Solutions to SQA examination

Higher Grade Physics


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Section A

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

Section B

21.a.i. Fhor= FResultantcosq Fhor= 4cos26o Fhor= 3.6N a.ii. Fhor = Fun = 3.6N a = Fun/m m = 18kg a = 3.6/18 a = ? a = 0.2m/s2 a.iii.u = 0m/s s = ut + 1/2at2 a = 0.2m/s2 s = 0x7 + 0.5x0.2x72 t = 7s s = 4.9m s = ? b. By decreasing the angle the cosine of the angle will increase. This makes the horizontal force greater as :Fhor= FResultantcosq The accelerating force, and the consequent acceleration, is therefore greater. Substituting a greater acceleration into the equation used in part a.iii. will result in a greater distance being calculated. 22.a.i.
Experiment Number 1 2 3 4 5 6
Mass of flask and air/kg 0.8750 0.8762 0.8748 0.8755 0.8760 0.8757
Mass of evacuated mask/kg 0.8722 0.8736 0.8721 0.8728 0.8738 0.8732
Mass of air removed/kg 0.0028 0.0026 0.0027 0.0027 0.0022 0.0025
  a.ii.	Mean = Total/Number of readings
	Mean = (0.0028+0.0026+0.0027+0.0027+0.0022+0.0025)/6
	Mean = (0.0155/6)
	Mean = 0.00258kg

	Random uncertainty = (max - min)/N
	Random uncertainty = (0.0028 - 0.0022)/6
	Random uncertainty = 0.0001kg

	Mass of air = (0.0026+-0.0001)kg

 a.iii.	mass(mean) = 0.0026kg
	volume = 2.0x10-3m3

	density = mass/volume
	r = m/V
	r = 0.00258/2.0x10-3
	r = 1.29kg/m3

 a.iv.	A flask of larger volume is better because this increases the 
	mass and volume of air and used in the experiment. This should 
	result in a smaller percentage error in the measurements of 
	both mass and volume of the gas. This will in turn reduce the 
	percentage error in the calculated density of air.	Pressure and volume are the important variables.
	This means Boyle's law requires to be used. However, as
   	it is the length of the cylinder that is given and not
	the volume the fact that:
	Volume(V) = Cylinder cross sectional area(A) x Length of the piston(L)
	must be used in the solution.

	V1 = L1A			V1/V2 = P1/P2
	V2 = L2A			V1/V2 = L1A/L2A = L1/L2
	P1 = 1x10-5Pa          =>L1/L2 = P1/P2
	P2 = ?			P2 = (L2/L1)P1
				P2 = (360/160)x1x10-5
				P2 = 2.25x10-5Pa

  b.ii.	The mass of the gas trapped is constant.

  b.iii.Pressure is caused by the gas particles exerting a force on the
	walls of the container. When the volume of the container decreases
	there is an increase in the collision rate, meaning that more force
	is exerted on the container walls. This increases the pressure as
	pressure is a measure of force per unit area (P = F/A).

    (A)	Contact time(tc) = 3.0x10-3s
	Favg = Fun = 0.5N

	NB/ Favg = Fun because there are no unbalanced forces.

	Impulse = Favg x tc
	Impulse = 0.5 x 3.0x10-3
	Impulse = 1.5x10-3Ns (kgm/s)  

    (B)	Impulse = Change in Momentum
	Impulse = mv-mu
	Impulse = m(v-u)

	u = 0m/s  	As the bead of water is initially at rest.
	m = 2.5x10-5kg

    => 	v = Impulse/m
	v = 1.5x10-3/2.5x10-5
   	v = 60m/s

  a.ii.	The impulse can be calculated from the area under the force 
	time graph.

	A = 1/2bh (Area of a triangle formula)
	A = 0.5 x 3.0x10-3 x 0.5
	A = 0.75

     =>	Actual impulse = 0.75x10-3Ns

	v = Impulse/m  (from above)
	v = 0.75x10-3/2.5x10-5
	v = 30m/s

     b.	The bead gains kinetic energy as it is accelerated in the
	electric field between the plates. The gain in the kinetic 
	energy of the bead, with charge q, accelerated by a 
	potential difference V, can be calculated using:

	Ek(gain) = qV

	This is equal to the gain in kinetic energy:

	Ek(gain) = mv2/2

      =>mv2/2 = qV		q = 6.5x10-6C
	v2 = 2qV/m		V = 5x103V	
	v = SQRT(2qV/m)		m = 4.0x10-5kg

	v = SQRT(2x6.5x10-6x5x103)/4.0x10-5
	v = SQRT(1625)
	v = 40.3m/s

24.a.i. The emf can be found by projecting the graph line back until it 
	cuts the voltage axis.

	emf = 4.8V

  a.ii.	The internal resistance is equal to the negative of the gradient
	of the line given.

	m = (y1-y2)/(x1-x2)
	m = (4.0-2.0)/(0.4-1.4)
	m = 2.0/-1.0
	m = -2

	r = 2W

	To justify the above consider:

	y = mx + c  ...1
	V = mI + c  ...2
	Vtpd= E - Ir   ...3
	Vtpd= -Ir + E  ...4

	From equation (3) 
	When I = 0A :Vtpd=emf
	Comparing (2) and (4)
	m = -r

24.b.i.	emf = Vtpd+Vlost
	emf = IR+Ir
 	The condition for a short circuit is: R=0W
     =>	emf = Ir
	I = emf/r
	I = 12/0.050
	I = 240A
  b.ii.	When the lamp is connected: R=2.5W
	I = emf/(R+r)
	I = 12/(2.5+0.05)
	I = 4.71A

	P = I2R
	P = (4.71)22.5
	P = 55.46W

25.a.i.	The initial charging current(Imax) occurs when all of the supply 
	voltage(Vsupply) is across the 1.5kW resistor(R).

