9702 June 2010 Paper 21 Worked Solutions | A-Level Physics

9702 June 2010 Paper 21 Worked Solutions | A-Level Physics

Question 1 (Units)

Unit is often expressed with prefix. For example, gram may be written with prefix ‘kilo’ as kilogram. Prefix represents a power-of-ten. In this case, power-of-ten is 10³.

Complete Fig to show each prefix with its symbol and power-of-ten:

Prefix Symbol Power-of-ten

Kilo k 10³

Nano n 10^-9

Centi c 10^-2

Mega M 10⁶

Tera T 10¹²

Question 2

(a)

Complete Fig to show whether each of the quantities listed is vector or scalar:

Distance moved: scalar

Speed: scalar

Acceleration: vector

(b)

Ball falls vertically in air from rest. Variation with time t of distance d moved by ball is shown in Fig.

(i)

Reference to Fig, how it can be deduced that

Ball is initially at rest:

The gradient (of the graph) represents the speed / velocity (can be scored here or in 2). The initial gradient {gradient at t = 0} is zero {so, the initial velocity is zero}.

Air resistance is not negligible:

The gradient (of the line / graph) becomes constant {terminal velocity is reached}

(ii)

Use Fig to determine speed of ball at time of 0.40 s after it has been released:

{The gradient of the tangent at t = 0.4s should be calculated (the tangent only touches the point on the curve at t = 0.4s, and no other points on the curve). This represents the instantaneous speed.

[By drawing the tangent at t = 0.4s, I obtained a straight line passing through points (0.21, 0) and (0.6, 1.1) which I used to calculate the gradient. Gradient = (1.1 – 0) / (0.6 – 0.21) = 2.82ms^-1]

Note that calculating the ratio of 0.55 m / 0.40 s would give the average speed over the first 0.4s, which is not what is required here.}

Speed of ball at time t = (2.8 ± 0.1) ms^-1

(iii)

On Fig, sketch graph to show variation with time t of distance d moved by ball for negligible air resistance:

The graph is a curved line which is never below the given line {air resistance opposes motion, so in the absence of air resistance, the speed is greater} and starts from zero. It is a continuous curve with increasing gradient {speed increases under gravity} and the line never becomes vertical {a vertical line would correspond to an infinite velocity which is not possible} or straight {a straight line corresponds to a constant velocity – if air resistance was present, this would be the terminal velocity}

{A graph with increasing gradient is shown. Fit it with the description given. It should look a bit like the second graph given (but drawn better [with no straight line]– I could not draw a proper graph on the computer)}

Question 3

{Detailed explanations for this question is available as Solution 627 at Physics 9702 Doubts | Help Page 124 - http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-124.html}

Question 4

(a)

The diffraction of a wave occurs when a wave (front) passes by / incident on an edge / slit. The wave bends / spreads (into the geometrical shadow)

(b)

Laser produces narrow beam of coherent light of wavelength 632 nm. Beam is incident normally on diffraction grating, as shown.

Spots of light are observed on screen placed parallel to grating. Distance between grating and screen is 165 cm. Brightest spot is P. Spots formed closest to P and on each side of P are X and Y. X and Y are separated by distance of 76 cm. Number of lines per metre on grating:

{Do not confuse the equation for diffraction grating and that for double slit.

Note that the equation for diffraction grating (not for the double slit) is dsinθ = nλ where d is the slit separation, θ is the angle made by the n^th maxima and λ is the wavelength. The diffraction grating formula does not directly contain the separation of the slit and the screen (D, which is 165cm), but to calculate the angle, we can use this D (this D is not the same as the slit separation, d).

