Monday, August 19, 2019

The diagram shows an experiment which has been set up to demonstrate two-source interference. Microwaves of wavelength λ pass through two slits S1 and S2.


Question 25
The diagram shows an experiment which has been set up to demonstrate two-source interference. Microwaves of wavelength λ pass through two slits S1 and S2.


The detector is moved from point O in the direction of the arrow. The signal detected decreases until the detector reaches point X, and then starts to increase again as the detector moves beyond X.

Which equation correctly determines the position of X?
A OX = λ
B OX = λ / 2
C S2X – S1X = λ
D S2X – S1X = λ / 2





Reference: Past Exam Paper – June 2014 Paper 12 Q27





Solution:
Answer: D.


A minima is found at X since the signal keeps on decreasing until it reaches X (so at X, the signal has the lowest amplitude). X is the first minima.

S2X and S1X are distances.


The condition for destructive interference is that the path difference should be nλ / 2.

For the first minima (n = 1),
Path difference = λ / 2
S2X – S1X = λ / 2

This distance (λ/2) is equal to the path difference between S2X and S1X which is (S2X – S1X). [D is correct]


The distance between the central maxima (O) and the first maxima is called the fringe separation x. It can be obtained from
x = λD / a                    (double-slit experiment)


The distance between the central maxima (O) and the first minima (at X) is half the fringe separation (= x/2            and x is given by the above formula).

So, OX is NOT equal to λ / 2.

Sunday, August 18, 2019

The force diagram shows an aircraft accelerating. At the instant shown, the velocity of the aircraft is 40 m s-1.


Question 31
The force diagram shows an aircraft accelerating. At the instant shown, the velocity of the aircraft is 40 m s-1.

At which rate is its kinetic energy increasing?
A 2.4 MW                    B 8.0 MW                    C 12 MW                     D 20 MW





Reference: Past Exam Paper – March 2016 Paper 12 Q15





Solution:
Answer: C.

The ‘rate of increase of kinetic energy’ has unit J/s. This is the unit for power.

The forward acceleration is due to a resultant force horizontally.

Horizontally, resultant force = engine thrust – air resistance

Resultant force = 500 – 200 = 300 kN


Power = Force × velocity
Power = 300 000 × 40 = 12 000 000 W
Power = 12 MW

Saturday, August 17, 2019

A satellite of mass mS is in a circular orbit of radius x about the Earth. The Earth may be considered to be an isolated uniform sphere with its mass M concentrated at its centre.


Question 4

A satellite of mass mS is in a circular orbit of radius x about the Earth.

The Earth may be considered to be an isolated uniform sphere with its mass M concentrated at its centre.


(a) (i) Show that the kinetic energy EK of the satellite is given by the expression
EK = GMmS / 2x

where G is the gravitational constant. Explain your working. [3]

(ii) State an expression, in terms of G, M, mS and x, for the potential energy EP of the
satellite. [1]

(iii) Using answers from (i) and (ii), derive an expression for the total energy ET of the
satellite. [2]


(b) Small resistive forces acting on the satellite cause the radius of its circular orbit to change.

Use your answers in (a) to state, for the satellite, whether each of the following quantities
increases, decreases or remains constant.

(i) total energy [1]

(ii) radius of orbit [1]

(iii) potential energy [1]

(iv) kinetic energy [1]





Reference: Past Exam Paper – November 2015 Paper 41 &42 Q1






Solution:
The gravitational force provides the centripetal force.
{Gravitational force = Centripetal force}
GMmS / x2 = mSv2 / x
{msv2 = GMms / x
Multiply by half on both sides,
½ msv2 = GMms / 2x
EK = GMms / 2x}

(ii) EP = – GMmS / x   

(iii)
ET = EK + EP
ET = GMmS / 2x GMmS / x
ET = – GMmS / 2x                   


(b)
(i) decreases
{The total energy decreases as work needs to be done against the resistive forces.}

(ii) decreases
{Since the total energy decreases, the radius of orbit also decreases.
ET = – GMmS / 2x
The total energy decreases, that is, it becomes more negative. The value of (GMmS / 2x) should be greater so that the total energy (with its negative sign) decreases. Since the value of (GMmS / 2x) increases, the radius of orbit, x, decreases.}

(iii) decreases
{The smaller the radius of orbit, the lower the potential energy. Recall that the potential energy is maximum (and equal to zero) at infinity.}

(iv) increases
{EK = GMms / 2x
From the formula, as the radius x decreases, the EK increases.}
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