Higher Physics

Unit 1 Multiple Choice Questions

Q1

Q2 12kmNorth

4kmSouth

8kmdisplacementNORTH

v = st

=84

= 2km/h north

v = dt

=16

4= 4 km/h

Q3

Q4

Remember redraw diagram nose to tail.

Q5

Q6

Q7

Displacement 50 km

v = dt

=70

2= 35 km/h v = s

t =

502

= 25km/h

Q8

Q9

Q10

Q11

u =v =a =s =t =

- 3m/s?- 9.8 m/s2

???5.0s

s = ut + ½ at2

s = -3x5 + ½ (-9.8) x (5)2

s = -15 + (-4.9 x 25)s = -15 - 122.5s = -137.5 m

+ -

Q12horizontal vertical

VH = ??d = 5.1mt = ?

v = dt

u = 0 m/sv = ?a = -9.8 m/s2s = -2.0 mt = ?

S = ut + ½ at2

-2 = 0 + ½ (-9.8)t2

-2 = -4.9 t2

t2 = -2 / (-4.9)t = 0.64s

v =5.1

0.64

v = 7.97m/s

Q13

Q14

Q15 TMB = TMA pnefa

m1u1 = m2v2 +m3v3

1215 x 0 = 1200v2 + 15 x 60

0 = 1200v2 + 9001200v2 = - 900v2 = - 900 /1200v2 = - 0.75m/s

-ve answer means cannon will be travelling in the opposite direction (i.e. west)

Q16

Q17

40m/s

30m/s

40m/s

50m/s

Q18

Between 1 to 3 secondsa = 4m/s2

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 secondsa = -2m/s2

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.

Q19

Fun = maFun = 400 x 2Fun = 800 N

Q20

Q21 W = mgW = 15 x 9.8W = 147N

Fun = 180 –147 = 33N

Fun = ma

a = Fun / m

a = 33 / 15

a = 2.2 m/s2

Q22

friction

Fun = ma

Fun = 10 x 10

Fun = 100N

Q23

Constant height = balanced forces

Upthrust = weight = mgUpthrust = 1.5 x 9.8Upthrust = 14.7 N

Q24Fun = ma

a = Fun / m

a = 18 / 10

a = 1.8 m/s2

Fun = ma

Fun = 4 x 1.8

Fun = 7.2N

Q25

EK at start

EK = ½ mv2

EK = ½ x 1000 x402

EK = 800,000 J

EK at end

EK = ½ mv2

EK = ½ x 1000 x 102

EK = 50,000 J

EK lost = 800,000 – 50,000 = 750,000 J = 750 kJ

Q26

FH

opphyp

adj

cos 40o = adj / hypcos 40o = FH / 100FH = 100 cos 40o

FH = 76.6 NEw = FdEw = 76.6 x 10Ew = 766 J

Q27 Fun = ma

Fun = 700 x 2

Fun = 1400 N

Q28

Force up is greater than force down

So lift is either accelerating upwards OR decelerating downwards.

Q29

Accelerating downwards so

Force down (weight) greater than force up (apparent weight)

Q30

TMB = TMAm1u1 = m2v2 + m3v3

2005 x 0 = 2000v2 + 5 x 500 = 2000v2 + 250V2 = -250 / 2000V2 = -0.125m/s

The answer is negative because it is travelling in opposite direction to the cannon ball.

Q31

Momentum = mvv = momentum / m v = 12 / 4v = 3 m/s

EK = ½ mv2

EK = ½ x 4 x 32

EK = 36 J

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.

Q33

Q34Ft = mv - mu

F = (mv – mu) / t

F = change in momentum per second.

Q35Ft = mv - mu

Change in momentum hasn’t changed therefore Ft is constant.

Q36

Q37

Q38

Q39

Q40

Q41

Q42

Q43

Q44

Q45

Q46

Q47

Q48

Q49

Q50

Q51

Q52

Q53

Q54

Q55

Q56