Solutions to SQA examination

Higher Grade Physics

2002

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


Section B

21.a.i.  Initial displacement(s1) from sensor = 0.2m 

  a.ii.  Final displacement(s2) from sensor   = 1.8m

	Ds =  s2 - s1
	Ds =  1.8 - 0.2
	Ds =  1.6m

  a.iii. s = 1.6	s = ut + (at2)/2
	 u = 0m/s	s = (at2)/2
	 t = 0.6s	a = 2s/t2
	 a = ?		a = 2x1.6/0.62
			a = 3.2/0.36
			a = 8.9m/s2   (as required)

    b.	Mean = total/N
	Mean(a) = (8.9+9.1+8.4+8.5+9.0)/5
	Mean(a) = (43.9)/5
	Mean(a) = 8.78m/s2 

	Random error = (max-min)/N
	Random error(a) = (9.1-8.4)/5
	Random error(a) = (9.1-8.4)/5
	Random error(a) = (0.7)/5
	Random error(a) = 0.14m/s2

	a = (8.8+-0.1)m/s2

	NB: It is only useful to quote the final mean and 
	    error to the same number of decimal places as 
	    the least accurate individual measurement.

	
    d.	


	i. The contact time is greater with the softer surface.	
	ii.The rebound height will be less because the sponge
	   will absorb more of the balls kinetic energy.
	   The reduced rebound height can also be explained 
	   as a result of the average upward force exerted by 
	   the sponge on the ball being less.
      iii. As the ball will sink more into the sponge the
	   maximum displacement from the sensor will be 
	   greater.
	

	
	
22.a.	P1 = 109kPa
	T1 = (15+273)K = 288K
	
 	P2 = ?
	T2 = (45+273)K = 318K

	P1/T1 = P2/T2 
	P2 = P1T2/T1
	P2 = 109x318/288
	P2 = 120.35kPa


   b.   As the temperature increases the nitrogen gas molecules gain
	kinetic energy. With increased kinetic energy the atoms are moving 
	faster and collide with the container walls more frequently
	and forcefully. The pressure(force/area) therefore increases.

  c.i.	P = 1.75x105Pa
	A = 4.0x10-6m2

	F = PA
	F = 1.75x105x4.0x10-6
	F = 0.7N

  c.ii. Read from the graph the length of the spring when the force
	is 0.7N.

	Length = 35mm

  d.	The assumption in the original set up is that the temperature
	of the water is the same as the temperature of the gas inside 
	the flask. Placing the thermometer inside the flask will give 
	a more direct and accurate reading of the gas temperature.



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

	Before collision
	Pbefore = mvehicleuvehicle + mprobeuprobe
	Pbefore = 2500x0.5 + 1500xuprobe
	Pbefore = (1250+1500uprobe)kgm/s


	After collision
	Pafter = (mvehicle+mprobe)v
	Pafter = (2500 + 1500)0.2
	Pafter = 800kgm/s

	(1250+1500uprobe) = 800	(by conservation of momentum)
	1500u = 800-1250
	1500u = -450
	u = -450/1500
	
	u = -0.3m/s		The negative indicates the probe 
				is initially moving in the opposite
				direction to the vehicle.




  b.i.	The direction of thrust from the engine must be opposite 
	to the direction of motion. This means it must be the 
	probe rocket engine that was switched on. (Favg = -500N)
  NB: 	The average force has a negative value because
	it is acting towards the left.
	
  b.ii.	Favgt = DP
	Favgt = (mv-mu)
	Favgt = m(v-u)
	t = m(v-u)/Favg
	t = [4000(0-0.2)]/-500
	t = 1.6s

  c.	The initial acceleration to the RHS must be produced 
	by the space vehicle rocket engine. To then decelerate
	the combined mass of the probe and vehicle the space 
	probe rocket engine must fire. In the maneuver the 
	space probe rocket must fire for twice the time of the 
	vehicle rocket because it only produces half the thrust.

