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Higher Physics Unit 1 Multiple Choice Questions

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

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Q4 Remember redraw diagram nose to tail.

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

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Q11 u = v = a = s = t = - 3m/s ? - 9.8 m/s 2 ??? 5.0s s = ut + ½ at 2 s = -3x5 + ½ (-9.8) x (5) 2 s = -15 + (-4.9 x 25) s = -15 - 122.5 s = -137.5 m + -

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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 = 5.1 0.64 v = 7.97m/s

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Q15 TMB = TMA pnefa m 1 u 1 = m 2 v 2 +m 3 v 3 1215 x 0 = 1200v 2 + 15 x 60 0 = 1200v 2 + 900 1200v 2 = - 900 v 2 = - 900 /1200 v 2 = - 0.75m/s -ve answer means cannon will be travelling in the opposite direction (i.e. west)

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

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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.

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

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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

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

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

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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

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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

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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

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Q27 F un = ma F un = 700 x 2 F un = 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 = TMA m 1 u 1 = m 2 v 2 + m 3 v 3 2005 x 0 = 2000v 2 + 5 x 50 0 = 2000v 2 + 250 V 2 = -250 / 2000 V 2 = -0.125m/s The answer is negative because it is travelling in opposite direction to the cannon ball.

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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

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

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

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