
4
Resolving perpendicular to the plane:
R – 5cos 30 ° = 0
R =5 3
2
Resolving parallel to the plane:
5sin 30 ° – F = 0
F = 5sin 30 °
F = 2.5 N
100g
P
R
F 60°
100g
P
R
F60°
30°
5
RF
Essential Specialist Mathematics Complete Worked Solutions 39
5
Resolving parallel to the plane:
sin 20 ° – F = 0 F = sin 20 °
Resolving perpendicular to the plane:
R – cos 20 ° = 0 R = cos 20 °
As F R
F
R
sin 20 °
cos 20 °
tan 20 °
Hence the particle will slide down the plane
when < tan 20 °
6
Resolving perpendicular to the plane:
R – wcos = 0 R = wcos
Resolving parallel to the plane:
wsin – F = 0 wsin = F wsin = 0.4 R wsin = 0.4 wcos tan = 0.4
= tan–1
(0.4 ) = 21.80 °
7
a Resolving perpendicular to the plane:
R – 3cos 25 ° = 0 R = 3cos 25 °
Resolving parallel to the plane:
3sin 25 ° – F = 0 F = 3sin 25 °
3cos 25 ° = 3sin 25 °
= tan 25 °
= 0.47 (correct to two decimal places)
20°
1
RF
°
w
RF = 0.4
25°
3 kg wt
RF
Essential Specialist Mathematics Complete Worked Solutions 40
b
As the particle is in limiting equilibrium,
P = F + 3sin 25 ° kg wt P = 3tan 25 ° cos 25 ° + 3sin 25 ° kg wt P = g(3tan 25 ° cos 25 ° + 3sin 25 °) N P = 9.8 2.5357 N P = 24.85 N
Hence a force of 24.85 N or more, when
applied up the plane, will cause the particle to
move.
8
On the point of moving down:
Resolving perpendicular to the plane:
R – 24 gcos = 0 R = 24 gcos
Resolving parallel to the plane:
24 gsin – F – 100 = 0 F = 24 gsin – 100 (1)
On the point of moving up:
Resolving perpendicular to the plane:
R = 24 gcos
Resolving parallel to the plane:
120 – F – 24 gsin = 0 F = 120 – 24 gsin (2)
a (1) + (2):
2F = 20 F = 10 N
b
Substitute F = 10 into (1):
24 gsin = 110
sin =55
12 g
= sin–1
55
12 g
27.88 °
c
As F = R
=F
R
=10
24 gcos = 0.0481019 . . . 0.05
25°
3 kg wt
RP
F
°
24g
R
F
100 N
°
24g
R
F
120 N
Essential Specialist Mathematics Complete Worked Solutions 41
Solutions to Exercise 13H
1
F1 = 2i and F2 = – 3j
resultant force F = F1 + F2 = 2i – 3j
a
Use F = ma
2i – 3j = a = x ..
b
magnitude of acceleration
| a | = 4 + 9 = 13
c
x ..
= 2i – 3j
x . = 2ti – 3tj + c
When t = 0, x . = 0
c = 0
x . = v = 2ti – 3tj
d
speed = | x .
| = 4t2 + 9t2 = t 13
When t = 1, speed = 13
e
velocity gives direction of motion ≈ 303.69°
2
F = 4i + 6j
a
F = ma implies 2a = 4i + 6j
a = 2i + 3j
b
r ..
= 2i + 3j
r . = 2ti + 3tj + c
When t = 0, r . = 0
c = 0
r .
= 2ti + 3tj
c
r = t2i + 3 t 2
2j + c1
As r = 0 when t = 0, c1 = 0
r = t2i + 3 t 2
2j
d
x = t2 and y = 3 t 2
2
y = 3 x
2, x ≥ 0
3
r (t) = 5t2i + 2(t2 + 4)j
a
When t = 0, r (0) = 8j
b
x = 5t2, y = 2t2 + 8
5
2 xt
y = 2 x
5 + 8, x ≥ 0
c
F = ma
r . = 10ti + 4tj
r ..
= 10i + 4j
F = 2(10i + 4j)
= 20i + 8j N
4
r (t) = 5(5 – t2)i + 5(t2 + 2) j
a
r (0) = 25i + 10j
b
x = 5(5 – t2) and y = 5(t2 + 2)
x = 25 – 5t2
5t2 = 25 – x
t2 =
25 x
5
Essential Specialist Mathematics Complete Worked Solutions 42
y = 5(
25 x
5 + 2)
= 25 – x + 10
= 35 – x
Also x = 5(5 – t2) x ≤ 25
y = 35 – x for x ≤ 25
c
r .
= –10ti + 10t j
r ..
= –10i + 10j
F = m r ..
= 5(–10i + 10j)
= – 50i + 50j N
5
F1 = 2i + j and F2 = i – 2j
resultant force F = F1 + F2 = 3i – j
a
Using F = ma
3i – j = 2a
3
2i – 1
2j = a
acceleration is 3
2i – 1
2j m/s2
b
r ..
