1. Year 12 Physics Gradstart

2. 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

3. 2.2 Vector Components Part 2

4. 1a) What is the northerly and easterly component of the velocity below v N = 35sin28 = 16.43150ms -1 E v E = 35cos28 = 30.90317ms -1

5. 2.2 Vector Components Part 2 1b) What is the northerly and easterly component of the velocity below v N = 12cos15 = 11.59111ms -1 E v E = – 12sin15 = – 3.10583ms -1

6. 2.2 Vector Components Part 2 2. Work out the horizontal and vertical components of the force below: y x F x = 6cos30 = 5.19615N F y = 6sin30 = 3.0N

7. 2.2 Vector Components Part 2 3. Work out the horizontal and vertical velocity components of the golfball below: y x v x = 30cos60 = 15ms -1 v y = 30sin60 = 25.98076ms -1

8. 2.2 Vector Components Part 2 4(a) Work out the weight into the slope and down the slope. y x W x = 15sin22 = 5.61910N W y = 15cos22 = 13.90776N W=mg = 15N 22 o

9. 2.2 Vector Components Part 2 4(b) What is the acceleration down the slope if it is frictionless. y x W x = 15sin22 = 5.61910N W y = 15cos22 = 13.90776N W=mg = 15N 22 o

10. 2.2 Vector Components Part 2 y x W x = 15sin22 = 5.61910N W y = 15cos22 = 13.90776N W=mg = 15N 4(c) What is the normal reaction force on the mass. In y direction N = W y N = 13.90776N N  14N N 22 o

11. 2.2 Vector Components Part 2 5(a) Work out an expression for the net force down the frictionless slope below. y x W x = mgsin  W=mg F x = mgsin  

12. 2.2 Vector Components Part 2 5(b) What is the formula for the acceleration of a mass down a frictionless slope? y x W x = mgsin  W=mg F x = mgsin  

13. 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 sin20 o = 10sin20 o = 3.40201 x = ut + ½ at 2 0.4 = ½ × 3.40201 × t 2 0.233904 = t 2 0.483637 = t t = 0.48m A Save in memory A of your calculator

14. 2.2 Vector Components Part 2 7 If Fr max = 4.0N and a 15N is applied to the 3.0kg block at 30 o to the horizontal (a) What is the acceleration on the mass? In x direction F net = F x - Fr ma = 12.99038 - 4 3a = 8.99038 a = 2.99679 a  3.0 ms -2 y x F x = 15cos30 = 12.99038N F y = 15sin30 = 7.5N Fr max = 4.0N B It is not worth saving 7.5 in the memory since it is so easy to enter in the calculator

15. 2.2 Vector Components Part 2 7 If Fr max = 4.0N and a 15N is applied to the 3.0kg block at 30 o to the horizontal (b) What is the normal reaction on the mass? In y direction N + F y = W N + 7.5 = 30 N = 22.5N N  23N F x = 15cos30 = 12.99038N F y = 15sin30 = 7.5N Fr max = 4.0N y x W = 3 × 10 = 30N N

16. 2.2 Vector Components Part 2 8 The 2.0kg mass below is held at rest on the 40 o slope below by friction. What is the magnitude of the friction holding the mass? In x direction Fr = W x Fr = 12.85575 N Fr  13N y x W x = 20sin40 = 12.85575 N W=mg = 20N Fr 40 o

17. 2.2 Vector Components Part 2 9 What will be the acceleration of the 1.6kg block on the 30 o slope below if Fr max = 3.0N? In x direction F net = W x – Fr ma = 8 – 3 1.6a = 5 a = 3.125 a  3.1 ms -2 y x W x = 16sin30 = 8 N W=mg = 16N Fr max = 3.0N 40 o

18. 2.2 Vector Components Part 2 10 If Fr max = 2.0N and a 12N is applied to the 1.0kg block at 30 o to the horizontal (a) What is the acceleration on the mass? In x direction F net = 12 – W x – Fr ma = 12 – 2 – 5 1a = 5 a = 5.0 ms -2 y x W x = 10sin30 = 5 N W y = 10cos30 = 8.66025N W=mg = 10N Fr max = 2.0N A 30 o

19. 2.2 Vector Components Part 2 10 If Fr max = 2.0N and a 12N is applied to the 1.0kg block at 30 o to the horizontal (b) What is the normal reaction on the mass? In y direction N = 8.66025N N  8.7N W x = 10sin30 = 5 N W y = 10cos30 = 8.66025N W=mg = 10N Fr max = 2.0N A y x N 30 o

20. 2.2 Vector Components Part 2 11 If Fr max = 8.0N and a 20N is applied to the 3.0kg block at 30 o to the horizontal What is the acceleration on the mass? 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. y x W x = 30sin30 = 15 N W=mg = 30N Fr max = 8.0N 30 o

21. 2.2 Vector Components Part 2 12A 55kg cyclist is riding up a 4.0 o 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.

22. 2.2 Vector Components Part 2 12A 55kg cyclist is riding up a 4.0 o 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 direction since F net = 0 F = 38.36606 N F  38N W x = 550sin4 = 38.36606 N W=mg = 55 × 10 = 550N F 4o4o y x

23. 2.5 Classic Lift Problems

24. 1If 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

25. 2.5 Classic Lift Problems 1If 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 = W T= 500N kg W = 500N T

26. 2.5 Classic Lift Problems 1If 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? kg W = 500N T

27. 2.5 Classic Lift Problems 2If the lift is accelerating upwards at 2.0ms -2 : (a) Draw a force diagram for the 50kg mass. kg W = 500N T

28. 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? F net = T – W ma = T - 500 50 × 2 = T – 500 100 = T – 500 600 = T T = 600N kg W = 500N T a = 2.0ms -2 +

29. 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? kg W = 500N T a = 2.0ms -2 +

30. 2.5 Classic Lift Problems 3If the lift is moving upwards and decelerates at 3.0ms -2 : (a) Draw a force diagram for the 50kg mass. kg W = 500N T

31. 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? F net = T – W ma = T - 500 50 × -3 = T – 500 -150 = T – 500 450 = T T = 450N kg W = 500N T a = 3.0ms -2 +

32. 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? kg W = 500N T + a = 3.0ms -2