Solutions to SQA examination

1993 Higher Grade Physics


Paper II Solutions


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1.a.	The ball is in free fall for the first 0.5s.

	u = 0m/s
	v = -4.5m/s
	t = 0.5s
	a = ?

	a = (v-u)/t 
	a = (-4.5-0)/0.5
	a = -9.0m/s/s
	
  b.	DEk = Ekafter-Ekbefore
	DEk = 1/2(mv2-mu2)
	2DEk = m(v2-u2)
	2DEk/(v2-u2) = m
	m = 2DEk/(v2-u2)	
	m = 2x-1.7/(42-[-4.52])
	m = -3.4/(16-20.25)
	m = 0.8kg

  c.	DP = mv-mu
	DP = m(v-u)
	DP = 0.8(4-[-4.5])
	DP = 6.8kgm/s

  d.	The time that the ball is in contact with the ground(tc) is 0.5s.
	F = DP/tc
	F = 6.8/0.5
	F = 13.6N

2.a.	
	
	
	w = ?
	m = 15000
	g = 1.6N/kg
	
	w = mg
	w = 15000x1.6
	w = 24000N

	The upward force(Fup) is equal in magnitude to the weight
	and acts in the opposite direction. There is no unbalanced force
	which means the craft must be moving at a constant speed and be moving
	in a straight line.

  b.i.	Funbalanced = Fup - w
	Funbalanced = 25500 - 24000
	Funbalanced = 1500N

	a = Funbalanced/m
	a = 1500/15000
	a = 0.1m/s/s upwards

	This means that the craft has an upward deceleration of -0.1m/s/s.


  b.ii.	a = 0.1m/s/s
	t = 18s
	u = -2.0m/s
	v = ?

	v = u + at
	v = -2.0 + 0.1x18
	v = -0.2m/s

  b.iii.v2 = u2 + 2as
	s = (v2-u2)/2a
	s = ([-0.2]2-[-2]2)/2x0.1
	s = (0.04-4)/0.2
	s = -3.96/0.2
	s = -19.8m

	This displacement is equal to that which the craft has descended from.
	This means the craft must have been 19.8m above the surface of the moon
	to start with.

3.a. 	 The law of conservation of linear momentum states that :
	Pbefore = Pafter

  b.i.	Before collision
	Pbefore = mAuA + mBuB
	Pbefore = 1400xuA + 1000x8
	Pbefore = 1400uA + 8000

	After collision
	Pafter = mAvA + mBvB
	vA=vB=v
	Pafter = (mA + mB)v
	Pafter = (1400 + 1000)15
	Pafter = 36000kgm/s

	Equate: Pbefore and Pafter
	1400uA + 8000 = 36000kgm/s
	1400uA = (36000-8000)/1400
	uA = 20m/s
				
  b.ii.	Consider the instant after the collision to be the starting point
	for this calculation.

	u = 15m/s
	v = 0m/s
	s = 20m
	a = ?

	v2 = u2 + 2as
	a = (v2-u2)/2s
	a = (02-152)/2x20
	a = -225/40
	a = -5.625m/s/s

	Ffriction = Fun
	Fun = ma
	Fun = 2400x-5.625
	Fun = -13,500N 		The negative sign indicates that the force
				acts to the LHS.

	OR
	
	The kinetic energy lost by the car is equal to the work done by 
	the frictional force over the 20m stopping distance.

	Eklost = Ework
	
	Eklost = Ekbefore - Ekafter 
	Eklost = (mv2 - mu2)/2
	Eklost = m(v2 - u2)/2
	Eklost = 2400(152 - 02)/2
	Eklost = 27000J

	Ework = Fd
	F = Ework/d
	F = 27000/20
	F = 13500N

  b.iii.Most of the kinetic energy will be converted into heat energy.

4.a.	P1 = 400kPa
	V1 = 1000cm3
	P2 = 250kPa
	V2 = ?

	P1V1 = P2V2
	V2 = P1V1/P2 
	V2 = 400x1000/250
	V2 = 1600cm3

  b.	The pressure increases.
	The volume remains constant.
	The temperature increases.

  c.	The starting pressure is that indicated by point B on the graph 
	and the final pressure is that indicated by point C. Over this 
	period the volume is constant.
	
	T1 = 300K
	P1 = 200kPa
	T2 = ?
	P2 = 500kPa

	P1/T1 = P2/T2
	T2 = P2(T1/P1)
	T2 = 500(300/200)
	T2 = 750K

5.a.i.	Q = CV
	C = Q/V
	C = 24x10-6/1.5
	C = 16x10-6F
	C = 16mF

  a.ii.	Mean = total/N
	Mean(C) = (16+18+20+16+15)/5
	Mean(C) = 17mF

	Random error = (max-min)/N
	Random error(C) = (20-15)/5
	Random error(C) = +-1mF

	C = (17+-1)mF

  a.iii.Measure the voltage across the capacitor directly by connecting
	the voltmeter across it. Additionally, to prevent leakage through 
	the voltmeter it would be better to use an oscilloscope for this
 	purpose.

  b.	The effective resistance of the capacitor in the circuit decreases 
	as the frequency of the supply increase. As the resistance decreases 
	the current in the circuit increases and the lamp glows more brightly. 
	
