Solutions to SQA examination

1998 Higher Grade Physics

Paper II Solutions

1.a.i.	Component of the force acting parallel to the slope(F_s) is calculated using:
	F_s = mgsinq
	F_s = 2.0x9.8xsin20^o
	F_s = 6.7N

  a.ii.	


  a.iii.To calculate the acceleration use [N2]
	a = F_unbalanced/m

	F_unbalanced = F_Friction+F_s
	F_unbalanced = -1.3 + (-6.7)	Where vectors acting down the slope are taken as negative.
	F_unbalanced = -8N

	a = -8/2.0
	a = -4m/s²
	Which equates to a deceleration of 4m/s², as required.
	
  a.iv.	At its furthest point up the slope the trolley will be at rest.

	v = 0m/s			v = u + at
	u = 3m/s			t = (v-u)/a
	a = -4m/s²			t = (0-3)/-4
	t = ?				t = 0.75s

  a.v.	v² = u² + 2as
	s = (v² - u²)/2a
	s = (0² - 3²)/(2x-4)
 	s = -9/-8
	s = 1.125m

  b.i.


  b.ii.	The force vectors in the above diagram are acting in opposite directions. This will 
	produce a smaller unbalanced force than that calculated in part a.iii. This means
	that the magnitude of the acceleration will be less when the trolley moves down the 
	slope.

2.a.	The weight of the student is equal to the scale reading when the lift is moving
	at a steady speed. This weight is used to calculate the mass of the student.

	w = mg
	m = w/g
	m = 588/9.8
	m = 60kg

  b.	The boy is accelerating at the same rate as the lift. The unbalanced force producing 
	this acceleration is equal to the scale reading when accelerating minus the steady  
	state scale reading. The diagram below illustrates this with the scale reading 
	represented by the the upward force labelled T and downward force, weight,
	represented by W.
 

	F_unbalanced = 678-588
 	F_unbalanced = 90N

	a = F_unbalanced/m
	a = 90/60 
	a = 1.5m/s²

  c.	The lift is decelerating when the upward force is less than the weight.	
 
	F_unbalanced = 498 - 588
	F_unbalanced = -90N		The negative indicates that the unbalanced force
						acts in the downwards direction.

	a = F_unbalanced/m
	a = -90/60
	a = -1.5m/s²

	This can be thought of as an upward deceleration of 1.5m/s² 

  d.	


3.a.i.	V = d/t₁		d = diameter of ball = 24mm = 0.024m
				t = time for ball to pass through light gate = 0.060s


	V = 0.024/0.060
	V = 0.4m/s

  a.ii.	F_avg = change in momentum/contact time
	F_avg = (mv-mu)/t
	F_avg = m(v-u)/t
	F_avg = 0.045(0.4-0)/0.005
	F_avg = 3.6N

  b.i.	Percentage error in mass 		= (0.01/45)x100 	= 0.02%
	Percentage error in contact time	= (0.001/0.005)x100 	= 20%
	Percentage error in 		t₁	= (0.001/0.060)x100 	= 1.67%
	Percentage error in ball diameter	= (1/24)x100 		= 4.17%

	The greatest uncertainty is in the contact time measurement.

  b.ii.	The percentage error in the result can be taken as equal to the largest 
	individual error. 

	20% of 3.6N = (20/100)x3.6 = 0.72N

	F_avg = (3.6 +- 0.72)N

4.a.	Initial volume of gas (V₁) in the container = (8-5)litres = 3 litres = 3 x10^-3m³
	Density of gas (r) = 1.29kg/m³
	
	m = rV₁
	m = 1.293x(3x10^-3m³)
	m = 3.87x10^-3Kg

  b.i.	Pressure before pumping (P_before) = 1.01x10⁵Pa	
	Pressure after pumping (P₁) = 3.0x10⁵Pa	

	Volume of air before and after pumping(V₁) is fixed at 3x10^-3m³
	Area(A) = 7.0x10³m²

	To calculate the force, after pumping, use: P₁ = F/A 

	F = P₁A
	F = 3.0x10⁵x7.0x10^-3
	F = 2100N

  b.ii.	When the volume of the water is 2 litres, the volume of the gas in the container(V₂) 
	is 6 litres,or 6x10^-3m³. The pressure at this time is P₂.