	Mathematically:
	Fprobetprobe = Fvehicletvehicle 
	500tprobe=1000tvehicle
	tprobe = 2tvehicle


24.a.	An emf of 6V means the battery will supply 6J of energy 
	to each coulomb of charge passing through it.

  b.i.	emf = Vr + VR1 + VR2
	emf = Ir + IR1 + IR2

	I = 200mA = 0.2A

	R2  = (emf - Ir - IR1)/I
	R2  = (6.0 - 0.2x2.0 - 0.2x20)/0.2
	R2  = (1.6)/0.2
	R2  = (1.6)/0.2
	R2  = 8W

  b.ii. Vtpd = emf - Vr
	Vtpd = emf - Ir
	Vtpd = 6.0 - 0.2x2
	Vtpd = 5.6V

  c.	When the switch is closed a larger current is drawn from the
	battery. This increases the voltage drop across the 2.0W
	resistor. This voltage drop, called the "lost volts", therefore 
	increases. Consequently, the voltmeter reading, equal to: 
	emf - "lost volts", decreases.

25.a.i.	Initially all the supply voltage is across the resistor.
	VR =  Vsupply = 6V

	When the capacitor is fully charged: Vsupply = Vcapacitor
	Vcapacitor = 6V

  a.ii. E = CV2/2
	E = (2000x10-6x62)/2
	E = 0.072/2
	E = 0.036J

 a.iii.	Imax = Vsupply/R
	R = Vsupply/Imax
	R = 6/7.5x10-3
	R = 800W
	
   b.
			 

   

26.a.	Period(T) = 4cm x 2ms/cm
	T = 8ms

	f = 1/T
	f = 1/8x10-3
	f = 125Hz
 	
   b.i.	Inverting mode.

   b.ii.The output voltage from the amplifier is calculated using:

	Voutput/Vinput = -Rfeedback/Rinput
     =>	Voutput = -Vinputx(Rfeedback/Rinput)
	Voutput(peak) = -6x(3.0x103/2.0x103)
	Voutput(peak) = -9V
	
  (A)
	

  (B)	VRMS = Vpeak/SQRT2
	VRMS = 9/1.414
	VRMS = 6.36V

 b.iii. As Rfeedback is increased the amplifier saturates. This means the 
	peak output voltage increases to a maximum of about +/- 13V.



27.a.	nglass = sinqair/sinqglass
	nglass = sin20o/sin13o
	nglass = 1.52

   b.	The critical angle is the angle,measured between the ray
	and the normal, at which light striking the glass air
	boundary will be totally internally reflected.
 
   c.	qcritical = sin-1(1/n)
	qcritical = sin-1(1/1.52)
	qcritical = sin-1(0.658)
	qcritical = 41.1o	
        

  c.iii.
	

28.a.	Threshold frequency.
 	
28.b.i.	f0 = 3.33x1014Hz
	work function(f) = hf0
	f = 6.63x10-34x3.33x1014
	f = 2.21x10-19J

  b.ii.	 Ephoton = hf
	 Ephoton = 6.63x10-34x 5.66x1014
	 Ephoton = 3.75x10-19J

	 Ek(electron) = Ephoton - f
	 Ek(electron) = 3.75x10-19 - 2.21x10-19
	 Ek(electron) = 1.54x10-19J

  b.iii. Ek(gain) = qDV
	 q = e = 1.6x10-19C
	 DV = 2.00x104V

	 Ek(gain) = 1.6x10-19 x 2.00x104
	 Ek(gain) = 3.2x10-15J
		

29.a.	Adding the impurity will decrease the resistance.

   b.i.	When the electrons in the n-type material combine with
	holes in the p-type material,as they cross the junction,  
	they lose energy. This energy is emitted as quanta of 
	visible radiation.

    ii.	dsinq = nl
	
	d = 5.0x10-6m
	q = 11o
	n = 2
	l = ?

	l = dsinq/n
	l = 5.0x10-6sin11o/2
	l = 4.77x10-7m
	l = 477nm
  	

30.a.	Start by calculating the mass of pure torbernite(mtorbernite) in 
	the 0.6kg of material.

	mtorbernite = (40/100)x0.6
	mtorbernite = 0.24kg

	The problem can now be solved by proportion.

	MASS  :	ACTIVITY
	1.0kg : 5.9x106decays per second = 5.9x106Bq = 5.9Mbq
	0.24kg: 0.24x5.9Mbq

	0.24kg has an activity of 1.416Mbq

   b.	Alpha

	H = QD
	H = (20x150)mSv
	H = 3000mSv

	H/t = 3000/6
	H/t = 500mSv/h.............1

	Other
	H   = (Qx400)mSv
	H/t = (Qx400/8)mSv/h
	H/t = (Qx50)mSv/h..........2

	Equate 1 & 2

	500mSv/h = (Qx50)mSv/h
	Q = 500/50
	Q = 10

	This radiation must be fast neutrons.


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