Download presentation

Presentation is loading. Please wait.

Published byDraven Goodier Modified about 1 year ago

1
Higher Physics Unit 1 Multiple Choice Questions

2
Q1

3
Q2 12km North 4km South 8km displacement NORTH v = s t = 8 4 = 2km/h north v = d t = 16 4 = 4 km/h

4
Q3

5
Q4 Remember redraw diagram nose to tail.

6
Q5

7
Q6

8
Q7 Displacement 50 km v = d t = 70 2 = 35 km/h v = s t = 50 2 = 25km/h

9
Q8

10
Q9

11
Q10

12
Q11 u = v = a = s = t = - 3m/s ? m/s 2 ??? 5.0s s = ut + ½ at 2 s = -3x5 + ½ (-9.8) x (5) 2 s = (-4.9 x 25) s = s = m + -

13
Q12 horizontalvertical V H = ?? d = 5.1m t = ? v = d t u = 0 m/s v = ? a = -9.8 m/s2 s = -2.0 m t = ? S = ut + ½ at 2 -2 = 0 + ½ (-9.8)t 2 -2 = -4.9 t 2 t 2 = -2 / (-4.9) t = 0.64s v = v = 7.97m/s

14
Q13

15
Q14

16
Q15 TMB = TMA pnefa m 1 u 1 = m 2 v 2 +m 3 v x 0 = 1200v x 60 0 = 1200v v 2 = v 2 = /1200 v 2 = m/s -ve answer means cannon will be travelling in the opposite direction (i.e. west)

17
Q16

18
Q17 40m/s 30m/s 40m/s 50m/s

19
Q18 Between 1 to 3 seconds a = 4m/s 2 So every second the speed increases by 4m/s so if it starts from rest is speed will be 16m/s after 4 seconds. Then between 3 to 5 seconds a = -2m/s 2 So every second the speed decreases by 2m/s so if it starts from 16m/s is speed will be 12m/s after 2 seconds.

20
Q19 Fun = ma Fun = 400 x 2 Fun = 800 N

21
Q20

22
Q21 W = mg W = 15 x 9.8 W = 147N F un = 180 –147 = 33N F un = ma a = F un / m a = 33 / 15 a = 2.2 m/s 2

23
Q22 friction F un = ma F un = 10 x 10 F un = 100N

24
Q23 Constant height = balanced forces Upthrust = weight = mg Upthrust = 1.5 x 9.8 Upthrust = 14.7 N

25
Q24 F un = ma a = F un / m a = 18 / 10 a = 1.8 m/s 2 F un = ma F un = 4 x 1.8 F un = 7.2N

26
Q25 E K at start E K = ½ mv 2 E K = ½ x 1000 x40 2 E K = 800,000 J E K at end E K = ½ mv 2 E K = ½ x 1000 x 10 2 E K = 50,000 J E K lost = 800,000 – 50,000 = 750,000 J = 750 kJ

27
Q26 FHFH opp hyp adj cos 40 o = adj / hyp cos 40 o = F H / 100 F H = 100 cos 40 o F H = 76.6 N Ew = Fd Ew = 76.6 x 10 Ew = 766 J

28
Q27 F un = ma F un = 700 x 2 F un = 1400 N

29
Q28 Force up is greater than force down So lift is either accelerating upwards OR decelerating downwards.

30
Q29 Accelerating downwards so Force down (weight) greater than force up (apparent weight)

31
Q30 TMB = TMA m 1 u 1 = m 2 v 2 + m 3 v x 0 = 2000v x 50 0 = 2000v V 2 = -250 / 2000 V 2 = m/s The answer is negative because it is travelling in opposite direction to the cannon ball.

32
Q31 Momentum = mv v = momentum / m v = 12 / 4 v = 3 m/s E K = ½ mv 2 E K = ½ x 4 x 3 2 E K = 36 J

33
Q32 F x t = mv -mu Impulse = change in momentum Since mass and speed of impact are constant change in momentum is constant. Crumple zones increase the time of contact thereby reducing the force of impact.

34
Q33

35
Q34 Ft = mv - mu F = (mv – mu) / t F = change in momentum per second.

36
Q35 Ft = mv - mu Change in momentum hasn’t changed therefore Ft is constant.

37
Q36

38
Q37

39
Q38

40
Q39

41
Q40

42
Q41

43
Q42

44
Q43

45
Q44

46
Q45

47
Q46

48
Q47

49
Q48

50
Q49

51
Q50

52
Q51

53
Q52

54
Q53

55
Q54

56
Q55

57
Q56

Similar presentations

© 2016 SlidePlayer.com Inc.

All rights reserved.

Ads by Google