Thursday, October 23, 2014

Physics 9702 Doubts | Help Page 1

  • Physics 9702 Doubts | Help Page 1




Question 1: [Kinematics > Linear motion]
An aircraft is at rest at one end of a runway which is 2.2km long. The aircraft accelerates along the runway with an acceleration of 2.5ms-2 until it reaches its take-off speed of 75 ms-1. Calculate

(i) the time taken to reach take-off speed
(ii) the distance travelled in this time
(iii) Just as the aircraft reaches take-off speed a warning light turns on. The maximum possible deceleration of the aircraft is 4.0ms-2 and 2.5s elapses before the pilot takes any action, during which times the aircraft continues at its take off speed. Determine whether or not the aircraft can be brought to rest in the remaining length of runway.
Reference: Cambridge International AS and A Level Physics Coursebook – 2nd edition – by David Sang, Graham Jones, Gurinder Chadha and Richards Woodside



Solution 1:
(i)
Initial speed, u = 0ms-1 (aircraft is at rest)
Acceleration, a = 2.5ms-2
Final speed, v = 75ms-1 (take-off speed)
a = (v - u) / t
Time taken to reach take-off speed, t = (v – u) / a = (75 – 0) / 2.5 = 30s

(ii)
Consider the equation for uniformly accelerated motion:
s = ut + ½at2
Distance travelled during this time, s = 0(30) + 0.5(2.5)(302) = 1125m

Alternatively,
Consider the equation for uniformly accelerated motion:
v2 = u2 + 2as
(75)2 = 02 + 2(2.5)s
Distance travelled during this time, s = 752 / 2(2.5) = 1125m
{I think the first method is better since the question asks for the distance travelled during that time and the second method does not make use of the time}

(iii)
Remaining length of runway when warning light turns on = 2200 – 1125 = 1075m
Distance travelled during the 2.5s before the pilot takes any action:
Speed (= distance / time) = 75ms-1
Distance = Speed x Time = 75 x 2.5 = 187.5m

Remaining length of runway for deceleration = 1075 – 187.5 = 887.5m
Consider the deceleration to be maximum (= 4ms-2) [acceleration, a = – 4ms-2]
Initial speed, u = 75ms-1 and final speed, v = 0ms-1 (aircraft brought to rest)

Let’s try to calculate the distance required to bring the aircraft to rest, s and compare it with the remaining length of runway.
Consider the equation for uniformly accelerated motion:
v2 = u2 + 2as
02 = 752 + 2(-4)s
s = (752) / 8 = 703.125m

Since the distance required to bring the aircraft to rest (703.125m) is less than the remaining length of runway (887.5m), the aircraft can be brought to rest in the remaining length of runway.





Question 2: [Current of Electricity > Resistance]


(a) {Detailed explanations for this part of the question is available as Solution 391 at Physics 9702 Doubts | Help Page 72 - http://physics-ref.blogspot.com/2015/02/physics-9702-doubts-help-page-72.html}

(b) Some heaters each labelled 230V, 1.0kW. Heaters have constant resistance. Determine total power dissipation for heaters connected as shown in each diagram below.

Reference: Past Exam Paper - November 2010 Paper 21 Qu6(b)



Solution 2:
(a) {Detailed explanations for this part of the question is available as Solution 391 at Physics 9702 Doubts | Help Page 72 - http://physics-ref.blogspot.com/2015/02/physics-9702-doubts-help-page-72.html}

(b)


{The heaters are each 230V, 1.0kW. This means that when there is a potential difference of 230V across a heater, there is a power consumption of 1.0kW.}
(i)
{When electrical components are connected in parallel to a power supply, the same potential difference (as the emf of the power supply) would occur through each of the components.
Here, there is a potential difference (p.d.) of 230V across each heater. So, each of them would have a power consumption of 1.0kW.}
Total power dissipation (= 1.0 + 1.0) = 2.0kW

Alternatively,
Let resistance of a heater = R
Power dissipation = V2 / R
{For a power supply of 230V, there is a power dissipation of 1.0kW}
The total resistance in the circuit = [(1/R) + (1/R)]-1 = R/2
Since R is now halved,
Total Power dissipation = V2 / 0.5R = 2 (V2 / R) = 2(1.0kW) = 2.0kW

