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

vN = 35sin28 = 16.43150ms-1

E

vE = 35cos28 = 30.90317ms-1

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

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

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

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

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

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

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

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

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

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

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

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

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

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

yx

Wx = 10sin30 = 5 N

Wy = 10cos30 = 8.66025N

W=mg = 10N

Frmax = 2.0NA

30o

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

yxN

30o

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.

yx

Wx = 30sin30 = 15 NW=mg

= 30N

Frmax = 8.0N

30o

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.

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

2.5 Classic Lift Problems

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

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

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

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

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

Ta = 2.0ms-2 +

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

Ta = 2.0ms-2 +

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

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

+

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