Monday, September 26, 2016

Physics 9702 Doubts | Help Page 242

  • Physics 9702 Doubts | Help Page 242



Question 1121: [Gravitation]
(a) By reference to the definition of gravitational potential, explain why gravitational potential is a negative quantity. [2]

(b) Two stars A and B have their surfaces separated by a distance of 1.4 × 1012 m, as illustrated in Fig. 1.1.


Fig. 1.1
Point P lies on the line joining the centres of the two stars. The distance x of point P from the surface of star A may be varied.
The variation with distance x of the gravitational potential φ at point P is shown in Fig. 1.2.


Fig. 1.2
A rock of mass 180 kg moves along the line joining the centres of the two stars, from star A towards star B.
(i) Use data from Fig. 1.2 to calculate the change in kinetic energy of the rock when it moves from the point where x = 0.1 × 1012 m to the point where x = 1.2 × 1012 m.
State whether this change is an increase or a decrease. [3]

(ii) At a point where x = 0.1 × 1012 m, the speed of the rock is v.
Determine the minimum speed v such that the rock reaches the point where x = 1.2 × 1012 m. [3]

Reference: Past Exam Paper – June 2016 Paper 43 Q1



Solution 1121:
(a)
The (gravitational) potential at infinity is defined as/is zero. The (gravitational) force is attractive so work got out/is done as the object moves from infinity.
(so the potential is negative)

(b)
(i)
ΔE = mΔφ
{Change in KE = Change in PE. If PE is lost, then KE has been gained and vice versa. Since in this case, PE has been lost, it means that KE has been gained.
At x = 1.2 × 1012 m, φ = – 14 × 108 J kg-1
At x = 0.1 × 1012 m, φ = – 10 × 108 J kg-1
Δφ = final φ – initial φ = – 14 × 108 – (– 10 × 108) = – 4 × 108 J kg-1
The calculations below are considering only the magnitude since KE is gained.}
ΔE = 180 × (14 – 10) × 108
ΔE = 7.2 × 1010 J        
The change is an increase {in KE}.

(ii)
{The rock needs to reach only the point where the field is zero. In the graph, this is where the change in gravitational potential is zero. Value of φ = – 4.4 × 108 J kg-1. The rock needs only to reach this point. The rock would be attracted by the field of the other planet afterwards.}
Energy required = 180 × (10 – 4.4) × 108       OR Energy per unit mass = (10 – 4.4) × 108
½ × 180 × v2 = 180 × (10 – 4.4) × 108            OR ½ × v2 = (10 – 4.4) × 108
Minimum speed v = 3.3 × 104 m s–1









Question 1122: [Measurement > Accuracy and Precision]
Four students each made a series of measurements of the acceleration of free fall g. The table shows the results obtained.
Which set of results could be described as precise but not accurate?

Reference: Past Exam Paper – June 2008 Paper 1 Q5



Solution 1122:
Answer: D.
A precise set of results have values that are very close to each other. An accurate set of results is one in which the mean is close to the actual value (g = 9.81) of the quantity under consideration.

Choice A: Mean = (9.81 + 9.79 + 9.84 + 9.83) / 4 = 9.8175
Choice B: Mean = (9.81 + 10.12 + 9.89 + 8.94) / 4 = 9.69
Choice C: Mean = (9.45 + 9.21 + 8.99 + 8.76) / 4 = 9.1025
Choice D: Mean = (8.45 + 8.46 + 8.50 + 8.41) / 4 = 8.455 [precise but not accurate]









Question 1123: [Current Electricity > Internal resistance]
A battery of electromotive force (e.m.f.) 12 V and internal resistance r is connected in series to two resistors, each of constant resistance X, as shown in Fig. 5.1.


Fig. 5.1
The current Ι1 supplied by the battery is 1.2 A.
The same battery is now connected to the same two resistors in parallel, as shown in Fig. 5.2.


Fig. 5.2
The current Ι2 supplied by the battery is 3.0 A.
(a) (i) Show that the combined resistance of the two resistors, each of resistance X, is four times greater in Fig. 5.1 than in Fig. 5.2. [2]

(ii) Explain why Ι2 is not four times greater than Ι1. [2]

(iii) Using Kirchhoff’s second law, state equations, in terms of e.m.f., current, X and r, for
1. the circuit of Fig. 5.1,
2. the circuit of Fig. 5.2.
[2]

(iv) Use the equations in (iii) to calculate the resistance X. [1]

(b) Calculate the ratio
power transformed in one resistor of resistance X in Fig. 5.1 / power transformed in one resistor of resistance X in Fig. 5.2
[2]

(c) The resistors in Fig. 5.1 and Fig. 5.2 are replaced by identical 12 V filament lamps.
Explain why the resistance of each lamp, when connected in series, is not the same as the resistance of each lamp when connected in parallel. [2]

Reference: Past Exam Paper – November 2014 Paper 22 Q5



Solution 1123:
(a) (i)
in series 2X     or in parallel X / 2
{In series, combined resistance RS = X + X = 2X
In parallel, combined resistance RP = [1/X + 1/X]-1 = [2/X]-1 = X/2}
other relationship given and 4× greater in series (than in parallel)
{RS / RP = 2X / 0.5X = 4}

(ii) Due to the internal resistance, the total resistance for the series circuit is not four times greater than the resistance for the parallel circuit

(iii)
1.
{e.m.f = sum of p.d.}
E = I1(2X + r)              or 12 = 1.2(2X + r)

2. E = I2(X/2 + r)         or 12 = 3.0(X/2 + r)

(iv)
{From 12 = 1.2(2X + r),} 2X + r = 10            and {From 12 = 3.0(X/2 + r),} X/2 + r = 4
{Subtracting the 2 equations,}
X = 4.0 Ω

(b)
P = I2R            or V2 / R          or VI  
ratio = [(1.2)2 × 4] / [(1.5)2 × 4] = 0.64

(c) The resistance (of a lamp) changes with V (or I).
V (or I) is greater in the parallel circuit (or circuit 2)              OR V (or I) is less in the series circuit (or circuit 1) 

  
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