(ii)
{When electrical components are connected in series to a power supply, the total e.m.f of the source is divided into p.d. across the different components (depending on their resistance) so that the total p.d. in the circuit is equal to the e.m.f of the power supply (given there is no loss volts)
Here, since both heaters have the same resistance, the e.m.f is divided equally into 2 as the p.d. across each heater. That is, there is a p.d. of 115V across each heater.
Power dissipation across a heater (= V2 / R) is proportional to the (p.d.)2 across it. So, if the p.d. across a heater is halved, the power dissipation is (= (½)2 = ¼) one quarter. Therefore, the power dissipated in each heater in the diagram is (1.0kW) / 4 (= 0.25kW).}
Total power dissipation (= 0.25 + 0.25) = 0.5kW

Alternatively,
Total resistance in circuit = R + R = 2R
Since R is now doubled,
Total power dissipation = V2 / 2R = 0.5 (V2 / R) = 0.5(1.0kW) = 0.5kW

(iii)
{The first method would be too time-consuming here. So, consider only the alternative method as the 2 above}
The circuit can be simplified as a series connection of a power supply and a heater, (both of which are) connected in series to a parallel combination of 2 heaters.
Total resistance in circuit = R + [(1/R) + (1/R)]-1 = R + R/2 = 3R / 2
Since total resistance is now 3R /2 (= 1.5R)
Total Power dissipation = V2 / 1.5R = (2/3) (V2 / R) = (2/3)(1.0kW) = 0.67kW









Question 3: [Deformation of Solids > Young modulus]
Wire of length 1.70 m hangs vertically from fixed point, as shown. Wire has cross-sectional area 5.74 × 10–8 m2 and is made of material that has Young modulus of 1.60 × 1011 Pa. Load of 25.0 N is hung from wire.


(i) Extension of wire
(ii) Same load is hung from second wire of same material. This wire is twice the length but same volume as first wire. Explain how extension of second wire compares with that of first wire

Reference: Past Exam Paper – June 2011 Paper 21 Q4(b)



Solution 3:
(i)
Young modulus, E = Stress / Strain
Young modulus, E = [F / A] / [e / l] (= Fl / Ae)
Extension, e (= Fl / AE) = (25 x 1.70) / ([5.74x10-8] x [1.6x1011])
Extension, e = 4.6x10-3m

(ii)
(Volume = Area x length)
(Since wire is of the same material, the Young modulus is the same. Also, same load = same F)
The area A becomes A / 2 OR the stress is doubled. Since the extension, e is proportional to (l / A) OR Substitute into the full formula {e = Fl / AE}. So, the total extension increase {for the second wire} is 4e {since l becomes 2(l) and A becomes (A/2) – both the change in l and A should be considered}





Question 4: [Current of Electricity > Variable resistor]
Variable resistor in (a) now connected as potential divider, as shown. Maximum possible and minimum possible current I2 in ammeter:


Reference: Past Exam Paper – June 2011 Paper 21 Q5(b)



Solution 4:









(The total current from the battery is divided at the junction connecting the fixed resistor and the variable resistor. I2 is the current flowing through the 6.0Ω resistor. )
{How the circuit works? Consider the point A and B. 

The resistance contributed by the variable resistor changes as the contact is moved. At A (which is connected to the +ve terminal of the battery), the resistance of the variable resistor is zero.

In this second circuit, the fixed resistor is connected in parallel to the variable resistor while for the first diagram, the fixed resistor is connected in series with the variable resistor. 

For a parallel connection, current splits up at a junction. From Ohm’s law, I = V / R. That is, the lower the resistance, the greater is the current flowing through it. Since the variable resistor has zero resistance at A, all current would flow through it. Thus, current I2 through the fixed resistor is zero.

In this case, it’s as if the fixed resistor has been short-circuited by the variable resistor of zero resistance.}
Minimum possible current = 0


{The resistance of the fixed resistor (6.0Ω) is known. Note that by choosing a resistance for the variable resistor, the circuit becomes a parallel combination of the fixed resistor and the resistance chosen for the variable resistor. 

