Solutions to SQA examination

1997 Higher Grade Physics

Paper II Solutions

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1.a.	Squaring the equation v = at gives:
	v2 = a2t2	equation 1
	The equation s = 1/2(at2) can be rearranged to give:
	t2 = 2s/a which can be substituted into equation 1

	v2 = a2(2s/a)
      =>v2 = 2as	

  b.i.	Funbalanced = Fthrust - Ffriction
	Funbalanced = 3150 - 450
	Funbalanced = 2700N

	a = Funbalanced/m
	a = 2700/1000
	a = 2.7m/s/s
  b.ii.	To calculate the length of the track use : v2 = u2 + 2as

	s = (v2 - u2)/2a
	s = (332- 02)/2x2.7
	s = 201.67m

	Use the cosine rule to calculate the resultant velocity,or,
	draw an accurate scale diagram.

	a2 = b2 + c2 -2bcCosA
	a2 = 362 + 122 -2x36x12Cos140
	a2 = 1296 + 144 - 864Cos140
 	a2 = 1296 + 144 - 864x-0.766
	a2 = 2101.9
	a = 45.8m/s

	Use the sine rule to calculate angle C.

	a/sinA = c/sinC
	sinC = csinA/a
	sinC = 12sin140o/45.8
	sinC = 0.168
	angle C = 9.7o

	Resultant velocity = 45.8m/s with a bearing of 350.3o

2.a.i.	Vhorizontal = Vcos36o
	Vhorizontal = 41.7cos36o
	Vhorizontal = 33.74m/s

  a.ii.	Vvertical = Vsin36o
	Vvertical = 41.7sin36o
	Vvertical = 24.5m/s

  b.	The time for the ball to reach Q(ttotal) can be calculated by summing
	the time to travel from O to P(t1) and the time to travel from
	P to Q(t2).

	ttotal = t1 + t2

	When P is reached the vertical component of the velocity is 0m/s.
	This velocity is the final velocity(v) of the first part of the journey.
	The initial component of the vertical velocity(u) for this part of the
	journey is 24.5m/s.

	v = 0m/s		v = u +at1
	u = 24.5m/s		t1 = (v-u)/a
	a = -9.8m/s/s		t1 = (0-24.5)/-9.8
	t1 = ?			t1 = 2.5s

	When falling from P to Q take the initial velocity(u) as 0m/s for
	this part of the journey.

	u = 0m/s		s = ut2 + 1/2(at22)
	s = -19.6m		s = 1/2(at22)
	a = -9.8m/s		t22 = 2s/a
	t2 = ?			t22 = 2x-19.6/-9.8
				t22 = 4
				t2 = 2s

	ttotal = t1 + t2
	ttotal = 2.5 + 2
	ttotal = 4.5s (As required)

  c.	Shorizontal = Vhorizontalxttotal
	Shorizontal = 33.74x4.5
	Shorizontal = 151.8m/s

3.a.	Use the law of conservation of momentum to solve this problem.
	Pbefore = Pafter

	Before collision
	Pbefore = mPuP + mQuQ
	Pbefore = 0.2x0.5 + 0.3x0
	Pbefore = 0.1kgm/s

	After collision
	Pafter = mPvP + mQvQ
	0.1 = 0.2vP + 0.3x0.4
	vP = (0.1 - 0.3x0.4)/0.2
	vP = -0.1m/s			The negative sign indicates the direction
					is to the left.

  b.	DPp = Pp(after) - Pp(before)
	Pp(after) = mPvP
	Pp(after) = 0.2x-0.1
	Pp(after) = -0.02kgm/s

	Pp(before) = mPuP
	Pp(before) = 0.2x0.5
	Pp(before) = 0.1kgm/s

	DPp = -0.02-0.1
	DPp = -0.12kgm/s
  c.i.	Faveraget = mP(uP - vP)
	Faverage  = mP(uP - vP)/t
	Faverage  = 0.2[(0.5-(-0.1)]/0.06
	Faverage  = 2N

	Note: the direction of this force is to the left.


