Solutions to SQA examination

1999 Higher Grade Physics

Paper II Solutions

1.a.i.	u = 0m/s			s = ut + 1/2at2
	a = 4m/s2			s = 0x7 + 1/2x4x72
 	t = 7s
	s = ?				s = 98m

  a.ii. First Test
	To calculate the final speed in the first test(v1)use:v1 = u1 + at.
		
	=>v1 = 0 + 4x7
	=>v1 = 28m/s
	

	Second Test
	u2 = 0m/s		v22 = u22 + 2as
	a = 4m/s2		v22 = 02 + 2x4x196
	s = 196m		v22 = 1568m
	v2 = ?			v2 = 39.6m/s	
					

	Increase in speed = v2 - v1
	Increase in speed = 39.6m/s - 28m/s
	Increase in speed = 11.6m/s


a.iii.	Third Test		  
	u = 40m/s			v2 = u2 + 2as
	a = -2.5m/s2			=>s = (v2 - u2)/2a
	v = 0m/s			=>s = (02-402)/2x-2.5 
	s = ?				=>s = -1600/-5
					=>s =  320m

b.i.To calculate the acceleration of the trolley the student needs to measure the:
 
length of the card(d);
time it takes the trolley to travel from the first light gate to the second(t1).


The computer must measure the time the card cuts the light beam produced by the 
light gate at:
the top of the slope(t2);
the bottom of the slope(t3).


b.ii.	The speed at the top of the slope(u) is calculated using:
	u=d/t2
	
	The speed at the bottom of the slope(v) is calculated using:
	v=d/t3
	
	The acceleration is calculated using:

	a=(v-u)/(t1)


2.a.	The bungee rope is at its maximum length at 3.6s. At this time the sign of
	the velocity of the bungee jumper changes, indicating a change in their direction 
	from down to up.
 
  b.	To calculate the unbalanced force acting on the jumper, the jumper's acceleration 
	must be calculated and then Newtons second law of motion applied.	

	u = -18m/s		a = (v-u)/t
	v =  16m/s		a = (16--18)/3
	t =  3s			a = 34/3
	a = ?			a = 11.33m/s2

	[N2] F = ma
	     F = 55x11.33
	     F = 623.3N
  
  c.	An elastic rope must be used to ensure the change in velocity does not take place 
	over a very short time interval. If the time interval was short the acceleration of 
	the jumper and hence the unbalanced force acting on the jumper would be much greater.
	This force could be large enough to injure the jumper.

3.a.	Total momentum before Pbefore = Total momentum after Pafter

  b.i.	Change in momentum of vehicle A(DP(A)) = P(A)after-P(A)before
	Change in momentum of vehicle B(DP(B)) = P(B)after-P(B)before


	DP(A) = MA(VA-UA)
	DP(A) = 0.75(0.82-0.4)
	DP(A) = 0.315kgm/s
	
	DP(B) = MB(VB-UB)
	DP(B) = 0.5(0-0.63)
	DP(B) = -0.315kgm/s

	Notes: 	i. Vector quantities to the right have been assigned positive values. 
		ii.Vector quantities to the left must be assigned negative values.

	The above calculations therefore show that the change in momentum of vehicle A
	is equal and opposite to that of vehicle B.

  b.ii.	If kinetic energy is conserved the collision is elastic, and if it is not
	the collision is inelastic.

	Ekbefore = 1/2MAUA2 + 1/2MBUB2
	Ekbefore = 0.5X0.75X0.822 + 0.5X0.5X02

4.a.	
 

  b.i.	The lower surface area of the submarine will experience a greater pressure 
	than the top surface because it is a greater depth below the surface.
	This pressure difference means that there will also be a greater force
	exerted on the lower surface than the top surface as the force(F) on each
	surface is equal to the product of the pressure(p) and area(A).

	Fupthrust = Fbottom - Ftop

	

    ii.	mass = density x volume
	mass = 1.02x103 x 14.5
	mass = 14790Kg

   iii.	The effective mass of the submarine is decreased when water in the tanks is 
	replaced with air, however, the upthrust force does not change. Therefore, as the 
 	acceleration is calculated using newtons second law, a = Fupthrust/m, the acceleration 
	increases as mass decreases.

5.a.	When the bridge is balanced:
	Rvariable/Rthermistor = R1/R2 

	Where 	R1 = 500W
	 	R2 = 900W
		Rvariable = 450W

		Rthermistor = (RvariablexR2)/R1
		Rthermistor = (450x900)/500)
		Rthermistor = 810W


  b.i.	Mean = Total/Number of readings
	Mean = (852+854+848+851+853)/5
	Mean = 4258/5 
	Mean = 851.6W

  b.ii.	Random Error = (max-min)/Number of readings
	Random Error = (854-848)/5
	Random Error = 6/5
	Random Error = 1.2W

  c.	A systematic error is one that will make all measurements either too high or too
	low.

6.a.	Vlost = emf - (V1+V2)
	Vlost = 1.60 - (1.20 + 0.30)
	Vlost = 0.10V

  b.	Vlost = Ir
	r = Vlost/I
	r = 0.1/0.04
	r = 2.5W

  c.i.	I = emf/(R1+R2+r)

	Where 	R1 = 30	
		R2 = Variable resistor
		r = 2.5W
	
	=>IR1 + IR2 + Ir = emf 
	=>IR2 = emf - IR1 - Ir
	=>R2  = (emf - IR1 - Ir)/I
	=>R2  = (1.60 - 0.02x30 - 0.02x2.5)/0.02
	=>R2  = 0.95/0.02
	=>R2  = 47.5W

  c.ii.	Vtpd = emf - Vlost
	Vlost = Ir
	
	By decreasing the value of the variable resistor the current in the circuit increases.
	Consequently, the "lost volts"  increase, leading to a drop in the Vtpd. 