From Kirchhoff’s law, there will be the same p.d. across both the fixed resistor and the variable resistor. This p.d. is equal to the e.m.f. in the circuit (that is 12V).

From Ohm’s law, I = V / R        
Current I2 through fixed resistor = 12 / 6.0 = 2.0A.}
Maximum possible current (= 12 / 6) = 2.0A







Question 5: [Sound Waves > CRO]
Loudspeaker produces sound wave of constant frequency. How cathode-ray oscilloscope (c.r.o.) may be used to determine frequency:

Reference: Past Exam Paper – November 2010 Paper 23 Q3



Solution 5:

Below is a diagram for the c.r.o:




A microphone / (terminals of) the loudspeaker is connected to the Y-plates of the c.r.o.
The c.r.o. is adjusted to produce a steady wave of 1 (or 2) cycles / wavelengths on the screen.
The length of a cycle / wavelength λ is measured and the time-base b is noted.
The frequency f is given by f = 1 / λb.

Source for 1st diagram: http://jjphysics.pbworks.com/f/10_11_H1_Wave+Motion_notes.pdf
Source for 2nd diagram: Cambridge International AS and A Level Physics Coursebook – 2nd edition – by David Sang, Graham Jones, Gurinder Chadha and Richards Woodside, Page 180







Question 6: [Kinematics > Linear motion]
Student takes measurements to determine value for acceleration of free fall. Some of apparatus used is illustrated in Fig. Student measures vertical distance d between base of electromagnet and bench. Time t for iron ball to fall from electromagnet to bench is also measured. Corresponding values of t2 and d are shown.

(a) Draw line of best fit
(b) Explain why there is a non-zero intercept on the graph
(c) Determine student’s value for (i) Diameter of ball            (ii) Acceleration of free fall

Reference: Past Exam Paper – November 2010 Paper 23 Q4



Solution 6:
(a)
Line of best fit for points:
An acceptable straight line (touching every point) should be drawn

(b)
Explain why there is a non-zero intercept on graph:
The distance fallen by the iron ball is not d. Distance d is the distance fallen by the ball plus the diameter of the ball

(c)
Determine student’s value for
(i)
Diameter of ball:
(The diameter is the value of y-intercept)
Diameter of ball: (allow) 1.5 ± 0.5cm

(ii)
Acceleration of free fall, g:
Gradient of graph = 4.76 ± 0.1 with evidence that the origin {point (0,0)} has not been used {since the line intercepts the y-axis at a value greater than 0}
{A graph of distance against (time)2 is plotted. Consider the equation ofr uniformly accelerated motion: s = ut + ½at2 where d = s and a = g here. Compare with y = mx + c where m is the gradient and c the y-intercept. From equation of motion, gradient of a s-t2 graph would be equal to ½g [just like the gradient m is multiplied by x, the gradient should be multiplied by the x-axis quantity (which is t2). So ½ g represents the gradient]}
Gradient of graph = g / 2
So, the acceleration of free fall, g = 9.5ms-2


  





Question 7: [Units]

Cylindrical tube rolling down slope of inclination θ moves distance L in time T. Equation relating these quantities is L (3+a2/P) = QT2sinθ where a is internal radius of tube and P and Q are constants.

Which line gives the correct units for P and Q?



            P                      Q
A         m2                    m2s-2
B         m2                    ms-2
C         m2                    m3s-2
D         m3                    ms-2


Reference: Past Exam Paper – November 2011 Paper 11 Q4 & Paper 13 Q5







Solution 7:

{Ans: B.} Expanding the left hand side gives 3L + a2L/P. 

For equation to be homogeneous (and hence correct), the units should be the same on both sides of the equation and the units for 3L = units for a2L/P. 

L has unit m (since it is a distance), so all the different parts should have an overall unit of m. 

Considering the a2L/P: [m2][m] / [P] = m. So, [P] = [m3] / [m] = m2

Now for QT2sinθ: sinθ has no units. So, [Q][s2] = m giving [Q] = ms-2.

 

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