Higher Physics Unit 1 Multiple Choice Questions. Q1.
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Higher Physics
Unit 1 Multiple Choice Questions
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Q1
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Q2 12kmNorth
4kmSouth
8kmdisplacementNORTH
v = st
=84
= 2km/h north
v = dt
=16
4= 4 km/h
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Q3
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Q4
Remember redraw diagram nose to tail.
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Q5
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Q6
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Q7
Displacement 50 km
v = dt
=70
2= 35 km/h v = s
t =
502
= 25km/h
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Q8
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Q9
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Q10
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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
+ -
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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
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Q13
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Q14
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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)
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Q16
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Q17
40m/s
30m/s
40m/s
50m/s
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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.
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Q19
Fun = maFun = 400 x 2Fun = 800 N
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Q20
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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
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Q22
friction
Fun = ma
Fun = 10 x 10
Fun = 100N
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Q23
Constant height = balanced forces
Upthrust = weight = mgUpthrust = 1.5 x 9.8Upthrust = 14.7 N
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Q24Fun = ma
a = Fun / m
a = 18 / 10
a = 1.8 m/s2
Fun = ma
Fun = 4 x 1.8
Fun = 7.2N
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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
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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
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Q27 Fun = ma
Fun = 700 x 2
Fun = 1400 N
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Q28
Force up is greater than force down
So lift is either accelerating upwards OR decelerating downwards.
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Q29
Accelerating downwards so
Force down (weight) greater than force up (apparent weight)
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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.
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Q31
Momentum = mvv = momentum / m v = 12 / 4v = 3 m/s
EK = ½ mv2
EK = ½ x 4 x 32
EK = 36 J
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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.
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Q33
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Q34Ft = mv - mu
F = (mv – mu) / t
F = change in momentum per second.
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Q35Ft = mv - mu
Change in momentum hasn’t changed therefore Ft is constant.
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Q36
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Q37
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Q38
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Q39
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Q40
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Q41
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Q42
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Q43
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Q44
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Q45
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Q46
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Q47
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Q48
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Q49
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Q50
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Q51
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Q52
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Q53
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Q54
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Q55
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Q56