= 3
2i – 1
2j
r .
= 3
2t i – t
2j + c
The particle starts at rest and therefore c = 0
r .
= 3
2t i – t
2j
the velocity is 3
2t i – t
2j m/s
c
r = 3
4t2 i –
t 2
4j + c2
When t = 0, r = 2i – 2j
c2 = 2i – 2j
r = (3
4t2 + 2) i – (
t 2
4 + 2)j
6
a
For the acceleration
r ..
= 3
)3(927 jiji
27i + 9j = 3i + j + 3 r ..
r ..
= 1
3(24i + 8j)
r ..
= 8i + 3
8j m/s
2
b
i.
F = m r ..
=
ji
3
8810
= ji3
8080 N
ii.
| F | =
2
2
3
8080
= 80
310 N
7
r (t) = 2t2i + (t2 + 6) j
a
x = 2t2 and y = t2 + 6
y = x2
+ 6 as 2t2 ≥ 0 for all t, x ≥ 0
y = x2
+ 6 for x ≥ 0
b
r . (t) = 4ti + 2tj
Essential Specialist Mathematics Complete Worked Solutions 43
c
speed = | r .
(t) | = 16t2 + 4t2
= 20t2
= 2t 5
2t 5 = 16 5
852
516 t
Hence, the speed of the particle is 16 5 m/s
after 8 seconds.
d
r ..
(t) = 4i + 2j
F = 2(4i + 2j)
= 8i + 4j N
8
F = 1
10(15i + 25j)
a
Using F = m r ..
1
2(3i + 5j) = 10 r
..
1
20(3i + 5j) = r
..
b
r .
(t) = t
20(3i + 5j) + c
r .
(0) = 3i + 5j
r .
(t) = t20
(3i + 5j) + 3i + 5j
= (3 t
20 + 3) i + ( t
4 + 5) j
c
r (t) = (3 t 2
40 + 3t) i + (
t 2
8 + 5t) j + c1
When t = 0, r (t) = 0i + 0j
c1 = 0
r (t) = (3 t 2
40 + 3t) i + (
t 2
8 + 5t) j
r (6) = ( 3 36
40 + 3 6) i + ( 6 2
8 + 5 6) j
= ji2
69
10
207
= 20.7i + 34.5j
d
x = 3 t 2
40 + 3t, y = t
2
8 + 5t
= 3
40(t2 + 40t) = 1
8(t2 + 40t)
t2 + 40t = 40 x
3
thus y = 1
8( 40 x
3)
= 5 x
3
and
t2 + 40t = 40 x
3
t2 + 40t + 400 – 400 = 40 x
3
400
2)20(t = 40 x
3
0 ,203
120040
3
120040
xxt
30 x
Hence,
y =
5x
3 , x – 30
9
y = 3x
r . = 5i + a j a
5 = 3 as velocity gives direction of motion
a = 15
r . = 5i + 15j
The speed in the direction of the y axis is 15
m/s.
The speed of the particle = | r . | = 22
155
= 250
= 5 10 m/s.
Essential Specialist Mathematics Complete Worked Solutions 44
Chapter review: multiple-choice questions 1 P = 3(6 i + 8 j ) P = 18 i + 24 j
|P | = (18 )2
+ (24 )2
|P | = 900 |P| = 30
Answer is D
2 R – mg = ma R = ma + mg R = 10 4 + 10 g R = 40 + 10 g R = 138 N
Answer is E
3
R = (8)2
+ (6)2
R = 100 R = 10 N
Answer is B
4
For the 5 kg mass:
5g – T = 5a (1)
For the 3 kg mass:
T – 3g = 3a (2)
(1) + (2):
8a = 2g
a =g
4 m/s
2
Answer is B
5
*Note the positioning of θ
Resolving in the j direction: N + Tcos – mg = 0
Answer is D
6
Resolving down and parallel to the plane: mgsin = ma a = gsin
a =4g
5 m/s
2
Answer is B
8 N
6 N
R
mg
NT
mg
N
= 0
Essential Specialist Mathematics Complete Worked Solutions 45
7
Using the cosine rule:
R2
= 102
+ 102
– 2(10 )(10 )cos (120 °)
R2
= 200 – 200 cos (120 °)
R2
= 300
R = 10 3 N
Answer is B
8 Resolving perpendicular to the plane: N + Tsin – W = 0 N = W – Tsin Resolving parallel to the plane: Tcos – F = 0 F = Tcos N = Tcos
=Tcos
N
=Tcos
W – Tsin
Answer is B
9
When an object moves in a circle, it is
constantly changing direction. Because of this
direction change the body is accelerating. This
acceleration is directed inwards towards the
centre of the circle. And according to Newton’s
second law of motion a body experiencing
acceleration must also be experiencing a net
force. This force is referred to as the centripetal
force.