	P₁V₁ = P₂V₂
	P₂ = P₁V₁/V₂
	P₂ = 3x10⁵x3x10^-3/6x10^-3
	P₂ = 1.5x10⁵Pa



5.a.i.	V_tpd = emf - Ir

	I = emf/(R+r)
	I = 1.5/(3+0.75)
	I = 1.5/3.75
	I = 0.4A

	V_tpd = 1.5 - 0.4x0.75
	V_tpd = 1.2V

  a.ii.	V_lost = Ir		or	emf = V_tpd+V_lost
	V_lost = 0.4x0.75		V_lost = emf - V_tpd
	V_lost = 0.3V		 	V_lost = 1.5 - 1.2 = 0.3V	

  b.i.	R = 1W
	V_lost = 2V

	V_tpd = emf - V_lost
	V_tpd = 6 - 2
	V_tpd = 4V

  b.ii.	I = V_tpd/R_variable
	I = 4/1
	I = 4A

	V_lost = Ir		V_lost = emf - V_tpd
	r = V_lost/I		V_lost = 6 - 4 = 2V
	r = 2/4
	r = 0.5W

6.a.	A capacitor of value 5mF will store 5 coulombs of charge 
	per volt across it.

  b.i.	Initially all the supply voltage is across the resistor in the circuit. 
	The current in the circuit at this point(I_initial) is 30mA, as read
	from the graph.

	I_initial = 30mA = 30x10^-6A
	V = 6V
	R = ?

	R = V/I
	R = 6/30x10^-6
	R = 2x10⁵W

  b.ii.	Halving the value of the resistor will double the initial charging current. Lowering 
	the resistance will also decrease the charging time.

	

  c.i.	The resistance of the variable resistor must be decreased to keep the charging current
	constant.

  c.ii.	To calculate the total charge transferred use: Q = It
	The values for I and t are read from the graph.

	Q = ?			
	I = 100mA = 100x10^-6A		
	t = 10s

	Q = Ixt 
	Q = 100x10^-6x10
	Q = 100x10^-5C = 1x10^-3C = 1mC
  
  c.iii.To calculate the capacitance use: Q = CV_capacitor
	At 10s V_capacitor = 5V

	Q = 1x10^-3C
	V_capacitor = 5V
	C = ?

	C = Q/V_capacitor
	C = 1x10^-3/5
	C = 2x10^-4F = 200mF


7.a.i.	To calculate the kinetic energy(E_k) use : E_k = mv²/2

	E_k = ?
	m_e = 9.11x10^-31kg
	v = 4.2x10⁷m/s

	E_k = mv²/2
	E_k = 9.11x10^-31x(4.2x10⁷)²/2
	E_k = 8.03x10^-16J 

7.a.ii.	The gain in the kinetic energy of the electron, with charge e, is a result of it being
	accelerated by a potential difference V.

	E_k(gain) = eV	
	V = E_k(gain)/e
	V = 8.03x10^-16/1.6x10^-19
	V = 5021.9V

  b.	Plate P can be made more positive to attract the electron. This will move the
	spot up vertically towards X. 
	Making Q more positive will attract the electron and move it horizontally
	towards X.If plate Q is made twice as positive as plate P the combined effect 
	will be to move the spot to position X.
	

8.a.i.	The amplifier is working in inverting mode. 
  
a.ii.	Gain =  -R_feedback/R_input
	
	R_feedback = 2x10⁶W
	R_input = 100x10³W

	Gain = -2x10⁶/100x10³ 
	Gain = -20

  a.iii.


  b.i.	Between 2ms and 8ms the gain of the amplifier is said to be saturated. This is due to 
	the fact that the output voltage from the amplifier cannot be greater than that supplied
	to it.

  b.ii.	To produce saturation you can:increase the size of the feedback resistor;
decrease the size of the input resistor;
increase the size of the input voltage.

