1.12 a) At t = 0, the object is at x = A. v = B + 2Ct + 3Dt 2.
a = 2C + 6Dt .
b) At t = 0, the object is at x = A .
v = - ωA sin(ω t) .
a = - ω
2A cos(ω t).
1.13 a)
v = c t + v0 .
x = l
2 c t 2 + v0t + x0 .
b) v = -
l
2 kt 2 + v0 .
x = - l
6 kt 3 + v0t + x0 .

v
2
R ,
so if v is doubled, F will be four times as great.
String breaks Object falls to ground
This is a top view. In fact the object
falls to the ground.
b) The molecular forces that hold the wheel together.
c) The gravitational force exerted on the satellite by the earth.
2.5 a)
F = 20.00 N
N
W
Horizontal component of total force = F to the right
Horizontal acceleration = F
m
= 10.00 m. s
-2 .
b) Vertical component of total force = N - W upwards
The vertical acceleration is zero so the vertical component of the total force must be zero,
i. e. N - W = 0 or N = W.
c)
F = 20.00 N
N
W Friction = 10.0 N
Horizontal component of total force = 20.00 N - 10.00 N
= 10.00 N.
Horizontal acceleration = 5.00 m. s
-2.

c)
W
W cosθ
W sinθ
The downhill component is W sinθ .
2.7 a)
F = 20.00 N
N
W
θ = 30°
Horizontal component of total force,
FH = F cos(θ )
= 20.00 N cos(30°)
= 17.32 N .
Vertical component of total force, FV = N + F cos (90°- θ ) - W .
b) Horizontal component of acceleration =
FH
m = 8.66 m. s
-2.
The vertical acceleration is zero because the vertical component of the pulling force,
20.00 N cos (90° - 30°) = 10.00 N, is smaller than the weight, 2.00 × 9.8 N = 19.6 N. The contact
force N adjusts as FV changes and is just sufficient to prevent a downward acceleration.
( ... continued over)
Answers 110
c) If F = 40.00 N in the same direction, the vertical component of the pulling force would be 20.00 N
which is greater than the weight, 19.6 N. The upward acceleration would be 0.2 m. s
-2. The contact
force disappears.
d) Only the horizontal motion would be changed. The horizontal acceleration would be less.
2.8 a) Total force component ↑ = 4.00 N + 8.00 N cos (90° + 30°) = 0 N.
Total force component → = 8.00 N cos 30° = 6.93 N.
b) A single force of 6.93 N in the ← direction or any number of forces which combine to give a single
force equal to this.
2.9 a)
T
θ
θ
T
P
The magnitude of the force on the car equals the tension in the rope, T (and so is the magnitude of
the force on the tree). If the car has not quite started to move, the resultant force on the piece of rope at
the bend in the middle is zero. Taking force components perpendicular to the applied force:
P - T cos(90°-θ ) - T cos(90°-θ ) = 0 .
This gives T = P
2cos(90°-θ )
.
.
3.4 a)
N
x
300 N
500 N
200 N
1.8 m

3.8 a) About 0.75 m. b) About 10 m.
3.9 Buoyant force = weight of displaced air
= density of air × volume × g
= ρa
Vg .
Weight of balloon = density of helium × volume × g
= ρh
Vg .
Downward force required = ρa
Vg - ρ h
Vg = (ρa - ρh)Vg
= 10.8 N.
3.10

-l.
4.2 i) The other vertical forces are the buoyant force and the combined weight of the person and parachute.
ii)
Weight Drag force
Buoyant force
At terminal velocity, the total downward force is zero.
so weight - buoyant force - λv
T
2 = 0
vT = weight - buoyant force
λ
Answers 114
(... continued over)
iii) In the sky diving position: vT = 62 m. s-l ( = 220 km. h-l).
With parachute opened: vT = 5.7 m. s-l (= 20 km. h-l).
iv) Average downward acceleration = 5.7 m. s
- l - 62 m. s
-l
l s
≈ - 0.06 × 103 m. s
-2.

detailed shape is unimportant as long as the acceleration component decreases.)
Answers 115
Downward acceleration
Time
g
The downward velocity component as a function of time looks like this.
A
B
C
D
E
Time
Downward velocity
Since the downward acceleration is continually decreasing, the slope of the velocity curve decreases
with time.
The area ABC in the graph represents the height reached by the object. The area CDE will, of
necessity, equal the area ABC. (Why?)
graph, the time
taken to fall (CE) is greater than the time taken to rise (AC).
4.5 i) As t → ∞, e
-(λ/m )t → 0 so v → mg
λ .
ii) At t = 0 , e
-(λ/m )t = 1 and v = 0 .
iii) Now mg
λ - v = mg
λ e
-(λ/m )t
= e
-l 
  mg 
λ
when t = m
λ .
iv)
mg
λ
v
t
mg
λ e
-1×
m
λ
Answers 116
v) Since d
d t
e
-(λ/m )t = - λ
m e
-(λ/m )t
then a = mg
λ λ
m e
-(λ/m )t = g e
-(λ/m )t .
vi)
a
g
e
m
λ t
vii) mg - λv = mg e
-(λ/m )t
= ma = m
dv
dt .
viii) Terminal velocity occurs when
mg - λv
T = 0 ;
i. e when vT = mg
λ as in (i).
ix) The difference, v
T - v = mg
λ e
-(λ/m )t
x) d a
d t = - λ
m g e
-(λ/m )t = - λ
m a .
∴ m
d a
d t = - λ a .