For double slit, x = λD /a where x = fringe separation, λ = wavelength, D = distance between slits and screen and a = slit separation.}

{distance PX = PY = 76 / 2 = 38cm. Let angle made by first maxima (spots at X and Y) to the horizontal = θ}

tan θ = 38 / 165 [= 0.2303]

Angle θ = 13^o [= 12.969]

dsinθ = nλ

{For the first maxima, n = 1. So, n is known here}

(Slit separation,) d (= [632x10^-9] / sin13) = 2.81x10^-6m

{One line corresponds to a separation of 2.81x10^-6m. So, the number of lines in 1m is obtained as follows:}

Number of lines per metre (= 1/d) = 3.6x10⁵

(c)

Grating in (b) is now rotated about axis parallel to incident laser beam, as shown. State what effect, if any, rotation will have on positions of spots P, X and Y:

The spot at P remains in the same position. The spots at X and Y rotate through 90^o.
{Since the grating has been rotated by 90^o, the lines are now perpendicular to the positions they were previously. So, diffraction occurs in a plane perpendicular to the previous one.

P is the 0^th (central) maxima. It is always found on the same line of incident of the light. So, it remains in the same position even if the grating is rotated}

(d)

In another experiment using apparatus in (b), student notices that distances XP and PY, as shown, are not equal. Reason for difference:

EITHER The screen is not parallel to the grating OR The grating is not normal to the (incident) light

Question 5

(a)

An electric field is a region / area where a charge experiences a force.

(b)

Electric field between earthed metal plate and 2 charged metal spheres is illustrated.

(i)

On Fig, label each sphere with (+) or (-) to show its charge:

{Direction of electric field is from +ve to –ve}

Sphere on left-hand side is (+) and sphere on right-hand side is (-)

(ii)

On Fig, mark region where magnitude of electric field is

Constant (label region C):

A correct region labeled C is within 10mm of the central plate

(otherwise within 5mm of plate)

Decreasing (label region D):

A correct region labeled D is the area of the field not included for (b)(ii)1

(c)

Molecule has its centre P of positive charge situated distance of 2.8 × 10^–10 m from its centre N of negative charge, as illustrated. Molecule is situated in uniform electric field of field strength 5.0 × 104 V m^–1. Axis NP of molecule is at angle of 30° to uniform applied electric field. Magnitude of charge at P and at N is 1.6 × 10^–19 C.

(i)

On Fig, draw arrow at P and arrow at N to show directions of forces due to applied electric field at each of these points:

{Electric field is from +ve to –ve. So, the arrow at P is in the same direction as the applied field (towards the right) and the arrow at N is in the opposite direction of the applied field (towards the left)}

Arrows through P (towards the right) and N (towards the left) in correct directions

(ii)

Torque on molecule produced by forces in (i):

Torque = Force x Perpendicular distance (between the forces)

Torque = [(1.6x10^-19)(5.0x10⁴)] [(2.8x10^-10)sin30] = 1.1x10^-24Nm

Question 6

An electric heater is to be made from nichrome wire. Nichrome has a resistivity of 1.0 × 10-6 Ω m at the operating temperature of the heater.

Question 7

One of isotopes of uranium is uranium-238 ( ²³⁸₉₂U).

(a)

Isotopes are nuclei / atoms {of the same element} with the same proton / atomic number but contain different numbers of neutrons / different atomic mass

(b)

For nucleus of Uranium-238,

(i)

Number of protons: 92

(ii)

Number of neutrons: (238 – 92 =) 146

(c)

A uranium-238 nucleus has radius of 8.9 × 10^–15 m. For uranium-238 nucleus,

(i)

Mass:

{1u = 1.66x10^-27kg}

Mass (= 238u) = 238 (1.66x10^-27) = 3.95x10^-25kg

(ii)

Mean density:

{Assuming the nucleus is of a spherical shape, volume of sphere = (4/3)πr³}

Volume of nucleus = (4/3) π (8.9x10^-15)³ [= 2.95x10^-42]

Density (= mass/volume) = (3.95x10^-25) / (2.95x10^-42) = 1.3x10¹⁷kgm^-3

(d)

Density of lump of uranium is 1.9 × 10⁴ kg m^–3. Using answer to (c)(ii), what can be inferred about structure of atom:

{Since the density of the nucleus is large} The nucleus contains most of the mass of the atom. EITHER The nuclear diameter is very much less than that of the atom OR The atoms is mostly (empty) space.