Therefore if the external resultant force on a
body is zero then the body cannot be moving. in a circle.
Answer is B
10
Resolving perpendicular to the plane: R = 9g Resolving parallel to the plane: 54 – FR = ma 54 – R = 18 R = 36
=36
9g
=4
g 0.41
Answer is C
10
10
R
120°
54 NF R
9g
R
Essential Specialist Mathematics Complete Worked Solutions 46
Chapter review: short answer questions
1 mass of man = 75 kg
mass of the lift = 500 kg
acceleration of the lift = 2 m/s2
acceleration due to gravity = g m/s2
a The force R exerted by the floor on the man is
given by
R – 75 g = 75 2
or R = 75(g + 2) N
= 885N
b total mass of the lift and the man = 575 kg
T – Mg = Ma
i.e. T – 575 g = 575 2
T = 575(g + 2) N
= 6785 N
2 For the 3 kg mass,
T – 3g = 3a (Newton’s 2nd law) 1
For the 5 kg mass,
5g – T = 5a 2 a Adding 1 and 2 gives 2g = 8a
a =
1
4 g m/s2
b 5 1 gives 5T – 15g = 15a
3 2 gives 15g – 3T = 15a
Subtracting yields 8T – 30g = 0
T = 30
8 g
= 15
4 g (g = acceleration due to gravity)
3 m = mass of the skier
Resolving in i direction: R – mg sin = ma 1
Resolving in j direction: R = mg cos 2
Substituting 2 into 1 gives
mg cos – mg sin = ma
a = g(sin – cos )
R
mg
T
3 kg 5 kg
T T
mg
i
j R R
Essential Specialist Mathematics Complete Worked Solutions 47
4 By Newton’s second law,
a 100 – R = ma
i.e. 100 – 0.4(10g) = 10a
a = (10 – 0.4g) m/s2
b If another block of mass 10 kg is placed on top of the first one, then
m = 20 kg
100 – (20g) = 20a
a = (5 – 0.4g) m/s2
5 m = 5 kg, F = 20
(t + 1)2 N at t seconds
a F = ma
at t seconds, a = 20
(t + 1)2 1
5
= 4
(t + 1)2 m/s2
b velocity v = a
1
1dt
= 4
(t + 1)2
1
1dt
= 4(t + 1)–2
1
1dt
= 4(t + 1)1
1 + c1
= 4
t + 1 + c1
Since the body starts from rest, v (0) = 0
4
t + 1 + c1 = 0
or c1 = 4
v = 4
t + 1 + 4
= 4t
t + 1 m/s
c displacement d = v
1
1dt
= 4
t + 1 + 4
1
1dt
= – 4 loge (t + 1) + 4t + c2, t + 1 0
At t = 0, d = 0,
– 4 loge (0 + 1) + 4(0) + c2 = 0
c2 = 0
d = (4t – 4 loge (t + 1)) m
6 mass of the car m = 1000 kg
initial velocity u = 60 km/h
mg
R
R 100 N
Essential Specialist Mathematics Complete Worked Solutions 48
= 60
3.6 m/s
final velocity v = 24
3.6 m/s
t = 5 s
a = v u
t
= 24
3.6 60
3.6
5
= 24 60
3.6 5
= – 2 m/s2
the retarding force using F = ma is
1000 – 2 = – 2000 N
i.e., the retarding force acting in a direction opposing motion is 2000 N.
7 Since the body is in limiting equilibrium, all the
forces balance each other. Resolving mg in the
directions of i and j, we see that mg sin acts
along the inclined plane and mg cos acts
perpendicular to it. for equilibrium, R = mg cos
and R = mg sin ( = coefficient of friction)
= mg sin
R
= mg sin
mg cos
= tan
Then the inclination of the plane is increased to
the body will begin to slide down.
Let the acceleration be a.
mg sin – tan mg cos = ma
(since = tan and R = mg cos )
mg sin – sin
cos mg cos = ma
a = g sin cos g cos sin
cos
= g
cos sin( – )
8 When the car is cruising down the slope, the velocity
is constant.
Let resistance = R,
a = 0
a 1000 g sin – R = ma = 0
R = 1000 g 1
20
= 50g N
= 490 N
R
mg
R
i j
R
mg
R
N
1000g
R
Essential Specialist Mathematics Complete Worked Solutions 49
b maximum acceleration = 1 m/s2
Let the force be F when going up.
F – mg sin – R = ma
F = 1000g 1
20 + 50g + 1000 1
= (1000 + 100g) N
= 1980 N
9 The forces acting on the parcel as the conveyor belt
moves upwards are as shown.
a At the instant it is about to slip down,
a = g
4
R – mg sin 30° = ma R = mg cos 30°
mg 3
2 – mg
1
2 = m
g
4
3
2 =
1
4 +
1
2
= 3
4
2
3
= 3
2
b When the belt is stopped, the parcel is moving at 7 m/s.
