• 9702 November 2009 Paper 22 Worked Solutions | A-Level Physics

Paper 22

Question 1

Simple pendulum may be used to determine value for acceleration of free fall g. Measurements are made of length L of pendulum and period T of oscillation. Values obtained, with uncertainties, are:

T = (1.93 ± 0.03) s

L = (92 ± 1) cm

(a)

Percentage uncertainty in measurement of

(i)

Period T:

% Uncertainty in T = (0.03/1.93) x 100% = EITHER 1.55%    OR 1.6%

(ii)

Length L:

% Uncertainty in L = (1/92) x 100% = EITHER 1.09%   OR 1.1%

(b)

Relationship between T, L and g is given by              g = 4π²L / T². Using answers in (a), percentage uncertainty in value of g:

% uncertainty in g = 1.09% + 2(1.55%) = EITHER 4.2%   OR 4.3%

(c)

Values of L and T used to calculate value of g as 9.751ms^-2.

(i)

Reference to measurements of L and T, why it would not be correct to quote value of g as 9.751ms^-2:

EITHER The value has more significant figures than the data available

OR The uncertainty of ± 0.4 renders more than 2 s.f. meaningless.

(ii)

Use answer in (b) to determine absolute uncertainty in g. State value of g, with its uncertainty, to appropriate number of significant figures:

Δg = ([4.2/100] x 9.751 =) ± 0.41 / ± 0.42

g = (9.8 ± 0.4) ms^-2 

Question 2

(a)

(i)

1 similarity between the processes of evaporation and boiling:

Choose any 1:

Both involve a (phase) change from liquid to gas / vapour

Thermal energy is required to maintain constant temperature

(ii)

2 differences between the processes of evaporation and boiling:

Evaporation takes place at the surface while boiling takes place in the body of the liquid.

Evaporation occurs at all temperatures while boiling occurs at one temperature.

(b)

Titanium metal has density of 4.5gcm^-3. Cube of titanium of mass 48g contains 6.0x10²³ atoms.

(i)

Volume of cube:

Volume = (mass/density = 48/4.5 =) 10.7cm³

(ii)

Estimate

1.

Volume occupied by each atom in cube:

Volume occupied by each atom = 10.7 / (6.0x10²³) = 1.8x10^-23cm³

2.

Separation of atoms in cube: 

{The centres of the atoms are found at the corners of the cube. So, the separation between the atoms is equal to the length of a side of the cube.

For more details, check Question332 at Physics 9702 Doubts | Help Page 58 - http://physics-ref.blogspot.com/2015/02/physics-9702-doubts-help-page-58.html} 
Separation of atoms = (1.8x10^-23)^1/3 = 2.6x10^-8m

Question 3

{Detailed explanations for this question is available as Solution 517 at Physics 9702 Doubts | Help Page 100 - http://physics-ref.blogspot.com/2015/03/physics-9702-doubts-help-page-100.html}

Question 4

(a)

Strain energy (elastic potential energy) is the ability to do work as a result of a change of shape of an object / stretched etc

(b)

Spring that obeys Hooke’s law has spring constant k. Show that energy E stored in spring when it has been extended elastically by amount x is given by E = ½ kx²:

Work done = average force x distance moved (in direction of the force)

EITHER Work = ½ Fx           OR Work is area under the F/x graph which is ½ Fx

(Hooke’s law) F = kx

So, work / energy = (½ (kx) x =) ½ kx²

(c)

Light spring of unextended length 14.2cm suspended vertically from fixed point, as illustrated. Mass of weight 3.8N hung from end of spring, as shown. Length of spring is now 16.3cm. Additional force F then extends spring so that its length becomes 17.8cm, as shown. Spring obeys Hooke’s law and elastic limit of spring not exceeded.

(i)

Show spring constant of spring is 1.8Ncm^-1:

k = (F/x =) 3.8 / 2.1 = 1.8Ncm^-1

(ii)

For extension of spring from length of 16.3cm to length of 17.8cm,

1.

Change in gravitational potential energy of mass on spring:

ΔE_p = mgΔh    or W Δh = 3.8 x (1.5x10^-2) = 0.057J

2.

Show that change in elastic potential energy of spring is 0.077J:

(At length 16.3cm, extension, x = 16.3 – 14.2 = 2.1cm = 0.021m. At length 17.8cm, extension, x = 17.8 – 14.2 = 3.6cm = 0.036m)

ΔE_s = ½ (1.8x10²) (0.036² – 0.021²) = 0.077J

3.

Work done by force F:

Work done = 0.077 – 0.057 = 0.020J

Question 5

A uniform string is held between a fixed point P and a variable-frequency oscillator, as shown in Fig. 5.1. ...

Question 6

{Detailed explanations for this question is available as Solution 661 at Physics 9702 Doubts | Help Page 132 - http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-132.html}

Question 7

Tungsten-184 (¹⁸⁴₇₄W) and tungsten-185 (¹⁸⁵₇₄W) are 2 isotopes of tungsten. Tungsten-184 us stable but tungsten-185 undergoes beta-decay to form rhenium (Re).

(a)

Isotopes are EITHER forms of the same element OR atoms / nuclei with number of protons where the atoms / nuclei contain different numbers of neutrons.

(b)

β-decay of nuclei of tungsten-185 is spontaneous and random.

(i)

Spontaneous decay is a decay which is not affected by environmental factors

(ii)

Random decay is a decay in which EITHER the time of decay (of a nucleus) cannot be predicted OR the nucleus has a constant probability (to decay) in a given a time.

(c)

Complete nuclear equation for β-decay of tungsten-185 nucleus:

¹⁸⁵₇₄W - - ->     ¹⁸⁵₇₅Re             +          ⁰_-1β