4.a.	The assumption in this experiment is that the gas in the container
	is the same as the water temperature. To facilitate this the can
	must be fully immersed to allow the gas and water temperature to 

  b.	P1 = 100kPa
	T1 = (17+273)K = 290K
 	P2 = ?
	T2 = (75+273)K = 348K

	P1/T1 = P2/T2 
	P2 = P1T2/T1
	P2 = 100x348/290
	P2 = 120kPa

  c.	P = F/A
	F = PA
	F = 120x103x0.001
	F = 120N

  d.	The mass and the volume of gas are fixed in this experiment,therefore,
	the density(mass/volume) must also remain constant. 

5.a. 	A battery  emf of 12V will supply 12J of energy to each coulomb 
	of charge passing through the cell.

  b.i.	P = V2/R
	R = V2/P
	R = 122/48
	R = 3W

  b.ii.	When S1 is closed the circuit can be treated as a simple series
	circuit with the internal resistor(r) in series with the headlamp 

	Rtotal = Rh + r
 	Rtotal = 3 + 0.05
	Rtotal = 3.05W

	I = emf/Rtotal
	I = 12/3.05
	I = 3.93A

	Vheadlamp = IRh
	Vheadlamp = 3.93x3
	Vheadlamp = 11.8V (as required)

  c.i.	The bulb and the starter motor are in parallel. The resistance of this
	network Rp is calculated as shown below:

	1/Rp = 1/Rbulb + 1/Rmotor
	1/Rp = 1/3 + 1/0.12
	1/Rp = 8.66...
	Rp  = 0.1154W

	Rtotal = Rp + r
	Rtotal = 0.115 + 0.05
	Rtotal = 0.165W

  c.ii.	I = emf/Rtotal
	I = 12/0.165
	I = 72.7A

6.a.i.	Gain = Voutput/Vinput
	Gain = -9x30mv/5x2mV
	Gain = -270mV/10mv
	Gain = 27

  a.ii.	Vpeak = 1.414Vrms
	Vrms = Vpeak/1.414
	Vrms = 190.9mV

  a.iii.Gain = Voutput/Vinput =-Rf/R1
	To produce a gain of magnitude 27 choose:

	Rf = 270kW
	R1 = 10kW
	Other values are acceptable as long as the ratio of Rf to R1
	is 27.

  b.i.	The bridge is balanced when the digital voltmeter reads 0V.
	This means the potential at X and Y are equal and there is 
	no potential difference between these points.

  b.ii.	Voutput = (-Rf/R-)(V--V+)
	(V--V+) = -Voutput(R-/Rf)
	(V--V+) = -(-0.18)(100x103/1x106)
	(V--V+) = 0.18(0.1)
	(V--V+) = 0.018V

	VY-VX = 0.018V

  b.iii.At 20oC the potential at Y(VY) falls to a value less than it was
	when the bridge was balanced at 23oC.

	As VY is less than VX =>  VY-VX is negative.

	     =>	Voutput=(-Rf/R-)(VY-VX)
		Voutput= positive value.

	Note: 	If the resistance of the thermistor varies linearly with temperature in
		range 20oC to 26oC the output voltage at 20oC will be +0.18V. 

7.a.i.	All the supply voltage is across the resistor at the instant the switch
	is closed. At this instant Vsupply = Vresistor.

	I = 100x10-6
	R = 150,000W
	Vresistor = ?

	Vresistor = IRresistor
	Vresistor = 100x10-6x150000
	Vresistor = 15V

	Vsupply = Vresistor = 15V

  a.ii.	The current in the circuit must first be calculated when the voltage
	across the resistor is 6V.