Now frictional force R acts down the plane and the parcel slides up some distance
before coming to stop.
mg sin 30° + R = ma
i.e. ma = mg
2 +
3
2 mg
3
2
a = g
1
2 +
3
4
= 5
4 g
Now using v2 – u2 = 2as, find the distance s.
0 – 49 = 2
5g
4 s
s = 49 2
5g
= 98
5g m
= 2 m
10 The maximum possible tension is 400 kg wt.
a When the particle is hauled upwards,
T – 320g = 320a
i.e. 400g – 320g = 320a
80g = 320a
a = 80g
320
= g
4 m/s2
maximum acceleration= g
4 m/s2.
mg 30°
R R
mg
T
Essential Specialist Mathematics Complete Worked Solutions 50
b If a particle of mass 480 kg is to be lowered by the same rope, the maximum tension is
again 400 kg wt.
Now 480g – T = 480a
or 480g – 480a = T
480g – 480a ≤ 400g
i.e. 80g ≤ 480a
a ≥ 80g
480
i.e. a ≥ g
6 m/s2
11 Given that F = 3 + 6x
F = ma implies 3 + 6x = 3x..
1 + 2x = x..
1 + 2x = d(
1
2 v2 )
dx since a =
d( 1
2 v2 )
dx
Integrating both sides:
x + x2 + c = 1
2 v2
When x = 0, v = 2
c = 2
1
2 v2 = x + x2 + 2
v = 2(x + x2 + 2)
When x = 2, v = 2(2 + 4 + 2)
= 16
= 4 m/s
12 m = 3 kg, F = 3i + 6j N, v (0) = i + 2j
a F = ma implies a = 3i + 6j
3
= i + 2j
b i a = i + 2j
v (t) = t i + 2t j + c1
Now v (0) = i + 2j,
i + 2j = c1
v (t) = t i + 2t j + i + 2j
= (t + 1)i + (2t + 2)j
= (t + 1)(i + 2j)
ii Speed = | v |
= (t + 1) 1 + 4
= 5 (t + 1)
Essential Specialist Mathematics Complete Worked Solutions 51
c v (t) = (t + 1)(i + 2j)
r (t) =
t2
2 + t (i + 2j) + c2
Since the particle is initially at the origin, r (0) = 0,
c2 = 0
i.e. r (t) =
t2
2 + t (i + 2j)
d The equation of the straight line in which the particle moves is given by
r = k(i + 2j), k ≥ 0
or y = 2x, x 0
13 Using s = ut + 1
2 at2, t = 20 and s = 500,
500 = 20u + 200a
25 = u + 10a 1
Using s = ut + 1
2 at2, t = 50 and s = 1000,
1000 = 50u + 1250a
20 = u + 25a 2
1 2 gives 5 = 15a
a = 1
3 (which shows deceleration)
Substituting in 1 , 25 = u 10
3
u = 85
3
Using v = u + at, and a = 1
3 , u =
85
3 and t = 50,
v = 85
3
1
3 50 =
35
3
Using v2 = u2 + 2as, and a = 1
3 , u =
35
3 and v = 0,
0 =
35
3
2
2 1
3 s
s = 1225
9
3
2 = 204
1
6
The train will travel a further 204 1
6 kilometres before coming to rest.
14 a = v u
t
= 15 0
60
= 1
4 m/s2
F = ma
= 9000 1
4
= 2250 N
Essential Specialist Mathematics Complete Worked Solutions 52
15 The train is travelling with uniform velocity on level ground. As it begins the ascent, the
initial velocity is u = 20 m/s.
F = R, where F = force of engine, R = resistance.
Also, on the incline, N = mg cos , since N and mg cos balance each other.
F – R – mg sin = ma
3
50 mg = ma
or a = 3
50 g
Now v2 = u2 + 2as
gives 0 = (20)2 – 2 3
50 g s
s = 50 202
6g
= 10 000
3g m
16 Let the mass of the body in the lift be m kg and the force
of the lift on the body be F.
Then F – mg = mf
(since the lift moves upwards)
i.e. F = m (g + f) N
17 Initial velocity of the bullet = 200 m/s
At s = 10 cm (= 0.1 m), its velocity is 0
Using v2 – u2 = 2as
gives 0 – 2002 = 2 a 0.1
a = 2002
2 0.1
= –200 000
= –2 105 m/s
If the board is 5 cm thick, then s = 0.05 m
and v2 – 2002 = 2 –2 105 0.05
v2 = 2002 – 2 2 105 0.05
= 20 000
v = 20 000 since v 0
= 100 2 m/s
18 Let the mass of the body in the lift be m kg.
When the lift accelerates upwards,
T – mg = ma
10g – mg = ma 1
mg
N
R F R
F
N0
mg
F
mg
T
mg
Essential Specialist Mathematics Complete Worked Solutions 53
When the lift accelerates downwards,
mg – T = m (2a)
mg – 7g = 2ma 2
a 2 – 2 1 gives
– 27g + 3mg = 0
m = 27g
3g
= 9 kg
weight of the particle is 9 kg wt.
