2 Force and Motion
6
Chapter 6 Work, Energy and Power
Work, Energy and Power
New Senior Secondary Physics at Work (Second Edition)
Oxford University Press 2013
0
2 Force and Motion
Chapter 6 Work, Energy and Power
Practice 6.1
(p. 213)
The distance
work
2 Force and Motion
Chapter 8 Projectile Motion
Revision exercise 8
9
B
Concept traps (p.321)
Take upwards and rightwards as positive.
1
F
Consider the vertical direction.
The object moves horizontally when it reaches
By sy = uyt +
the maximum height.
2
T
2 Force and Motion
Chapter 7 Momentum
Practice 7.1 (p.268)
1
KE of C =
C
Take the moving direction of the shell as
7
(a)
1
m(2v)2 = 4E
2
No, it does not contradict the law of
positive.
conservation of momentum. The
By conservation of momentum,
momentum of
2 Force and Motion
Chapter 8 Projectile Motion
Practice 8.2 (p.318)
The dog takes 0.441 s to reach its
1
D
maximum height.
2
B
3
D
Time of flight = 2 0.4414 = 0.883 s
(c)
5
Consider an object being projected at an initial
Take upwards and the initial hori
2 Force and Motion
Chapter 7 Momentum
Practice 7.2 (p.280)
much shorter time. The force of impact
1
D
would be much larger.
Take upwards as positive.
(Or other reasonable answers)
Change in momentum = 0.5(6) 0.5(8)
= 7 kg m s
2
(c)
The crumple zones at th
2 Force and Motion
Chapter 10 Gravitation
Practice 10.1 (p.376)
1
B
2
B
3
C
4
B
5
A
m=
= 1850 kg
10
1
2
rY
1
2
rX
(R + H)2 =
=
2
rX
2
rY
=
6.67 10 (7)7
0.2182
2
=
R
(R H )2
= 6.88 108 N
(b)
a=
= 7 9.81
= 68.7 N
(c)
rM
=
rE
=
M E rM
2
M M rE
2
= 5.90 103 N
2 Force and Motion
Chapter 9 Uniform Circular Motion
Practice 9.1 (p.336)
1
C
(d)
2
A
= r 2
3
A
= 25(1.7453 103)2
4
A
= 7.62 105 m s2
Angular speed =
5
(a)
3
= 2 = 0.236 rad s1
t
20
Centripetal acceleration of Mary
= 17.19(1.7453 103)2
= 5.24 105 m s2
v
A
2 Force and Motion
Chapter 5 Moment of a Force
Practice 5.2 (p.198)
1
8
(a)
B
passengers stand on the upper deck. The
Take moment about the pivot.
c.g. may be shifted outside the base of
Sum of clockwise moment
support even if the bus is only tilted
= sum
2 Force and Motion
Chapter 9 Uniform Circular Motion
Practice 9.2 (p.353)
1
B
2
B
(b)
= T sin 20 = 8.35 sin 20 = 2.86 N
(c)
mv 2
= r
mg
centripetal force
weight
= 0.408
=
7
(a)
B
mv 2
,
r
(1.5 sin 20 ) 2.86
=
0.8
rF
=
m
1.35 m s1
2
=
= 0.7854 rad s1
8
Fri
2 Force and Motion
Chapter 8 Projectile Motion
Practice 8.1 (p.304)
1
B
2
D
t = 0.3497 s
Take downwards and rightwards as positive.
Consider the horizontal direction.
s
30
Minimum speed = x =
= 85.8 m
0
.
3497
t
Consider the horizontal direction.
s1
t=
s
2 Force and Motion
Chapter 10 Gravitation
Practice 10.2 (p.385)
1
B
2
D
3
A
7
M=
rD
=
rP
1
2
vD
1
2
vP
4
vP
2
vD
The Earths mass is 6.04 1024 kg.
RvP
vD
v=
8
2
2
GMm mv 2
=
,
r2
r
GM E
RE
2
R
E2
v
2
2
2r
GM = 2
r
g0 R E
v2
9.81(6.37 106 ) 2
=
3000 2
=
r3
2 Force and Motion
Chapter 2 Motion (II)
Practice 2.2 (p.68)
By v2 = u2 + 2as,
1
a=
C
Take the moving direction of the plane as
positive.
By v = u + 2as,
2
2
6
B
7
(a)
2
s=
2
km
By v2 = u2 + 2as,
v 2 u 2 0 152
a=
=
= 0.625 m s2
2 180
2s
The length of the
2 Force and Motion
Chapter 1 Motion (I)
Practice 1.4 (p.28)
1
Magnitude of acceleration =
B
Time take =
s2
60 20
25
9
C
3
A
v = 5 m s1
Its velocity is 5 m s1 downwards.
