Solutions to SQA examination

Higher Grade Physics

2003

Section A
1. D		11. D			
2. E		12. E			
3. A		13. C			
4. D		14. B			
5. A		15. B			
6. C		16. B			
7. E		17. D			
8. C		18. A			
9. B		19. E			
10.E		20. A			


Section B

21.a.i.  Vhorizontal=Vresultant x cosq 
         V=35xcos40o
         Vhor = 26.81(m/s)
	  

   a.ii. Vvertical=Vresultant x sinq
         Vver=35xsin40o
         Vver = 22.5(m/s)


   a.iii.Maximum height reached when the vertical component 
	 of velocity (Vver) is 0m/s.
	 The vertical component of velocity calculated in part a.ii.
	 is the initial velocity (u) for the purpose of this calculation.
	
	 vver = 0(m/s)
	 uver = 22.5(m/s)
	 a = -9.8(m/s/s)
	 t = ?

	 v = u + at
	 t = (v-u)/a
	 t = (0-22.5)/-9.8
	 t = 2.29(s)

    b. 	 Horizontal distance is calculated by multiplying the
	 total time of flight by the horizontal velocity, which
  	 is constant throughout the flight.

	 Total time = [(2x2.29) + 0.48]s
	 ttotal = 5.06(s)
	 Vhor=26.81(m/s)
	 
	 d = Vhor x ttotal
	 d = 26.81(m/s)x5.06(s)
	 d = 135.66(m) 

	 
  

  22.a.  DPs = ms(vs-us)
	 DPs = 38(4.6-2.2)
	 DPs = 91.2(kgm/s)
 
     b.	 Favgtc = DP
	 tc = DP/Favg
	 tc = 91.2/130
	 tc = 0.7(s)

     c.  DPR = -DPS
	 DPR = -91.2(kgm/s)

	 DPR = mR(vR-uR)
	 -91.2 = 54(vR-2.2)
	 vR = 0.51(m/s)

     d.   If kinetic energy is conserved the interaction is elastic.

	  Ekbefore = 1/2MRUR2 + 1/2MSUS2
	  Ekbefore = 0.5x54x2.22 + 0.5x38X2.22
	  Ekbefore = 222.64(J)
 
	  Ekafter = 1/2MRVR2 + 1/2MSVS2
	  Ekafter = 0.5x54x0.512 + 0.5x38x4.62
	  Ekafter = 409(J)

	  The gain in kinetic energy indicates that the collision is NOT ELASTIC. 
	  The interaction is more like an explosion.

23.a.i.  

	
   a.ii. Pliquid = rgh 
	 Pliquid = 1000x9.8x0.25
	 Pliquid = 2450(Pa)

   a.iii.As the depth increases the total pressure acting on the air inside
	 the tubing increases. Liquid enters the tube,compressing the air, until 
	 the total pressure acting on the air inside the tube is equal to the 
   	 pressure of the air inside the tube.


  b.	The pressure resulting from tank and water can 
	be calculated using: 
	
	P = F/A   where F is the weight of the tank and water.

	F = wtotal = mtotalg
	wtotal = (2.7x103+300)9.8
	wtotal = 3000x9.8
	wtotal = 29400(N)

	A = 2.0x1.5 = 3.0m2

	P = 29400/3
	P = 9800(Pa)

	Atmospheric pressure is pushing down on the tank. This means that
	the total pressure acting on the surface that the tank is resting
	on is the sum of the atmospheric pressure and the pressure resulting
	from the weight of the tank.

	

	Ptotal = Ptank+water + Patmospheric
	Ptotal = 9800 + 1.01x105
	Ptotal = 1.108x105(Pa)

24.a.	P1 = 1.56x105Pa
	T1 = (27+273)K = 300k
	
 	P2 = ?
	T2 = (77+273)K = 350k

	P1/T1 = P2/T2 
	P2 = P1T2/T1
	P2 = 1.56x105x350/300
	P2 = 1.82x105Pa

   b.i.	P = I2Relement

	I = V/Rtotal

	I = 30/(0.5+1.5) = 30/2
	I = 15A

	P = 152x0.5
	P = 112.5W

  b.ii.	The output power of the element would be less.
	This is because the current in the circuit would be 
	less because of the increase of the total series 
	resistance of the circuit.

	Other explanations involving Vtpd are also valid.