b 1 + 2 gives 3g = 3ma
a = 3g
3m
= g
9 m/s2
19 By Newton’s second law,
for particle A, m1g – T = m1a 1
and for particle B, T – m2g = m2a 2
a Adding 1 and 2 gives
m1g – m2g = m1a + m2a
a = (m1 m2)g
m1 + m2 m/s2
b From 1 and 2 , m1g T
m1 =
T m2g
m2
m1m2g – m2T = m1T – m1m2g
T(m1 + m2) = 2m1m2g
T = 2 m1 m2 g
m1 + m2 N
20 The forces involved are represented in the sketch.
Friction can be neglected as the surface is smooth.
We have T = m2a 1
and m1g – T = m1a 2
a Adding 1 and 2 gives
m1g = (m1 + m2)a
a = m1 g
m1 + m2 m/s2
b Therefore A is pulled along the table towards the pulley with acceleration a.
Now T = m2a
= m1 m2 g
m1 + m2 N
T T
A B
m2g
T
T
A
B
m1g
m2g
R
m1g
Essential Specialist Mathematics Complete Worked Solutions 54
21 Given that m1 > m2, then the tension will cause A to
be pulled up the plane.
Both A and B will move with the same acceleration,
say a.
For A, T – m2g sin = m2a 1
For B, m1g – T = m1a 2
a Adding 1 and 2 gives
m1g – m2g sin = (m1 + m2)a
a = g(m1 m2 sin )
m1 + m2
particle A will move up the plane with an acceleration
a = g(m1 m2 sin )
m1 + m2 m/s2
b From 1 and 2 we have
T m2g sin
m2 =
m1g T
m1
m1T – m1m2g sin = m1m2g – m2T
T(m1 + m2) = m1m2g + m1m2g sin
T = m1 m2 g (1 + sin )
m1 + m2 N
22 Resolving the weight mg along the plane and
perpendicular to the plane, we get:
R = mg cos
and by Newton’s second law:
mg sin – R = ma
ma = mg sin – mg cos
a = g(sin – cos )
23 Friction can be neglected as the table is smooth.
For B, 6g – T = 6a 1
For A, T = 10a 2
a Adding 1 and 2 gives
6g = 16a
a = 3
8 g m/s2
b From 2 we get T = 30
8 g
= 15
4 g N
A
T
T R
B
m1g m2g
mg
R
R
Essential Specialist Mathematics Complete Worked Solutions 55
c Two forces acting on the pulley are shown in the
diagram. Since the forces are equal in magnitude, they
are represented by the sides of an isosceles triangle.
Further they are at right angles.
By the triangle law, the resultant force is
T2 + T2 = 2T
in the direction of 45° to the horizontal, i.e., 15 2
4 g N.
d Using s = ut + 1
2 at2, 1 = 0 +
1
2
3
8 g t2
t2 = 16
3g
t = 4
3g since t 0
0.74 s
e From part d, it takes B, and also A, 4
3 g seconds to travel 1 metre. Now as B has the
reached the floor, there is no longer any tension in the string, so A will travel the next 1
metre with constant velocity (there are now no horizontal forces acting on A as the
horizontal table is smooth).
For the constant velocity, use 2 2
2v u as where 3
0, and 18
u a g s
2 30
4
3
2
v g
gv
Now using 3
12
2
3
s vt
gt
tg
So the time it takes A to reach the edge is
4 2 6
3 3 3g g g
24 Friction can be neglected for the smooth surface.
For particle A, forces along the surface are 10g sin
and T in opposite directions.
Let a m/s2 be the acceleration of A down the plane.
10g sin – T = 10a 1
For B, T – 3g = 3a 2
a If = 60°, 1 + 2 give
45°
T
T
T
45° 2T
T
A
3g 10g
T
T R
B
Essential Specialist Mathematics Complete Worked Solutions 56
10 sin 60° – 3g = 13a
a = 10g sin 60 3g
13
= g(5 3 3)
13 m/s2
b T – 3g = 3a
T = 3g + 3 g(5 3 3)
13
= 3g(5 3 + 10)
13 N
25 Given that
mA = 5 kg, mB = 3 kg, = 0.2
The forces acting on each body are shown in
the diagram.
For A, T – R = 5a
i.e. T – 0.2 5g = 5a
T g = 5a 1
For B, 3g – T = 3a 2
a 1 + 2 gives 3g – g = 8a
a = 2g
8
= g
4 m/s2
b B is hanging 1 metre from the ground.
Using v2 – u2 = 2as, we can find v when s = 1.
v2 – 0 = 2 g
4 1
v2 = g
2
or v = g
2 m/s
The velocity of A as B reaches the ground is g
2 m/s towards the pulley.
c Once B reaches the ground, T = 0. However, A continues to move on with
R N acting opposite to the direction of motion.