40 0
Deceleration = 3.6
5
= 2.22 m s
4
10
Magnitude of average acceleration
=
2
D
11
5
2 Force and Motion
Chapter 3 Force and Motion (I)
Practice 3.3 (p.114)
For Q,
1
C
7 T = ma
2
A
(1) + (2),
F = ma = 40 0.5 = 20 N
7 5 = 2ma
3
C
ma = 1 N
1
a
Slope of graph =
=
F m
Net force acting on P = ma = 1 N
Alternative solution:
1
1
Mass =
= 3 1 = 2
2 Force and Motion
Chapter 2 Motion (II)
Practice 2.3 (p.78)
The ball will hit the ceiling after 0.584 s.
1
D
8
Take upwards as positive. Consider the
2
C
upward journey.
3
A
(a)
Consider the downward journey. Take
By v2 = u2 + 2as,
u = v 2 2as
downwards
2 Force and Motion
Chapter 2 Motion (II)
Practice 2.1 (p.57)
(b)
Take the initial moving direction of the
1
A
cat as positive.
2
A
s/m
3
B
5
4
D
0
5
C
6
Car A
1
6
26
t/
s
5
(c)
Take the direction towards the left as
positive.
s/m
30
Car B
0
30
60
15
9
(a)
2 Force and Motion
Chapter 1 Motion (I)
Practice 1.3 (p.20)
1
=
D
Average speed =
4 3.64 1011
365 24 60 60
10
= 46.2 km s1
Magnitude of average velocity
731
=
= 0.937 m s1
(5 8)60
It travels back to the starting position after
each revolution.
Average ve
2 Force and Motion
Chapter 1 Motion (I)
Practice 1.2 (p.12)
= (16 10) 2 12 2 = 13.4 km
1
C
2
D
tan =
3
C
12
16 10
= 63.4
After the first swap, the coin is at A. After the
The total displacement is 13.4 km
second swap, the coin is at C. Therefore, the
S63
2 Force and Motion
Chapter 3 Force and Motion (I)
Practice 3.2 (p.107)
1
C
2
D
3
C
Since the ferry moves at a constant velocity,
the net force acting on it is zero.
Take forwards as positive.
Tf =0
T = f = 2000 N
4
The bowling ball is harder to stop since
2 Force and Motion
Chapter 1 Motion (I)
Practice 1.1 (p.7)
1
B
Maximum percentage error
10 6
=
100% = 1.16 109 %
24 60 60
2
Percentage error
0.005
=
100%
23.38
= 0.0214%
7
(a)
B
From 1 January 2014 to 10 January 2014, the
= 2.49%
watch runs slower than
Concept Q.: RLC Circuit w/ Light
bulb
As I slide the core into the inductor the light bulb
changes brightness. Why?
I am driving the circuit through resonance by
1.
2.
3.
4.
5.
continuously increasing the frequency of
current oscillations in the circuit
c
Concept Question: Capacitor
Consider a circular capacitor,
with an Amperian loop (radius
r) in the plane midway
between the plates. When the
capacitor is charging, the line
integral of the magnetic field
around the Amperian loop (in
direction shown) is
1.
Concept Question: Capacitors
Three identical capacitors are connected to a battery.
The battery is then disconnected.
A
How do the charge on A, B & C
compare before and after the battery
B
C
is removed?
BEFORE;
1.
2.
3.
4.
5.
6.
7.
QA = Q B = QC;
QA = Q B
Concept Question: Dipole in Field
From rest, the coil above will:
1.
2.
3.
4.
5.
6.
7.
8.
rotate clockwise, not move
rotate counterclockwise, not move
move to the right, not rotate
move to the left, not rotate
move in another direction, without rotating
b
Concept Question: LC Circuit
Consider the LC circuit at
right. At the time shown the
current has its maximum
value. At this time
1.
The charge on the capacitor has its maximum value
2.
The magnetic field is zero
3.
The electric field has its maximum value
Concept Question: Magnetic Field Lines
The picture shows the field lines outside a
permanent magnet The field lines inside the
magnet point:
1.
2.
3.
4.
5.
6.
Up
Down
Left to right
Right to left
The field inside is zero
I dont know
1
Concept Question Answ
Concept Question: Biot-Savart
0
The magnetic field at P points towards the
1.
2.
3.
4.
5.
6.
7.
+x direction
+y direction
+z direction
-x direction
-y direction
-z direction
Field is zero (so no direction)
1
Concept Question Answer: BiotSavart
Answer: 3.
Concept Question: Faraday Circuit
A magnetic field B penetrates this
circuit outwards, and is increasing
at a rate such that a current of 1 A
is induced in the circuit (which
direction?).
Th potential
The
t ti l difference
diff
VA
VA-VB
VB iis:
1.
2.
3.
4
Concept Question: Question
The integral expression
B ds
1.
is equal to the magnetic work done around a closed
path
2.
is equal to the current through an open surface
bounded by the closed path.
3.
is always zero.
4.
is equal to the magnetic potential ener
Measuring the Masses of Galaxies
1. Measuring the masses of Galaxies
a. Measure their rotation curves
i. Normal spirals and large ellipticals (about 1011-1012 solar
masses)
ii. Irregular (about 108-1010 solar masses)
iii. Dwarfs (106-107 solar masses)
b.