	I = ?
	Vresistor = 6V
	R = 150,000W

	I = V/R
	I = 6/150000
	I = 40x10-6A = 40mA 

	From the graph read the time when the current is 40mA.
	Time = 40s

  a.iii.Vsupply = Vresistor + Vcapacitor
	Vcapacitor = Vsupply - Vresistor
	Vcapacitor = 15 - 6
	Vcapacitor = 9V

  b.i.A	Q = CV

	C = 16mF = 16x10-6
	V = 6kV = 6000V	(This is the voltage across the capacitor when fully charged)
	Q = ?

	Q = 16x10-6 x 6000
	Q = 0.096C

  b.i.B	E = 1/2(QV)
	E = 1/2(0.096X6000)
	E = 288J	

  b.ii.	I = Q/t
	I = 0.096/2x10-3
	I = 48A

8.a.i.	A maxima occurs when two waves interfere constructively. This happens
	when waves are in phase.
	A minima occurs when two waves interfere destructively. This happens when 
	the two waves are 180o out of phase.

	The minima is produced when the path difference YQ-XQ is equal to 3/2l.

	YQ-XQ = (5.2-4.0)m = 1.2m
	YQ-XQ = 3/2l
	l = 0.8m

  b.i.	The speed of sound is constant in air, therefore increasing the frequency
	will decrease the wavelength. If the frequency is increased by a factor of 
	5 the wavelength will decrease by a factor of 5.

	Wavelength at 1000Hz = 1.2/5 = 0.24m

  b.ii.	If the path difference is fixed at 1.2m this will represent a whole 
	number of wavelengths for certain frequencies and produce maxima, however,
	for other frequencies this will represent a whole number of half wavelengths
	and produce minima. For this reason a series of maxima and minima are produced.

9.a.	nplastic = sinqair/sinqplastic	
	nplastic = sin15o/sin10o	(Note that it is the angle between the ray 
					and the normal that must be used.)

	nplastic = 1.49

  b.i.	qcrit = sin-1(1/nglass)
	qcrit = sin-1(1/1.44)
	qcrit = sin-1(0.694)
	qcrit = 44.0o


  b.ii.	Ray Q will be refracted and partially reflected.
	Ray P will be totally internally reflected because the incident angle
	of 45o is greater than the critical angle.	


  a.ii.	Individual electrons in the metal atom absorb the energy of a single 
	photon, the energy of which is dependent on the frequency of the photon.
	The photon energy can be calculated using the equation E = hf.   Below
	a certain frequency the energy absorbed by an electronin the metal, from 
	the photon, is below that required to escape from the metal atom, thus 
	the current in the circuit will be zero.

  b.i.	The threshold frequency will provide electrons with just enough 
	energy to overcome the work function of the metal and eject electrons 
	with no excess kinetic energy. Thus to estimate the work function from the 
	graph the line must be extrapolated until it cuts the x-axis of the 

	X-axis intercept = 6.7x1014Hz
	This is the threshold frequency.

  b.ii. The work function is calculated using:

	Ework function = hfthreshold	
	Ework function = 6.63x10-34x6.6x1014
	Ework function = 4.4x10-19J

	This is closest to the work function of the metal calcium

11.a.	Alpha particles have a positive charge. It was therefore concluded
	that electrostatic repulsion from a like charge, of very large mass,
	was required to produce a large angle deflection. However, as large 
	angle deflections were very rare, and most alpha particles were
	undeviated from their original path,  it led Rutherford to conclude 
	that most of the atom was empty space with the large positive mass
	concentrated at the centre of the atom.

  b.i.	The total dose equivalent received is calculated using the equation:
	H =QD.

	Hgamma = QgammaDgamma
	Hgamma = 1x200
	Hgamma = 200mSv

	Hneutrons = QneutronsDneutrons
	Hneutrons = 3x100
	Hneutrons = 300mSv

	Htotal = Hgamma + Hneutrons
  	Htotal = 200 + 300
	Htotal = 500mSv

	Three half value thicknesses are required.
	3t1/2 = 3x8 = 24mm



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