Now for A, u = g
2 , v = 0 and a =
R
m
= 0.2 5g
5
= – 0.2g
Using v2 = u2 + 2as, 0 = g
2 + 2 – 0.2g s
0.4s = 1
2
A R
T
3g
R
T 5g
B
1 m
Essential Specialist Mathematics Complete Worked Solutions 57
or s = 1
0.8
= 5
4 m
26 If the two forces each equal to P N act at a point, inclined at 120°, we can use the triangle law
to find the resultant.
It can be seen that the forces make an angle of 60° in
the triangle. But the two forces being equal make the
remaining two angles of the triangle equal 60°.
Thus, we have an equilateral triangles and hence the
resultant R must also have a magnitude of P N.
Alternative method
As before, we can resolve along the direction of one
force and a perpendicular direction.
This gives
P + P cos 120° = P(1 + cos 120°)
and P cos 30°
R = P2(1 + cos 120)2 + P2 cos2 30°
= P 1
= P
27 The body is in limiting equilibrium. Thus the
resolved part of its weight along the plane is just
balanced by the friction R.
i.e. 10g sin 30° = R
= 10g cos 30°
= 10g sin 30
10g cos 30
= tan 30°
If the plane is now raised to an incline of 60°, the
body will start sliding down the plane. Let F be the
force required to prevent sliding.
Then 10g sin 60° = R + F
= 10g cos 60° + F
F = 10g sin 60° – tan 30° . 10g cos 60°
= 10g[sin 60° – tan 30° cos 60°]
= 10g
3
2 –
1
3
1
2
= 10g
3 1
2 3
= 10g
3 N
R
P N
60°
P N
R P
P cos 120° + P
P cos 30°
120°
30° 10g
R R
60°
10g
R
R + F
Essential Specialist Mathematics Complete Worked Solutions 58
28 The particle rests in equilibrium.
Let T be the tension in the string, now inclined at
60° to the vertical. Hence, the angles between the
three forces are 90°, 150° and 120° as shown in the
diagram.
By Lami’s theorem:
P
sin 120 =
T
sin 90 =
5g
sin 150
P = 5g
sin 150 sin 120°
= 5g
3
2
1
2
= 5g 3 N
29 The final position of the particle in equilibrium is as
shown in the diagram. From the right triangle,
cos = 2
2.5
= 4
5
= cos1
4
5
The three angles between the forces are 90°, 180 and 90 + .
P
sin (90 + ) =
T
sin 90 =
2g
sin (180 )
P
cos = T =
2g
sin
P = 2g
sin cos
= 10g
3
4
5 since sin =
3
5
= 8g
3 N
and T = 2g
sin
= 10g
3 N
150° O
120°
5g
P N T
2g
P
2·5 m
2 m
O
Essential Specialist Mathematics Complete Worked Solutions 59
30 The forces may be represented by a triangle of
vectors.
T1 = 5g cos
= 5g 5
13
= 25g
13 N
T2 = 5g sin
= 5g 12
13
= 60g
13 N
The tension in the two strings is 25g
13 N and
60g
13 N.
T1
T2
12
5
13
5g
Essential Specialist Mathematics Complete Worked Solutions 60
Chapter review: extended-response questions
1 a 42 – T – 4g = 0
T = 42 – 4g
2.8 N
b 42 – 4g = 4x..
x..
= 42 4g
4
= 0.7 m/s2
T 4g
42 N
c i For constant acceleration, s = ut + 1
2 at2
5 = 0 t + 1
2
21 2g
2 t2
20
21 2g = t2
t = 20
21 2g since t 0
3.78
It takes approximately 3.78 seconds to reach the surface.
ii The velocity of the buoy as it reaches the surface is given by
v = u + at
= 0 + 21 2g
2
20
21 2g
= 5(21 2g) m/s
2.65
The velocity of the buoy is approximately 2.65 m/s when it reaches the surface.
d When the buoy leaves the water it is acting under only gravitational force, therefore use
v2 = u2 + 2as
Now u = 5(21 2g), a = –g, v = 0,
0 = 5(21 2g) – 2gs
s = 5(21 2g)
2g
0.357
The buoy reaches a height of approximately 0.357 metres.
2 a i For the 2.8 kg mass
T2 – 2.8g = 2.8a 1
For the 2.2 kg mass
T1 + 2.2g – T2 = 2.2a 2
For the 3 kg mass
3g – T1 = 3a 3
T1
2.8g
2.2g
3g
T2 T2
T1
Essential Specialist Mathematics Complete Worked Solutions 61
Substitute from 3 in 2 for T1 ,
3g – 3a + 2.2g – T2 = 2.2a
5.2g – T2 = 5.2a 4
Add 1 and 4 ,
T2 – 2.8g + 5.2g – T2 = 8a
2.4g = 8a
0.3g = a
ii Substitute in 3 ,
3g – T1 = 3 0.3g
3g – 0.9g = T1
T1 = 2.1g
b x..
