Year 12 Physics Gradstart
2.1 Basic Vector Revision/
Progress Test
You have 20 minutes to work in a
group to answer the questions on the
Basic Vector Revision/Progress Test
Worksheet
2.2 Vector Components Part 2
2.2 Vector Components Part 2
1a) What is the northerly and easterly
component of the velocity below
E
vN = 35sin28
= 16.43150ms-1
vE = 35cos28
= 30.90317ms-1
2.2 Vector Components Part 2
1b) What is the northerly and easterly
component of the velocity below
vE = – 12sin15
= – 3.10583ms-1
E
vN = 12cos15
= 11.59111ms-1
2.2 Vector Components Part 2
2. Work out the horizontal and vertical
components of the force below:
y
x
Fy = 6sin30
= 3.0N
Fx = 6cos30
= 5.19615N
2.2 Vector Components Part 2
3. Work out the horizontal and vertical velocity
components of the golfball below:
y
vy = 30sin60
= 25.98076ms-1
x
vx = 30cos60
= 15ms-1
2.2 Vector Components Part 2
4(a) Work out the weight into the slope and
down the slope.
Wy = 15cos22
o
= 13.90776N 22
Wx = 15sin22
= 5.61910N
W=mg
= 15N
2.2 Vector Components Part 2
4(b) What is the acceleration down the slope if
it is frictionless.
In x direction
a = ? Fnet = 5.61910N m = 1.5kg
𝑚
Fnet =
𝑎
a=
Wy = 15cos22
o
= 13.90776N 22
Wx = 15sin22
= 5.61910N
W=mg
= 15N
𝐹
𝑚
5.61910
1.5
a=
a = 2.80955
a  2.8 ms-2
2.2 Vector Components Part 2
4(c) What is the normal reaction force on the mass.
N
Wy = 15cos22
o
= 13.90776N 22
Wx = 15sin22
= 5.61910N
W=mg
= 15N
In y direction
N = Wy
N = 13.90776N
N  14N
2.2 Vector Components Part 2
5(a) Work out an expression for the net force
down the frictionless slope below.
Fx = mgsin

W=mg
Wx = mgsin
2.2 Vector Components Part 2
5(b) What is the formula for the acceleration of
a mass down a frictionless slope?
In x direction
a = ? Fnet = mgsin m = m
𝑚
Fnet =
Fx = mgsin

W=mg
Wx = mgsin
𝑎
𝑚𝑔𝑠𝑖𝑛𝜃
𝑚
a=
a = g sin
2.2 Vector Components Part 2
6 If the surface below is frictionless, how long will it take
the mass to reach the base of the slope if it starts from
rest?
t=?
u=0
x = 0.4m
a = g sin20o
= 10sin20o
= 3.40201
x = ut + ½ at2
0.4 = ½ × 3.40201 × t2
0.233904 = t2
0.483637 = t
t = 0.48m
A
Save in memory
A of your
calculator
2.2 Vector Components Part 2
7 If Frmax = 4.0N and a 15N is applied to the 3.0kg block at
30o to the horizontal
(a) What is the acceleration on the mass?
y
x
Fy = 15sin30
= 7.5N
Frmax = 4.0N
Fx = 15cos30
= 12.99038N
B
It is not worth
saving 7.5 in the
memory since it is
so easy to enter in
the calculator
In x direction
Fnet = Fx - Fr
ma = 12.99038 - 4
3a = 8.99038
a = 2.99679
a  3.0 ms-2
2.2 Vector Components Part 2
7 If Frmax = 4.0N and a 15N is applied to the 3.0kg block at
30o to the horizontal
(b) What is the normal reaction on the mass?
y
N
x
Fy = 15sin30
= 7.5N
Frmax = 4.0N
Fx = 15cos30
= 12.99038N
W = 3 × 10
= 30N
In y direction
N + Fy = W
N + 7.5 = 30
N = 22.5N
N  23N
2.2 Vector Components Part 2
8 The 2.0kg mass below is held at rest on the 40o slope
below by friction. What is the magnitude of the
friction holding the mass?
Fr
40o
Wx = 20sin40
W=mg
= 12.85575 N
= 20N
In x direction
Fr = Wx
Fr = 12.85575 N
Fr  13N
2.2 Vector Components Part 2
9 What will be the acceleration of the 1.6kg block
on the 30o slope below if Frmax = 3.0N?
Frmax = 3.0N
40o
Wx = 16sin30
W=mg
=8N
= 16N
In x direction
Fnet = Wx – Fr
ma = 8 – 3
1.6a = 5
a = 3.125
a  3.1 ms-2
2.2 Vector Components Part 2
10 If Frmax = 2.0N and a 12N is applied to the 1.0kg block at
30o to the horizontal
(a) What is the acceleration on the mass?