	Imax = Vsupply/R
	Imax = 6/1500
	Imax = 4x10-3A

  a.ii.	When fully charged the voltage across the supply voltage is equal to 
	the voltage across the capacitor. 
	Vsupply = Vc = 6V

	Ecapacitor = QVc/2
	Q = CVc
      =>Ecapacitor = CVc2/2
	Ecapacitor = 470x10-6x62/2
	Ecapacitor = 8.46x10-3J 

 a.iii.	Increasing the supply voltage would increase the energy 
	storing capacity of the capacitor. This is because the
	final voltage, across the fully charged capacitor, would
	be higher. 
  25.b.	Etotal = 6.35x10-3J

	fphoton = 5.80x1014Hz
	h = 6.63x10-34Js
	Ephoton = ?

	Ephoton = hfphoton
	Ephoton = 6.63x10-34 x 5.80x1014
	Ephoton = 3.84x10-19J

	Etotal = NEphoton
	N = Etotal/Ephoton
	N = 6.35x10-3/3.84x10-19
 	N = 1.65x1016

26.a.i.	The amplifier is acting in inverting mode. 

  a.ii.	The gain of the amplifier is calculated using the equation:

	Voutput/Vinput = -Rfeedback/Rinput
	Voutput = -VinputxRfeedback/Rinput
	Voutput = -18-3x1.6x106/20x103

	Voutput = -1.44V

 a.iii.	The output voltage is not affected. This is because the output
	voltage is still much lower than the supply voltage. Only when
	the calculated output voltage gets close to, or greater than,
	the supply voltage is the output voltage affected.



27.a.	dsinq = nl
	d = 2.16x10-6m
	n = 2
	l = 486x10-9m
	q = ?

	sinq = nl/d
	sinq = 2x486x10-9/2.16x10-6
	sinq = 0.45
	q = 26.74o

   b.i.	Angle i = 47o
	Angle r = 27o
	nglass = sin(i)/sin(r)
	nglass = sin47o/sin27o
	nglass = 0.731/0.454
  	nglass = 1.61 required

  b.ii.	qcritical = sin-1(1/n)
	qcritical = sin-1(1/1.61)
	qcritical = sin-1(0.613)
	qcritical = 38.4o

	At point X the incident angle of 63o is greater than
	the critical angle. This means that the light is totally 
	internally reflected at this boundary.


	A high voltage or other energy source can be  used to pump 
	electrons up into higher energy states. For example, an 
	electron can be pumped up to energy level (E2) and then
	fall into the metastable state E1, creating what is called 
	an inverted population. A passing photon, having an energy equal 
	to the energy gap E1 to E0 can encourage/stimulate an electron 
	to drop from energy state E1 to E0 with  the production of a 
	photon in phase, with the same frequency and travelling 
	parallel to the stimulating photon. Thereafter, photons 
	produced by stimulated  emission can cause further stimulated 
	emission. This is the basis for stimulated emission and 

  b.i.	P = IA

	I = 1020Wm-2

	A = pr2	
	A = px(5.00x10-4)2
	A = 7.85x10-7m-2

	P = 1020x7.854x10-7
	P = 8.01x10-4W

  b.ii.	The laser beam is non divergent. It does not spread out. 
	This means the radius of the spot is a constant.

29.a.i.	The reaction is induced fission. The reaction is
	described as this type because the Pu nuclei absorb
	neutrons and become unstable. The Pu nuclei then split
	and release energy.

  a.ii.	The energy released from the reaction is a result of
	the mass of the products being less than the mass of 
	the reactants.
	Mass of reactants = 3.9842x10-27kg
	Mass of products = 3.9825x10-27kg

	Mass defect = (3.9842-3.9825)x10-27kg
	Mass defect = 0.0017x10-27kg

  	This mass defect is converted into energy. 
	The energy is calculated using the equation: E = mc2

	E = 0.0017x10-27x(3x10-27)2
	E = 1.53x10-13J

   b.i.	D = E/m
	E = 9.6x10-5J
	m = 0.5kg

	D = 9.6x10-5/0.5
	D = 1.92x10-4J/kg (Gy)

  b.ii.	H = QD

	Q = 1
	D = 1.92x10-4J/kg

	H = 1x1.92x10-4
	H = 1.92x10-4Sv

 b.iii.	Three half value thicknesses of lead (120mm)
	are required.


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