25.a.i.	Y-gain = 5volts/div

   a.ii.f = 1/T

	T = 2.5ms = 2.5x10-3
	f = 1/2.5x10-3
	f = 400Hz

   b.i.	VRMS = VPeak/SQRT2
	VRMS = 12/1.414
	VRMS = 8.49V

  b.ii. E = 1/2(CVpk2)
	E = 1/2(220x10-6x122)
	E = 0.016J

 b.iii.	The reading on the ammeter increases because the
	current is directly proportional to the frequency.

  b.iv.	The capacitor is constantly charging and discharging,
	as the polarity of the supply voltage changes in this 
	ac circuit. The flow of charge during this process means 
	that the current in the circuit does not fall to zero.
	In a dc circuit the capacitor charges and opposes the 
	flow of charge from the supply. When fully charged no
	current flows in the circuit.


26.a.	 When the bridge is balanced:
	 R1/R2 = R3/R4 
 
	 Where 	R1 = 5500W
		R2 = RLDR
	 	R3 = 330000W
		R4 = 150000W

		RLDR = (R1xR4)/R3
		RLDR = (5500x150000)/330000)
		RLDR = 2500W

   b.i.	 MOSFET (n-channel)

   b.ii.V1 = VX = 1.28V
	V2 = VY = 1.50V
	 
	Vouput = (Rfeedback/R1)x(V2-V1)
	Vouput = (22.5/1.5)x(1.5-1.28)
	Vouput = 15x0.22
	Vouput = 3.3V

  b.iii.When water reaches the maximum level the beam of
	light is not totally internally reflected. This means
	that the light intensity incident on the LDR decreases,
	increasing its resistance and increasing VX.
	This decreases the output voltage from the differential
	amplifier. When the output voltage falls below 2.0V
	the MOSFET does not conduct and the current in the
	solenoid falls to zero, closing the valve. 

26.c.	The incident angle at the glass water boundary is less
	than the critical angle so there is no total internal 
	reflection.


27.a.i.	The energy of the emitted photon is equal to the
	energy difference between levels.The longest wavelength 
	radiation will have the lowest frequency and energy.

 	The smallest energy gap is beween E4 and E3
	 	
  a.ii.	Ephoton = E3-E2
	Ephoton = (-2.4x10-19)-(-5.6x10-19)
	Ephoton = 3.2x10-19J

	Ephoton = hf
	f = Ephoton/h
	f = 3.2x10-19/6.63x10-34
	f = 4.83x1014Hz

  b.i.	fair = fglass
	fglass = 4.74x1014Hz

 b.ii.	vair = fairlair 
	lair = vair/fair
	lair = 3x108/4.74x1014
	lair = 632.9nm

	lglass = lair/n
	lglass = 632.9/1.6
	lglass = 395.6nm

28.a.i.	Maxima are produced when the path difference of the microwaves,
	from each gap, to a point along AB is zero or a whole number of 
	wavelengths. This is because constructive interference occurs 
	at these points as the waves are in phase.

 	Minima are produced when the path difference of the microwaves,
	from each gap, to a point along AB is l/2, 3l/2, 5l/2, 7l/2 etc.  
	This is because destructive interference occurs at these points
	as the waves are P out of phase.

  a.ii.
		

	PD = (766 - 682)mm
	PD = 84mm

	Third Max PD = 3l
	3l = 84mm
	l = 84/3
	l = 28mm

  b.i.	Gain = -Rfeedback/Rinput
	Gain = -10kW/10kW
	Gain = -1

	


	Voutput/Vinput = -Rfeedback/Rinput
    	Voutput = -Vinputx(Rfeedback/Rinput)
	Voutput(peak) = -6x(3.0x103/2.0x103)
	Voutput(peak) = -9V

 b.ii.

29.a.i.	42He is is an alpha particle.

  a.ii.	A = N/t

	N = 7.2x105disintegrations
	t = 2 minutes = 120s

	A = 7.2x105/120
	A = 6000Bq = 6kBq

  b.i.	Half value thickness(t1/2)  =	thickness of barrier required to 
					reduce the count rate by half.

  	60counts/s -> 30counts/s requires 3cm
	30counts/s -> 15counts/s requires 3cm

	t1/2 = 3cm

 b.ii.	The bar indicates the uncertainty in the average reading.
	This results from the random nature of radioactive decay.

 c.i.	Htotal = 6.4x10-5Sv
	
	Alpha
	Qa = 20
	Ha = QaxDa
	Ha = 20Da

	Gamma
	Qg = 1
	Dg = 1.2x10-5Gy
	Hg = QxDg	
	Hg = 1x1.2x10-5Sv
	Hg = 1.2x10-5Sv
	
	Htotal = Ha + Hg
	6.4x10-5Sv = Ha + 1.2x10-5Sv
	Ha = 6.4x10-5Sv - 1.2x10-5Sv
	Ha = 5.2x10-5Sv

	Ha = QDa
	Ha = 20Da
	20Da = 5.2x10-5Sv 
	Da = 5.2x10-5/20
	
	Da = 2.6x10-6Gy = 2.6mGy



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