= 0.3g
x. = 0.3gt + c
When t = 0, x. = 0,
c = 0
x. = 0.3gt
When t = 1.5, x. = 0.3 1.5g
= 0.45g
2.8g – T = 2.8a
and T – 2.2g = 2.2a
0.6g = 5a
a = 0.12g
(in the opposite direction to the initial velocity)
2.2g
T T
2.8g Using v2 = u2 + 2as, 0 = u2 – 2 0.12g s
with u = 0.45g s = (0.45g)2
0.24g
= 8.26875
The 2.2 kg mass falls a further distance of 8.27 metres, correct to two decimal places.
3 Let F be the tractive force of the engine.
a i Resistance force for the engine = 50
1000 60 000
= 3000 N
Resistance force for the truck = 30
1000 12 000
= 360 N
Resolving parallel to the plane with the whole
system:
F – 72 000g sin – 3000 – 360 = 0
F = 72 000g 1
200 + 3000 + 360
= 360g + 3360 6888 N
F R2
T
360
R1
3000
T
60 000g
12 000g
Essential Specialist Mathematics Complete Worked Solutions 62
ii Let T N be the tension in the coupling.
Resolving parallel to the plane for the engine:
F – T – 60 000g 1
200 – 3000 = 0
T = (360g + 3360) – 60 000g
200 – 3000
= 60g + 360
948 N
b i F – 72 000g sin – 3000 – 360 = 72 000 0.1
F = 360g + 10 560
14 088 N
ii T = (360g + 10 560) – 60 000 0.1 – 60 000g
200 – 3000
= 60g + 1560
2148 N
4 a i 0.4g – T = 0.4x..
1
T – 0.4 0.2g = 0.2x..
2
1 + 2 gives
0.4g – 0.08g = 0.6x..
x.. =
0.32g
0.6
= 8g
15
5.23 m/s2
R
T
0.4g
R
T 0.2g
ii 0.4g – 0.4 8g
15 = T
T = 14g
75
1.83 N
b Consider the system before it strikes the floor.
v2 = u2 + 2as
v2 = 2 1.5 8g
15
v2 = 8g
5
For the particle on the table, after the 0.4 kg particle hits the floor,
T – R = mx..
T = 0, = 0.4, R = 0.2g,
–0.2g 0.4 = 0.2x..
–0.4g = x..
Essential Specialist Mathematics Complete Worked Solutions 63
Use v2 = u2 + 2as
When v = 0, 0 = 8g
5 – 2 0.4g s
s = 8g
5 0.8g
= 2
The particle goes a further two metres after the 0.4 kg particle hits the floor.
5 a i x..
= –(a + bv2)
v dv
dx = –(a + bv2)
dv
dx =
(a + bv2 )
v
ii From i, dx
dv =
v
a + bv2
Let w = a + bv2, dw
dv = 2bv and
dx
dv =
v
w
x = v
w
1
1dv
= 1
2b 1
w
1
1dw
= 1
2b loge (a + bv2 ) + c, a + bv2 0
When x = 0, v = u,
c = 1
2b loge (a + bu2 )
x = 1
2b (loge (a + bu2 ) – loge (a + bv2 ))
x = 1
2b loge
a + bu2
a + bv2
The train comes to rest when v = 0,
i.e. x = 1
2b loge
a + bu2
a
= 1
2b loge
1 +
bu2
a
b i dv
dt = –(a + bv2 )
dt
dv =
1
a + bv2
= 1
b( a
b + v2 )
= 1
ab
a
b
a
b + v2
t = 1
ab tan–1
b
a v + c
Essential Specialist Mathematics Complete Worked Solutions 64
When t = 0, v = u,
c = 1
ab tan–1
b
a u
t = 1
ab tan–1
b
a u –
1
ab tan–1
b
a v
When v = 0, t = 1
ab tan–1
b
a u
ii If b = 0.005, a= 2 and u = 25,
then t = 1
0.01 tan–1
0.005 25
2
= 10 tan–1 (1.25) 8.96
It takes 8.96 seconds for the train to stop.
6 a mg – 0.02mv2 = mx..
x..
= g – 0.02v2
x..
= g – v2
50 =
50g v2
50
v dv
dx =
50g v2
50
dv
dx =
50g v2
50v
dx
dv =
50v
50g v2
Let w = 50g – v2
dw
dv = –2v
x = –251
w
1
1dw
= –25 loge (50g – v2 ) + c, 50g – v2 0
When x = 0, v = 0,
c = 25 loge (50g)
and x = 25 loge
50g
50g v2
b e
x
25 =
50g
50g v2
v = 50g
1 e
x
25
c v
0 x
v = 50g
Essential Specialist Mathematics Complete Worked Solutions 65
7 a T – 200g sin 60° – 200g cos 60° = 0
200g – 100g 3 – 100g = 0
= 200 100 3
100
= 2 – 3
0.2679
60°
Mg
200g
T T
50 m
b If the value of M is 200, the crate is on the point of moving up the plane. Consider the
crate being on the point of moving down the plane.