9.a.	DE₃₂ = E₃-E₂
	DE₃₂ = -2.416x10^-19-(-5.524x10^-19)
	DE₃₂ = 3.006x10^-19J

	To calculate the wavelength of a photon of this energy use : E = hf = hc/l

	l = hc/E₃₂
	l = 6.63x10^-34x3x10⁸/3.006x10^-19
	l = 6.617x10^-7m = 661.7nm

	This corresponds to spectral line Z.

	This answer can also be deduced by the fact that the photon produced by the transition 
	between the energy levels E₃ and E₂has the lowest energy of the four in the diagram.
 	This lowest energy photon will also have the longest wavelength,which is Z.  

  b.	The more photons of a particular frequency emitted the brighter the spectral line
	will be. This means that there must be more electron transitions between the energy
	levels producing the lines Y and Z.

  c.i.  E = hf
	E = 6.63x10^-34x7.486x10¹³   
	E = 4.96x10^-20J

  c.ii.	DE₅₄ = E₅-E₄
	DE₅₄ = -0.864x10^-19-(-1.360x10^-19)
	DE₅₄ = 4.96X10^-20J

	The transition between energy levels E₅ and E₄ produces the infrared photon.

10.a.i. 


	Angle i = 50^o
	Angle r = 28^o
	
	n_glass = sin(i)/sin(r)
	n_glass = sin 50^o/sin 28^o
	n_glass = 0.766/0.496
  	n_glass = 1.63

  a.ii.	q_critical = sin^-1(1/n)
	q_critical = sin^-1(1/1.63)
	q_critical = sin^-1(0.613)
	q_critical = 37.8^o

  b.i.	n_quartz = l_air/l_quartz 
	l_quartz = l_air/n_quartz 
	l_quartz = 510/1.55
	l_quartz = 329nm

  b.ii.	As the wavelength of visible light increases the refractive index of the quartz(n_quartz)
	decreases. This means that (1/n_quartz) increases and therefore q_critical[sin^-1(1/n_quartz)]
	also increases.

  b.iii.Flint glass has a higher refractive index than crown glass for all wavelengths of light. 
	This would result in the spectrum being projected onto a lower position on the screen. 
	Additionally, because the difference in refractive index across the spectrum is greater 
	with flint glass, the distance between the red end and the violet end of the spectrum, 
	as appearing on the screen, would be greater.

11.a.i.	To conserve atomic and mass number in the decay of Thorium to Palladium a 
	beta particle(b, ⁰_-1e)is emitted.

  a.ii.	Number of neutrons(n) = mass number(A) - atomic number(Z)
	n = 238 - 92
	n = 146

  a.iii.²³⁴₉₂U -> ²³⁰₉₀Th + ⁴₂He

  b.i.	Soil and rocks containing natural radioactive isotopes, and cosmic rays,  contribute 
	to background radiation.

  b.ii.	The earths atmosphere absorbs cosmic radiation and reduces the effective dose equivalent 
	rate at sea level. At heights above 4km there is less atmosphere to absorb the radiation 
	which results in a higher dose equivalent rate at these rarefied altitudes.

  b.iii.
     (A)t = 7h
	H/t = 5mSv/h
	H = ?

	H = tx5mSv/h
	H = 7hx5mSv/h
	H(flight) = 35mSv
     
     (B)During 40 flights: H = 40x35mSv 

	H = 1400mSv (As a result of flights)
	
	In addition to this the traveller will be exposed to sea level background radiation. 
	Time at sea level t_sea = (365x24)-(7x40) = 8480h
	H(at sea level) = 8480hx0.2mSv/h
	H(at sea level) = 1696mSv

	H(total) = H(at sea level) + H(flight)
	H(total) = 1696mSv + 1400mSv
	H(total) = 3096mSv = 3.096(mSv/year)


					END OF PAPER


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