7.a.i.	Ipeak = 1.414Irms
	Ipeak = 1.414x200
	Ipeak = 282.8mA
  
  a.ii.	


  b.i	To calculate the charge stored on the capacitor, when the current is 2mA, the 
	voltage across the capacitor (Vc) must be calculated and then the equation
	Q = CVc used. 

	Vsupply = Vc + Vresistor
	Vc = Vsupply - Vresistor 
	
	Vresistor = IR
	Vresistor = 0.002x4000
	Vresistor = 8V

	Vc = 12 - 8
	Vc = 4V

	Q = CVc
	Q = 5000x10-6x4
	Q = 0.02C

  b.ii.	When the capacitor is fully charged the potential difference across it is 12V.

	E = CV2/2
	E = 5000x10-6x122/2
	E = 0.360J

8.a.i.	The amplifier is acting in inverting mode. The gain is therefore given by the equation:

	Voutput/Vinput = -Rfeedback/Rinput
	Voutput = -VinputxRfeedback/Rinput
	Voutput = -0.20x400/22

	Voutput = 3.64V

  a.ii.	The theoretical output voltage with a 10MW feedback resistor is calculated as above.

	Voutput = -VinputxRfeedback/Rinput
	Voutput = -2.0x10000/22
	Voutput = -909.1V

	However, the maximum output voltage from the amplifier is limited by the voltage 
	supplied to the amplifier. In this case the the output voltage is limited to a range
	of -13V to +13V, therefore the output voltage would be -13V.

  b.i.	The amplifier is operating in differential mode.
  b.ii.	When there is no difference between the voltages input into the inverting input(V-)
	and non inverting input(V+) the output voltage(Voutput) will be 0V.
	At this voltage the technician knows the light sources are equally bright.
  
  b.iii.V- = 1.65V
	V+ = 1.85V
	
	Vouput = -(Rfeedback/R1)x(V--V+)
	Vouput = -(100/10)x(1.65-1.85)
	Vouput = -10x-0.20
	Vouput = -2.0V

  b.iv.	The output voltage will:
become less negative as V- approaches 0.85V;
be equal to 0V when V- is equal to 0.85V;
become positive when V- is greater than 1.85V.
	The above statements can be explained by reference to the equation:
	Vouput = -(Rfeedback/R1)x(V--V+).

	As V- increases the difference between V- and V+ changes.
	The affect of this is as described above.

9.a.	qcritical = sin-1(1/nwater)
	qcritical = sin-1(1/1.33)
	qcritical = sin-10.752
	qcritical = 48.75o

	As the incident angle is greater than the critical angle the light is 
	totally internally reflected.

  b.	The refractive index of the new material is equal to the gradient of the graph.

	n = 0.4/0.28
	n = 1.43	(taking the answer to two decimal places)

  c.	lliquid = lair/n
	lliquid = 670nm/1.47
	lliquid = 455.8nm

10.a.i.	dsinq = nl
	
	d = 5.0x10-6m
	q = 14o
	n = 2
	l = ?

	l = dsinq/n
	l = 5.0x10-6sin14o/2
	l = 6.04x10-7m

  a.ii.	By decreasing the slit separation to 2.0x10-6m the distance between maxima, 
	as well as angle q, will increase. This will result in the measurement errors 
	producing smaller percentage errors, and therefore lead to a more accurate value of l.

  b.i.	White light is made up of many different wavelengths of light. At the position of the
	central maximum there is no path difference between any pair of waves of a particular
	wavelength. These waves are in phase and interfere constructively to produce a bright 
	band of white light.

  b.ii.	All other maxima are in the form of a continuous spectra because constructive
	interference, of each wavelength produced by the source, takes place at a different
	position on the screen. This leads to a bright band of each colour where that
	wavelength is interfering constructively.

11.a.i.	The addition of group III or V atoms to pure semiconductor is called doping.

   a.ii.Doped semiconductor has a lower resistance.

   b.i.	The diode is operating in photoconductive mode.
  
  b.ii.	Photons incident on the pn junction provide enough energy to excite electrons
	into the conduction band and leave holes in the valence band. The intensity of light 
	determines the number of charge carriers produced. At low intensity fewer charge carriers 
	are produced resulting in a smaller current flowing in the circuit for a given applied 
	potential difference.

  c.	E = hf	f = c/l  

	E = hc/l
  	l = hc/E
	l = (6.63x10-34x3x108)/2.3x10-19
 	l = 864.8nm

12.a.	To conserve mass number and atomic number the emmitted radiation must be beta(b).
	b = 0-1e

   b.i.	D = E/m 
	E = Dm	 	
	E = 750x10-6x0.04
	E = 3x10-5J or 30mJ
	
   b.ii.H/t = QxD/t
	12.5x10-6 = Qx750x10-6/120
	Q = 12.5x10-6x120/750x10-6
	Q = 2

  c.	Number of half value thicknesses(N(t1/2)) = Total thickness/half value thicknesses(t1/2)
 	N(t1/2) = 24/8
	N(t1/2) = 3
	
	     t1/2     t1/2    t1/2
	1200 --> 600 --> 300 --> 150

	The new count rate is 150cpm.




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