Frmax = 2.0N
30o
Wy = 10cos30 A
= 8.66025N
Wx = 10sin30
W=mg
=5N
= 10N
In x direction
Fnet = 12 – Wx – Fr
ma = 12 – 2 – 5
1a = 5
a = 5.0 ms-2
2.2 Vector Components Part 2
10 If Frmax = 2.0N and a 12N is applied to the 1.0kg block at
30o to the horizontal
(b) What is the normal reaction on the mass?
N
Frmax = 2.0N
30o
Wy = 10cos30 A
= 8.66025N
Wx = 10sin30
W=mg
=5N
= 10N
In y direction
N = 8.66025N
N  8.7N
2.2 Vector Components Part 2
11 If Frmax = 8.0N and a 20N is applied to the 3.0kg block at
30o to the horizontal
What is the acceleration on the mass?
Frmax = 8.0N
30o
Wx = 30sin30
W=mg
= 15 N
= 30N
In x direction
The force up the slope
is not enough to
overcome friction and
the weight force down
the slope (max = 23N)
so the mass will be
stationary.
2.2 Vector Components Part 2
12
A 55kg cyclist is riding up a 4.0o slope at a constant speed
of 3.0ms-1 ?
(a) What is the net force on the cyclist?
Since the cyclist is travelling
at constant speed the net
force is zero.
2.2 Vector Components Part 2
12
A 55kg cyclist is riding up a 4.0o slope at a constant speed
of 3.0ms-1 ?
(b) What is the horizontal force of the rear wheel
propelling the cyclist up the hill if there is no rolling
resistance (friction)?
F
4o
Wx = 550sin4
= 38.36606 N
W=mg
= 55 × 10
= 550N
In x direction
since Fnet = 0
F = 38.36606 N
F  38N
2.5 Classic Lift Problems
2.5 Classic Lift Problems
1
If the lift below is moving vertically at a constant speed
of 3.0 ms-1:
(a) Draw a force diagram for the 50kg mass.
T
kg
W = 500N
2.5 Classic Lift Problems
1
If the lift below is moving vertically at a constant speed
of 3.0 ms-1:
(b) What is the Normal Reaction on the mass?
T
Since the lift is travelling at
constant speed the net force
is zero and so
T=W
T= 500N
kg
W = 500N
2.5 Classic Lift Problems
1
If the lift below is moving vertically at a constant speed of
3.0 ms-1:
(c) How much would the mass register on a set of scales?
T
kg
Since apparent weight = T
𝑻
Apparent Mass =
𝒈
𝟓𝟎𝟎
𝟏𝟎
Apparent Mass =
Apparent Mass = 50kg
W = 500N
2.5 Classic Lift Problems
2
If the lift is accelerating upwards at 2.0ms-2:
(a) Draw a force diagram for the 50kg mass.
T
kg
W = 500N
2.5 Classic Lift Problems
2
If the lift is accelerating upwards at 2.0ms-2 :
(b) What is the Normal Reaction on the mass?
a = 2.0ms-2
+
T
kg
W = 500N
Fnet = T – W
ma = T - 500
50 × 2 = T – 500
100 = T – 500
600 = T
T = 600N
2.5 Classic Lift Problems
2
If the lift is accelerating upwards at 2.0ms-2 :
(c) How much would the mass register on a set of scales?
a = 2.0ms-2
+
T
Since apparent weight = T
𝑻
Apparent Mass =
𝒈
𝟔𝟎𝟎
𝟏𝟎
Apparent Mass =
Apparent Mass= 60kg
kg
W = 500N
2.5 Classic Lift Problems
3
If the lift is moving upwards and decelerates at 3.0ms-2:
(a) Draw a force diagram for the 50kg mass.
T
kg
W = 500N
2.5 Classic Lift Problems
3
If the lift is moving upwards and decelerates at 3.0ms-2 :
(b) What is the Normal Reaction on the mass?
+
T
a = 3.0ms-2
kg
W = 500N
Fnet = T – W
ma = T - 500
50 × -3 = T – 500
-150 = T – 500
450 = T
T = 450N
2.5 Classic Lift Problems
3
If the lift is moving upwards and decelerates at 3.0ms-2 :
(c) How much would the mass register on a set of scales?
+
T
a=
Since apparent weight = T
𝑻
Apparent Mass =
𝒈
𝟒𝟓𝟎
𝟏𝟎
Apparent Mass =
Apparent Mass= 45kg
3.0ms-2
kg
W = 500N