Sunday, February 7, 2016

Physics 9702 Doubts | Help Page 234

  • Physics 9702 Doubts | Help Page 234

Question 1097: [Current of Electricity]
Two heating coils X and Y, of resistance RX and RY respectively, deliver the same power when 12 V is applied across X and 6 V is applied across Y.
What is the ratio RX / RY?
A ¼                             B ½                             C 2                              D 4

Reference: Past Exam Paper – November 2007 Paper 1 Q29

Solution 1097:
Answer: D.
Power P = VI = V2 / R

Power delivered by X = Power delivered by Y
122 / RX = 62 / RY
RX / RY = 144 / 36 = 4

Question 1098: [Dynamics]
A stationary nucleus of mass 220u undergoes radioactive decay to produce a nucleus D of mass 216u and an α-particle of mass 4u, as illustrated in Fig. 3.1.

Fig. 3.1
The initial kinetic energy of the α-particle is 1.0 × 10–12 J.
(a) (i) State the law of conservation of linear momentum.
(ii) Explain why the initial velocities of the nucleus D and the α-particle must be in opposite directions.

(b) (i) Show that the initial speed of the α-particle is 1.7 × 107 m s–1.
(ii) Calculate the initial speed of nucleus D.

(c) The range in air of the emitted α-particle is 4.5 cm.
Calculate the average deceleration of the α-particle as it is stopped by the air.

Reference: Past Exam Paper – November 2009 Paper 21 Q3

Solution 1098:
(a) (i) EITHER The sum / total momentum (of system of bodies) is constant           OR total momentum before = total momentum after         
for an isolated system / no (external) force acts on system

(ii) There should be zero momentum before / after decay {since the nucleus if mass 220u is initially at rest} so the α-particle and nucleus D must have momenta in opposite directions {so that the total momentum is also zero}  

Kinetic energy = ½ mv2
{mass m = 4u since an α-particle is a helium nucleus and u is the unified atomic mass constant, not the rest mass of a proton}
1.0 × 10-12 = ½ × 4 × 1.66 × 10-27 × v2
Speed v = 1.7 × 107 m s-1

{In terms of magnitude, the momentum of the α-particle should be equal to the momentum of the nucleus D. Momentum = mv}
1.7 × 107 × 4u = 216u × V
V = 3.1 × 105 m s-1  

{The initial speed of the α-particle is 1.7 × 107 m s-1. When stopped, the speed is zero. Consider the equation: v2 = u2 + 2as    where a is the deceleration and s = 4.5cm = 4.5 × 10-2 m}
(1.7 × 107)2 = 2 × deceleration × 4.5 × 10-2
Deceleration, a = 3.2 × 1015 m s-2

Question 1099: [Waves > Diffraction]
(a) Explain what is meant by the diffraction of a wave.

(b) Light of wavelength 590 nm is incident normally on a diffraction grating having 750 lines per millimetre.
The diffraction grating formula may be expressed in the form
d sinθ = nλ.
(i) Calculate the value of d, in metres, for this grating.
(ii) Determine the maximum value of n for the light incident normally on the grating.
(iii) Fig. 5.1 shows incident light that is not normal to the grating.

Fig. 5.1
Suggest why the diffraction grating formula, d sinθ = nλ, should not be used in this situation.

(c) Light of wavelengths 590 nm and 595 nm is now incident normally on the grating. Two lines are observed in the first order spectrum and two lines are observed in the second order spectrum, corresponding to the two wavelengths.
State two differences between the first order spectrum and the second order spectrum.

Reference: Past Exam Paper – June 2005 Paper 2 Q5

Solution 1099:
(a) When a wave (front) is incident on an edge or an obstacle/slit/gap, the wave ‘bends’ into the geometrical shadow / changes direction / spreads

{d is the slit separation. There are 750 lines in 1 mm (= 1×10-3 m). So, slit separation d = 1×10-3 / 750}
d = 1 / (750 × 103)
d = 1.33 × 10-6 m

{The angle of diffraction, θ, cannot be greater than 90° (90° is the maximum value that can be considered). Substituting the values into the equation d sinθ = nλ gives}
1.33×10-6 × sin90° = n × 590×10-9
n = 2 (must be an integer)

EITHER The formula assumes no path difference of light before entering the grating        OR there is a path difference before the grating

The lines are further apart in second order
The lines fainter in second order
{The differences will be based on either angles of diffraction or intensity.
sinθ = nλ / d. For acute angles (which is the case here), sinθ increases with θ. If the wavelength λ is longer, sinθ and thus, θ will be greater – that is the lines are further apart.
Also, depending on the wavelength, the wave may need to travel more in one case, thus its amplitude will decrease by a greater amount. The lines are fainter.}

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