# Higher Physics Unit 1 Multiple Choice Questions. Q1.

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Higher Physics Unit 1 Multiple Choice Questions

Q1

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 + -

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)

Q16

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

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.

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

Q20

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

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

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

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

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

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

Q27 F un = ma F un = 700 x 2 F un = 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 = 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.

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

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

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

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

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