Part 1
What is a force?
Force and gravity
Other forces
Forces and vectors
Part 2
Forces and equilibrium
Kinematics
Position Vectors
Constant acceleration
Part 3
Newton’s first law
Newton’s second law
Newton’s third law
Mathematical modelling
Practice tests
Practice Test Paper 1
Solution to Test Paper 1
Practice Test Paper 2
Solution to Test Paper 2
Practice Test Paper 3
Solution to Test Paper 3
Part 4
Moments
Equilibrium
Momentum and impulse
Conservation of momentum
Part 5
Conservation of energy
Variable acceleration I
Variable acceleration II
Circular Motion
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Solution to Test Paper 1
From Mechanics
Revision as of 17:58, 18 January 2011 by
Ian
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Solutions
1 (a)
0
2
=
7
2
+
2
(
−
.
8)
s
s
=
49
19
.
6
=
2
.
5
Max Height
=
5
+
2
.5
=
7
.5 m
(3 marks)
1 (b)
0
=
7
−
9
.
8
t
t
=
7
9
.
8
=
0
.
714
s
(3 marks)
(6 marks)
2 (a)
T
−
8
00
9
.
8
=
8
00
0
.
2
T
=
7840
+
160
=
8000 N
(3 marks)
2 (b)
T
−
8
00
9
.
8
=
8
00
(
−
0
.
2)
T
=
7840
−
160
=
7680 N
(3 marks)
2 (c)
T
=
8
00
9
.
8
=
7
840
N
(1 mark)
(7 marks)
3 (a)
(1 mark)
3 (b)
F
=
2
1
.
8
=
3
.
6
N
(2 marks)
3 (c)
R
=
2
9
.
8
=
1
9
.
6
N
(2 marks)
3 (d)
3
.
6
=
1
9
.
6
=
3
.
6
19
.
6
=
0
.
184
(3 marks)
(8 marks)
4 (a)
(1 mark)
4 (b)
R
+
T
sin
3
0
=
2
00
9
.
8
R
+
0
.
5
T
=
1
960
R
=
1
960
−
0
.
5
T
(3 marks)
4 (c)
F
=
T
cos
3
0
T
cos
3
0
=
0
.
6(1960
−
0
.
5
T
)
T
(
cos
3
0
+
0
.
3)
=
1
176
T
=
1176
(
cos
3
0
+
0
.
3)
=
1
010
N
(4 marks)
(8 marks)
5 (a)
5
i
−
2
j
=
4
i
+
3
j
+
1
0
a
a
=
1
10
(
i
−
5
j
)
=
0
.
1
i
−
0
.
5
j
m
s
-2
(3 marks)
5 (b)
r
=
(
4
i
+
3
j
)
t
+
0
.
5(0
.
1
i
−
0
.
5
j
)
t
2
(2 marks)
5 (c)
r
=
(
4
t
+
0
.
05
t
2
)
i
+
(
3
t
−
0
.
25
t
2
)
j
3
t
−
0
.
25
t
2
=
0
t
(3
−
0
.
25
t
)
=
0
t
=
0
or
t
=
3
0
.
25
=
1
2
s
t
=
1
2
s
(3 marks)
5 (d)
v
=
(
4
+
0
.
1
t
)
i
+
(
3
−
0
.
5
t
)
j
3
−
0
.
5
t
=
0
t
=
6
s
(3 marks)
(11 marks)