year 12 physics gradstart. 2.1 basic vector revision/ progress test you have 20 minutes to work in a...
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Year 12 Physics Gradstart
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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
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2.2 Vector Components Part 2
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2.2 Vector Components Part 2
1a) What is the northerly and easterly component of the velocity below
vN = 35sin28 = 16.43150ms-1
E
vE = 35cos28 = 30.90317ms-1
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2.2 Vector Components Part 2
1b) What is the northerly and easterly component of the velocity below
vN = 12cos15 = 11.59111ms-1
E
vE = – 12sin15 = – 3.10583ms-1
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2.2 Vector Components Part 2
2. Work out the horizontal and vertical components of the force below:
y
x
Fx = 6cos30 = 5.19615N
Fy = 6sin30 = 3.0N
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2.2 Vector Components Part 2
3. Work out the horizontal and vertical velocity components of the golfball below:
y
x
vx = 30cos60 = 15ms-1
vy = 30sin60 = 25.98076ms-1
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2.2 Vector Components Part 2
4(a) Work out the weight into the slope and down the slope.
y
x
Wx = 15sin22 = 5.61910N
Wy = 15cos22 = 13.90776N W=mg
= 15N
22o
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2.2 Vector Components Part 2
4(b) What is the acceleration down the slope if it is frictionless.
In x directiona = ? Fnet = 5.61910N m = 1.5kg
Fnet = a = a = a = 2.80955 a 2.8 ms-2
y
x
Wx = 15sin22 = 5.61910N
Wy = 15cos22 = 13.90776N W=mg
= 15N
22o
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2.2 Vector Components Part 2
y
x
Wx = 15sin22 = 5.61910N
Wy = 15cos22 = 13.90776N W=mg
= 15N
4(c) What is the normal reaction force on the mass.
In y direction N = Wy
N = 13.90776NN 14N
N
22o
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2.2 Vector Components Part 2
5(a) Work out an expression for the net force down the frictionless slope below.
y
x
Wx = mgsin
W=mg
Fx = mgsin
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2.2 Vector Components Part 2
5(b) What is the formula for the acceleration of a mass down a frictionless slope?
In x directiona = ? Fnet = mgsin m = m
Fnet = a = a = g sin
y
x
Wx = mgsin
W=mg
Fx = mgsin
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2.2 Vector Components Part 26 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
= 3.40201x = ut + ½ at2
0.4 = ½ × 3.40201 × t2
0.233904 = t2
0.483637 = tt = 0.48m
A
Save in memory A of your calculator
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2.2 Vector Components Part 27 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?
In x directionFnet = Fx - Frma = 12.99038 - 43a = 8.99038a = 2.99679a 3.0 ms-2
y
x
Fx = 15cos30 = 12.99038N
Fy = 15sin30 = 7.5NFrmax = 4.0N
B
It is not worth saving 7.5 in the
memory since it is so easy to enter in
the calculator
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2.2 Vector Components Part 27 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?
In y directionN + Fy = WN + 7.5 = 30N = 22.5NN 23N
Fx = 15cos30 = 12.99038N
Fy = 15sin30 = 7.5NFrmax = 4.0N
y
x
W = 3 × 10 = 30N
N
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2.2 Vector Components Part 28 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?
In x directionFr = Wx
Fr = 12.85575 NFr 13N
y
x
Wx = 20sin40 = 12.85575 NW=mg
= 20N
Fr
40o
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2.2 Vector Components Part 29 What will be the acceleration of the 1.6kg block
on the 30o slope below if Frmax = 3.0N?
In x direction Fnet = Wx – Fr
ma = 8 – 31.6a = 5a = 3.125a 3.1 ms-2
y
x
Wx = 16sin30 = 8 NW=mg
= 16N
Frmax = 3.0N
40o
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2.2 Vector Components Part 210 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?
In x directionFnet = 12 – Wx – Frma = 12 – 2 – 51a = 5a = 5.0 ms-2
y
x
Wx = 10sin30 = 5 N
Wy = 10cos30 = 8.66025N
W=mg = 10N
Frmax = 2.0NA
30o
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2.2 Vector Components Part 210 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?
In y directionN = 8.66025NN 8.7N
Wx = 10sin30 = 5 N
Wy = 10cos30 = 8.66025N
W=mg = 10N
Frmax = 2.0NA
y
xN
30o
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2.2 Vector Components Part 211 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?
In x directionThe 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.
y
x
Wx = 30sin30 = 15 NW=mg
= 30N
Frmax = 8.0N
30o
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2.2 Vector Components Part 212 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.
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2.2 Vector Components Part 212 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)?
In x directionsince Fnet = 0
F = 38.36606 NF 38N
Wx = 550sin4 = 38.36606 NW=mg
= 55 × 10 = 550N
F
4o
y
x
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2.5 Classic Lift Problems
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2.5 Classic Lift Problems1 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.
kg
W = 500N
T
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2.5 Classic Lift Problems1 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?
Since the lift is travelling at constant speed the net force is zero and so
T = WT= 500Nkg
W = 500N
T
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2.5 Classic Lift Problems1 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?
Since apparent weight = TApparent Mass = Apparent Mass = Apparent Mass = 50kg
kg
W = 500N
T
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2.5 Classic Lift Problems2 If the lift is accelerating upwards at 2.0ms-2:
(a) Draw a force diagram for the 50kg mass.
kg
W = 500N
T
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2.5 Classic Lift Problems2 If the lift is accelerating upwards at 2.0ms-2 :
(b) What is the Normal Reaction on the mass?
Fnet = T – W ma = T - 50050 × 2 = T – 500 100 = T – 500 600 = T T = 600N
kg
W = 500N
T
a = 2.0ms-2 +
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2.5 Classic Lift Problems2 If the lift is accelerating upwards at 2.0ms-2 :
(c) How much would the mass register on a set of scales?
Since apparent weight = TApparent Mass = Apparent Mass = Apparent Mass= 60kg
kg
W = 500N
T
a = 2.0ms-2 +
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2.5 Classic Lift Problems3 If the lift is moving upwards and decelerates at 3.0ms-2:
(a) Draw a force diagram for the 50kg mass.
kg
W = 500N
T
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2.5 Classic Lift Problems3 If the lift is moving upwards and decelerates at 3.0ms-2 :
(b) What is the Normal Reaction on the mass?
Fnet = T – W ma = T - 50050 × -3 = T – 500 -150 = T – 500 450 = T T = 450N
kg
W = 500N
T
a = 3.0ms-2
+
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2.5 Classic Lift Problems3 If the lift is moving upwards and decelerates at 3.0ms-2 :
(c) How much would the mass register on a set of scales?
Since apparent weight = TApparent Mass = Apparent Mass = Apparent Mass= 45kg
kg
W = 500N
T+
a = 3.0ms-2