Solutions to SQA examination

1992 Higher Grade Physics

Paper II Solutions

1.a.	From 0s up to 1.5s(point B) the trolley is moving away from the sensor.
	At 1.5s the trolley stops, changes direction, and moves back towards the 
	sensor for the next 2s.
	
	This means the trolley is at its greatest distance from the trolley after 1.5s.
	
   b.	From OA			From AB			From BC
	
	u = 0m/s		u = 1.5m/s		u = 0m/s
	v = 1.5m/s		v = 0m/s		v = -1m/s
	t = 0.5s		t = 1s			t = 2s
	a = ?			a = ?			a = ?

	a = (v-u)/t  		a = (v-u)/t		a = (v-u)/t
	a = (1.5-0)/0.5		a = (0-1.5)/1		a = (-1.0-0)/1
	a0A = 3m/s/s		aAB = -1.5m/s/s	    	 aBC = -1m/s/s


	

   c.	
	
	Fslope = component of gravitational force acting down the slope.

   d. 	When moving up the slope: both Fslope and Ffriction are acting in 
	the same direction. This means that the unbalanced force causing 
	the acceleration between A and B is the sum of these two forces.

	Funbalanced =  Fslope + Ffriction


	When moving down the slope: Ffriction acts up the slope, in the
	opposite direction to Fslope. This means that the unbalanced force 
	causing the acceleration between B and C is the difference between 
	these two forces.

	Funbalanced =  Fslope - Ffriction

	The acceleration is therefore greater when the trolley is moving up the slope.

2.a.i.	Step 1 : Calculate the energy available from 0.7kg of fuel.

	Eincrease = mass of fuel x energy per kilogram
	Eincrease = 0.7 x 2.8x107
	Eincrease = 19.6x106J
	
	Ek(initial) = 0J
	Ek(final) = Ek(initial) + Eincrease
	Ek(final) = 19.6x106J
	Ek(final) = mv2/2

	v2 = 2Ek(vehicle)/m
 	v = [2Ek(vehicle)/m]1/2
	v = [2x19.6x106/500]1/2
	v = [78400]1/2
	v = 280m/s

	Note: 	It is not clear from the question whether the 500kg mass is that
		of the vehicle and dummy after, or before, burning the fuel. To 
		get an answer exactly equal to 280m/s you seem to have to assume
		that the 500kg is the mass after burning the fuel.

 a.ii.	u = 0m/s
	v= 280m/s
	t = 8s
	a = ?

	a = (v-u)/t
	a = (280 -0)/8
	a = 35m/s/s

	s = ut +at2/2
	s = 0 + 35x82/2
	s = 1120m

  b.i.	
	
	The horizontal velocity of the projectile is constant at 280m/s.
	The vertical component of the motion will be up to a maximum 
	height and then fall back down to earth.

  b.ii.	The maximum height reached by the dummy will occur when the
	vertical component(vver) of the velocity is zero.

	vver2 = uver2 + 2asver
	sver = (vver2 - uver2)/2a
	sver = (02 - 502)/2x(-10)
	sver = -2500/-20
	sver = 125m


3.a.	Step 1: calculate the time that the projectile takes to fall 0.8m.

	sver = -0.8m
	uver =  0m/s
	aver = -10m/s/s
	tver =  ?

	s = ut + at2/2
	t2 = 2(s - ut)/a
	t2 = 2(-0.8 - 0)/-10
	t2 = 0.16s
	t = 0.4s

	vhor = shor/t
	vhor = 0.2/0.4
	vhor = 0.5m/s

3.b.i.	Momentum before(Pbefore) = Momentum after(Pafter)
 
  b.ii.	mpelletupellet + mputtyuputty= (mpellet+mputty)v
	5.0x10-4upellet+0.1x0 = (5.0x10-4+0.1)0.5
	5.0x10-4upellet = 0.05025
	upellet = 0.05025/5x10-4
	upellet = 100.5m/s		

  c.	Reducing the mass of the lump of putty will mean that after the collision
	the pellet embedded in the putty will travel a greater horizontal displacement.
	This will give a more accurate horizontal velocity, which in turn will give
	a more accurate value of the pellets initial velocity.

  a.ii.	Take each pair of pressure and temperature reading.
	For each pair calculate P/T.

	89x103/253  = 351.8 Pa/K
	96x103/273  = 351.6 Pa/K
	103x103/293 = 351.5 Pa/K
	110x103/313 = 351.4 Pa/K
	117x103/333 = 351.3 Pa/K

	From these calculations it is clear that : P/T = Constant

  a.iii.As the temperature increases the helium atoms have more kinetic energy.
	With increased kinetic energy the atoms are moving faster and collide
	with the container walls, and the pressure probe, more frequently
	and forcefully. The pressure(force/area) therefore increases.

  b.i.	P/T = constant
	T = P/constant
	T = 24000Pa/(351.5Pa/K)
	T = 68.3K

5.a.i.	
	

	The variable resistor is connected across the audio output terminals.
	The voltmeter is connected across this resistor and the ammeter measures 
	the current in the circuit.

  a.ii.	The internal resistance has a value equal to the negative gradient of
	the graph.

	r = -m

	m = (y2- y1)/(x2- 1)
	m = (2-1.2)/(0-0.5)
	m = -1.6W

	=>r=1.6W

  a.iii.The emf of the supply is equal to the output voltage when there is no 
	current in the circuit.

	E = 2.0V
	
	I = E/(R+r)
	I = 2.0/(4+1.6)
	I = 0.357A

	Vloudspeaker = IR
	Vloudspeaker = 0.357x4
	Vloudspeaker = 1.4V

  b.	When the voltage drop across the internal resistor and the calibrated
	variable resistor is the same, as will be the case when the voltage 
	across the calibrated resistor is equal to half of the emf, the internal
	resistor and the calibrated resistor must have the same value.