6.a.	The electromotive force is the energy supplied by the power supply
	to each coulomb of charge.

  b.i.	P = I2R	
	I2 = P/R
	I = (P/R)1/2
	I = (8.0/0.32)1/2
	I = 251/2
	I = 5A

  b.ii.	V = IR
	V = 5x0.32
	V = 1.6V

  b.iii.Vr = emf - Vheater
	Vr = 2.0 - 1.6
	Vr = 0.4V

	Vr = Ir
	r = Vr/I
	r = 0.4/5
	r = 0.08W

  c.	Two heaters in parallel will reduce the resistance of the circuit.
	The current in the heater elements will increase which has the 
	effect of increasing the rate at which the water is heated.

	The resistance of the two heaters in parallel can be calculated as 
	shown below:

	Rp = Rheater/2 
	Rp = 0.32/2
	Rp = 0.16W

	Rtotal = Rp + r
	Rtotal = 0.16 + 0.08
	Rtotal = 0.24W

 	I = emf/Rtotal
	I = 2.0/0.24
	I = 8.33A

	Vr = Ir
	Vr = 8.33x0.08
	Vr = 0.67V

	Vheater = emf - Vr
	Vheater = 2.0 - 0.67
	Vheater = 1.33V

	This is the voltage across each heater element.

	Iheater = Vheater/Rheater
	Iheater = 1.33/0.32
	Iheater = 4.16A

	The power developed in each heater can be calculated using:
	P = I2R

	P = 4.162x0.32
	P = 5.5W

	The two heating elements develop a total power of 11W.
	This means that the water is heated about 1.4 times faster.

7.a.i.	The amplifier is working in inverting mode.

  a.ii.	Gain = -Rfeedback/R1
	Gain = -1x106/2x103
	Gain = 500

  a.iii.
	

  a.iv.	The LED will pulse ON/OFF.
	The LED will be on when Vout is positive.
	The LED will be off when Vout is zero. 

  b.i.	The op-amp is working in differential mode.
	The gain equation for the op-amp in this mode is:

	Vout = (Rf/R1)(V2-V1)

  b.ii.	Use the equation in b.ii. to calculate Vout. Take the 
	voltages V1 and V2 from the graphs given.
Time interval/ms V2/mV V1/mV Vout/V
0-50 0 0 0
50-100 -6 -6 0
100-150 +6 0 +3
150-200 0 -6 +3
200-250 0 0 0
250-300 -6 -6 0
300-350 +6 0 +3
350-400 0 -6 +3
	

8.a.	A ray of sunlight is made up of many different wavelengths that
	is perceived as white light. However, each individual wavelength
	represents a particular colour of light.
	The diffraction grating diffracts the different wavelengths in 
	sunlight by different amounts: the longer wavelengths being 
	diffracted the most and the shorter wavelengths the least. 
	At position X on the screen the short wavelength violet light
	has interfered constructively to produce a band of violet light.
	At position Y on the screen the long wavelength red light
	has interfered constructively to produce a band of red light.
	Between X and Y the other colours of the visible spectrum produce 
	bands of indigo, blue, green, yellow and orange light.

  b.i.	dsinq = nl
	
	d = 1/N = 1/6x105 = 1.667x10-6m
	q = ?
	n = 1
	l = 410nm = 410x10-9m

	sinq = nl/d
	sinq = 1x410x10-9/1.667x10-6
	sinq = 0.246
	q = sin-10.246
	q = 14.2o

  b.ii.	d = 1.667x10-6m
	q = 14.2o + 9o = 23.2o
	n = 1
	l = ?

	l = dsinq/n 
	l = 1.667x10-6sin(23.2o)/1
	l = 656.7x10-9m = 656.7nm
	
  c.	The diffraction grating will produce many spectra symmetrical about the 
	q = 0o line.
	The prism produces one spectrum by refraction.  
	The relative positions of each colour of light produced  
	by refraction will be the reverse of that produced by the 
	diffraction grating, as blue light is refracted more than 
	red light. 

9.a.	Monochromatic means the laser light has a single wavelength/frequency.
	Coherent means the waves are in phase/step.

  b.i.	DE = hf 	where DE = E2-E2
	f = E2-E2/h
	f = -4.67x10-20-(-6.55x10-20)/6.63x10-34
	f = 452.5nm

  b.ii.	The wavelength of the emitted radiation corresponds to the blue end of 
	the spectrum.

  c.	The beam is intense because the energy is spread over a small area and
	a large number of photons per second are emitted from the laser.

10.a.	Light incident on the pn-junction photodiode will increase the number
	of electron hole pairs and consequently increase the conductivity of 
	the diode.

  b.	The diode is operating in photoconductive mode.

  c.	I1d12 = I2d22
	I2 = I1d12/d22
	I2 = 3.0x12/0.752
	I2 = 5.33mA

11.a.i.	


  a.ii.	Take the background count to be 20cpm.
	Add this to the corrected count rate and plot as shown below.




  b.	Time/years	Corrected Count Rate/cpm
	0		520
	5.25		260
	11.5		130
	16.75		65
	21		32.5

	After 21 years the count rate is 32.5cpm.


		

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