Then Mg sin 60° – T – (2 – 3 ) R = 0
Mg 3
2 – 200g – (2 – 3 ) Mg
1
2 = 0
Mg
2 ( 3 – (2 – 3 )) = 200g
Mg
2 (2 3 – 2) = 200g
M = 100( 3 + 1) kg
The crate will remain stationary for 200 ≤ M ≤ 100( 3 + 1)
c i Let M = 150. The mass will move up the plane.
200g – T = 200x.. 1
T – 150gcos 30° – 150g sin 30° = 150x..
200g – 75g 3 – 75g = 350x..
200g – 75g 3 – 75g(2 – 3 ) = 350x..
200g – 150g = 350x..
200g 60°
150g
T T
50g = 350x..
g
7 = x
..
The acceleration is g
7 m/s2.
ii 200g – T = 200x.. (from 1 )
T = 200g – 200 g
7
= 1200g
7
The tension is 1200g
7 N.
iii When the rope breaks, the 200 kg mass has a speed of g
7 2 =
2g
7 m/s and has
travelled 1
2 at2 =
2g
7 metres in the two seconds.
Essential Specialist Mathematics Complete Worked Solutions 66
Using v2 = u2 + 2as, with u = 2g
7, a = g and s = 50 –
2g
7,
v2 =
2g
7
2
+ 2 g
50 –
2g
7
v2 = 4
49 g2 + 100g –
4
7 g2
v 30.54, since v 0
The speed of the 200 kg weight when it hits the ground is 30.54 m/s.
8 a v = 125(1 – e–0.1t )
dv
dt = 12.5e–0.1t
b i P – 20v = 250 12.5e–0.1t
P = 3125e–0.1t + 20v
= 2500(1 – e–0.1t ) + 3125e–0.1t
= 2500 + 625e–0.1t
= 625 (4 + e–0.1t)
ii P = 20v + 3125e–0.1t
= 20v + 3125
1 –
v
125
= 20v + 3125 – 25v
= 3125 – 5v
= 5(625 – v)
iii When v = 20, P = 3025 N
iv When t = 30, P = 2500 + 625e3
2531.117 N
c
0
3125
2500
t
P
9 a Resolving perpendicular to the plane:
T sin + R = Mg cos
R = (Mg cos – T sin
b Resolving parallel to the plane:
Mg
R T
T cos – Mg sin – 0.1R = 0
T cos – Mg sin – 1
10 (Mg cos – T sin ) = 0
T = 0.1 Mg cos + Mg sin
cos + 0.1 sin N
Essential Specialist Mathematics Complete Worked Solutions 67
c i T = Mg (0.1
3
5 + 4
5 )
cos + 0.1 sin
= Mg
5 4.3
cos + 0.1 sin
= 43 Mg
50(cos + 0.1 sin )
= 8.6 g
cos + 0.1 sin N
ii Let y = cos + 0.1 sin
dy
d = –sin + 0.1 cos
dy
d = 0 implies tan =
1
10
= 0.09967c
or = 5.71° = 5°43´
This maximises cos + 0.1 sin , and minimises 8.6g
cos + 0.1 sin .
iii Minimum T = 86 101
101 g N, since sin =
1
101 and cos =
10
101 .
d Resolving the forces produces the same expression to be maximised as before, i.e.,
clearly no effect. So θ = 5°43´
10 If sin = 5
13 , then cos =
12
13 .
a i N = 50g cos
= 50g 12
13
= 600g
13 N
50g
N
ii Friction force = N
= 0.1 N
= 60g
13 N
b 50g sin – 60g
13 = 50x
..
250g
13 –
60g
13 = 50x
..
19g
65 = x
..
The acceleration down the plane is 19g
65 m/s2.
Essential Specialist Mathematics Complete Worked Solutions 68
c i Constant acceleration
Use v2 = u2 + 2as
= 2 19g
65 10
= 76g
13
v 7.569 m/s
The speed of the particle at the bottom of the plane is 7.569 m/s.
ii v = at
76g
13 =
19g
65 t
t 2.64
The time taken to reach the bottom is 2.6 seconds.
d i 300 – 250t + 50g sin – 5g cos = 50x..
300 – 250t + 250g
13 –
60g
13 = 50x
..
x.. 8.86 – 5t
ii Let dv
dt = 8.86 – 5t
50g
F
v = 8.86t – 5t2
2 + c
When t = 0, v = 0,
c = 0
v = 8.86t – 5t2
2
i.e. dx
dt = 8.86t –
5t2
2
x = 8.86t2
2 –
5t3
6 + c
When t = 0, x = 0,
c = 0
x = 4.43t2 – 5
6 t3
When x = 10, 10 = 4.43t2 – 5
6 t3
Using a CAS calculator to find t, the particle reaches the bottom of the slope when
t = 1.865.