6.a.
	

  b. 	Vs = VR + Vc
	VR = Vs - Vc
	VR = 9-6
	VR = 3V

	I = VR/R
	I = 3/6800
	I = 441.2mA

  c.i.	When fully charged: Vc = 9V

	Energy stored in capacitor = Energy dissipated in resistor 
	E = CV2/2
	E = 1000x10-6x92/2
	E = 40.5mJ

  c.ii.	The energy dissipated in the 20W resistor will be the 
	same as that dissipated in the 10W. This is because
	the energy stored in the capacitor is still the same. The only difference 
	is that it will take longer for the capacitor to discharge when connected 
	across the 20W resistor.


7.a.i.	Equate the energy gained from the electric field(Eelect) to the final kinetic
	energy (Ek)of the electron.

	Eelect = Ek
	qV = mev2/2
	v2 = 2qV/me
	v2 = 2x1.6x10-19x5000/9.11x10-31
	v2 = 1.75x1015
	v = 41.9x106m/s

  a.ii.	v2 = (2qV/me)1/2 
	v2 = (2x1.6x10-19x2500/9.11x10-31)1/2
	v2 = 29.6x106m/s

	v1 = const x V11/2
	v2 = const x (V1/2)1/2

	v2/v1 = const x (V1/2)1/2/const x (V1)1/2
	v2/v1 = 1/sqrt2
	v2 = v1/sqrt2

  b.	Q = It		(Q= total charge)
	Nq = It		(N = number of electrons & q = electronic charge)
	N = It/q
	N = 15x10-3x1/1.6x10-19
	N = 9.375x1016
	
8.a.i.	Vpk = 1.5x0.05
	Vpk = 0.075V

  a.ii.	Rf/R1 = - Vout/Vin
	Rf = -R1Vout/Vin
	Rf = -10000x(0.9/-0.075)
	Rf = 120000W
	Rf = 120kW

  b.i.	Vout = (Rf/R1)(V2-V1)
	Vout = (5x106/10x103)(0.4x10-3)
	Vout = 500x0.40x10-3
	Vout = 0.2V

  b.ii.	The differential amplifier amplifies the difference between the inverting
	and non inverting input. The 50Hz mains signal will be input into both these
	inputs and will therefore be filtered out. This will leave only the amplified
	potential difference between the two pads.

9.a.i.	

	The high voltage pumps electrons up into energy state(E2) which 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 E2. This is the basis for stimulated emission. 

	Note: 	This energy level diagram is only an illustration to help with
		the understanding of the concept of energy levels. Each different
		laser will have a different energy level diagram.
	
  a.ii.	The fully reflecting mirror keeps all the photons in the laser cavity to  
	enable further stimulated emission. The partially reflecting  mirror not 
	only ensures some reflection to sustain the stimulated emission but also 
	allows a narrow beam of laser light to leave the cavity.

  b. 	Each bright spot occurs when the laser light interferes constructively.
	This happens when the path difference, from each adjacent slit in 
	the diffraction grating to the screen, is a whole number of wavelengths.

	In between the bright spots dark areas are produced  when the laser light 
	interferes destructively. The laser light will completely cancel when the 
	path difference, from each adjacent slit in the diffraction  grating to 
	the screen, is an odd number of wavelengths/2.

  c.	The energy from any laser light entering the eye is concentrated 
	into a very small area. This produces a very intense beam that will 
	produce significant damage to the retina.(I = P/A)

	The energy emitted from a 100W lamp is spread over a much larger area.
	This results in relatively low intensity light that will not result
	in any damage to the eye.

  d.	I = P/A
	P = E/t

      =>I = E/At
	I = 0.1/1.5x10-9x0.5x10-3
	I = 133x109W/m2
	

10.a.i. qcritical = sin-1(1/n)
	qcritical = sin-1(1/1.52)
	qcritical = sin-1(0.658)
	qcritical = 41.1o


   a.ii.
	

	Note: This diagram has not been drawn to scale.

  a.iii.
	

	Angles have been omitted for clarity.


  b.	The image would be inverted.


11.a.i.	Atomic number(Z) 95 = Number of protons in the nucleus.
	Mass number(A) 241  = Number of protons + number of neutrons in the nucleus.

  a.ii.	30kBq means that there are 30,000 nuclear disintegrations per second,
	or equivalently, 30000 nuclei decay every second.

  a.iii.Radiation can cause atoms or molecules to lose electrons and become
	positive ions. These atoms, or molecules, have been ionised by the 
	ionising radiation.

  a.iv.	Use conservation of atomic and mass number to find the type of radiation.

	Mass Number
	241 = 237 + A
	A = 4

	Atomic number
	95 = 93 + Z
	z = 2

	Radiation = 42He  (alpha radiation)

  a.v.	Smoke detectors are usually located on the roof of rooms. This means that
	any alpha radiation emitted will either be absorbed by the smoke detector 
	construction or the surrounding air, making it impossible for the radiation 
	to cause any biological damage.

  b.i.	Radioactive isotopes contained in natural rocks or soil contribute to
	background radiation.

  b.ii.	Total absorbed dose = dose per detector x number of detectors
	Total absorbed dose = 1.2x10-8x15000
	Total absorbed dose = 180mGy

   	Dose equivalent = Absorbed dose x Quality factor
	H = DQ
	H = 180x10-6x20
	H = 3.6mSv

	The permissible level of 5mSv per year is NOT exceeded.


 END OF QUESTION PAPER


Return to past paper index page.