Friday, October 10, 2014

9702 June 2014 Paper 22 Worked Solutions | A-Level Physics

  • 9702 June 2014 Paper 22 Worked Solutions | A-Level Physics

Paper 22

Question 1
Show SІ base units of power are kg m2 s–3:
Power = work done / time      OR energy / time
EITHER Units of force: kg ms-2
OR Units of kinetic energy (½mv2): kg (ms-1)2
(Unit distance: m        and (time)-1: s-1)
And hence Unit of power:  kgms-2 m s-1 = kgm2s-3

Rate of flow of thermal energy Q /t in material is given by Q / t = CAT / x where A is cross-sectional area of material, T is temperature difference across thickness of material, x is thickness of material, C is a constant.
SI base units of C:
Units of Q / t: kgm2s-3
Units of area, A: m2    distance, x: m              and temperature, T: K
Correct substitution into C = Qx / tAT
So, units of C: kg ms-3 K-1

Question 2
Coin is made in shape of thin cylinder, as shown. Fig. shows measurements made in order to determine density ρ of material used to make coin.
Quantity                      measurement               Uncertainty
Mass                            9.6g                             ±0.5g
Thickness                    2.00mm                       ±0.01mm
Diameter                     22.1mm                       ±0.1mm

Density ρ in kgm-3:
ρ = m / V
Volume, V = (πd2/4) t = 7.67x10-7m3
ρ = (9.6x10-3) / [π({22.1/2}x10-3)2 (2.00x10-3)] = 12513kgm-3

Percentage uncertainty in ρ:
Δρ/ρ x 100% = [Δm/m + Δt/t + 2(Δd/d)] x 100%
Δρ/ρ x 100% = 5.21% + 0.50% + 0.905% = 6.6%     (6.61%)

ρ with its actual uncertainty:
ρ = 12500 ± 800 kgm-3 

Question 3
Newton’s first law of motion states that a body / mass / object continues (at rest or) at constant / uniform velocity unless acted on by a resultant force.

Box slides down slope, as shown. Angle of slope to horizontal is 20°. Box has mass of 65 kg. Total resistive force R acting on box is constant as it slides down slope.
Names and directions of other 2 forces acting on box:
The weight: vertically down
The normal / reaction / contact (force): perpendicular / normal to the slope

Variation with time t of velocity v of box as it moves down slope is shown.
Use data from Fig to show that acceleration of box is 2.6ms-2:
Acceleration of box = gradient OR (v – u) / t            OR Δv/t
Acceleration of box = (6.0 – 0.8) / (2.0 – 0.0) = 2.6ms-2

Resultant force on box:
Resultant force, F = ma = 65 x 2.6 = 169N (allow to 2 or 3 sf)

Resistive force R on box:
Component of weight along slope = mgsinθ (= 218N)
218 – R = 169
Resistive force, R = 49N

Question 4
Gravitational potential energy is the energy of a mass due to its position in a gravitational field
Kinetic energy is the energy (a mass has) due to its motion / speed / velocity

Ball of mass 400g is thrown with initial velocity of 30.0 m s–1 at angle of 45.0° to horizontal, as shown. Air resistance negligible. Ball reaches maximum height H after a time of 2.16 s.
Initial kinetic energy of ball:
Initial Kinetic energy = ½ mv2 = ½ (0.4)(30)2 = 180J

Maximum height H of ball:
s = (30sin45)(2.16) + ½ (-9.81)(2.16)2 = 22.94 (22.9) m
s = (30sin45)2 / (2 x 9.81) = 22.94 (22.9) m

Gravitational potential energy of ball at height H:
Gravitational potential energy, GPE = mgh = 0.4(9.81)(22.88) = 89.8 (90) J

Kinetic energy of ball at maximum height:
KE at maximum height = initial KE – GPE = 180 – 90 = 90J

Why kinetic energy of ball at maximum height not zero:
The (horizontal) velocity is not zero / (object) is still moving / answer explained in terms of conservation of energy

Question 5
Young modulus is defined as the ratio of stress to strain

2 wires P and Q of same material and same original length lo are fixed so that they hang vertically, as shown. Diameter of P is d and diameter of Q is 2d. Same force F is applied to lower end of each wire.
Ratio of stress in P to stress in Q:
Stress = Force / Area
In P, stress = F / (πd2/4)
In Q, stress = F / (πd2)
Ratio of stress in P to stress in Q = 4 (or 4:1)

Ratio of strain in P to strain in Q:
Young modulus, E is the same for both wires (as they are of the same material) [e.g. EP = EQ]
Strain = stress / E
So, ratio of strain in P to strain in Q = 4 (or 4:1)

Question 6
Battery is connected in series with resistors X and Y, as shown. Resistance of X is constant. Resistance of Y is 6.0 Ω. Battery has electromotive force (e.m.f.) 24 V and zero internal resistance. Variable resistor of resistance R is connected in parallel with X.
Current І from battery is changed by varying R from 5.0 Ω to 20 Ω. Variation with R of І is shown.
Why potential difference (p.d.) between points A and C is 24 V for all values of R:
There are no lost volts / energy in the battery OR There are no lost volts / energy lost in the internal resistance.

Use Fig to explain variation of p.d. across resistor Y as R is increased:
The current / I decreases (as R increases). So, the p.d. across resistor Y decreases (as R increases).
The parallel resistance (of X and R) increases. So, the p.d. across the parallel resistors increases and so, the p.d. (across Y) decreases.

For R = 6.0 Ω,
Show that p.d. between points A and B is 9.6V:
(from graph) For R = 6.0 Ω, Current = 2.4A
So, p.d. across points A and B = 24 – (2.4 x 6) = 9.6V
Total resistance = (24V / 2.4A =)10 Ω
(parallel resistance = (10 – 6 =) 4Ω), p.d. = 2.4 x (4/10) = 9.6V  

Resistance of X:
Resistance through AB, R = 9.6 / 2.4 = 4.0Ω
1/6 + 1/X = 1/4
So, resistance of X = 12Ω
Current through R, IR = 9.6 / 6.0 = 1.6A
Current through X, IX = 2.4 – 1.6 = 0.8A
Resistance of X = (9.6 / 0.8 =) 12Ω

Power provided by battery:
Power provided by battery = VI or EI or V2/R or E2/R or I2/R
(total resistance through circuit = 6 + [1/12 + 1/6]-1 = 10)
Power = (24 x 2.4) or (24)2/10 or (2.4)2(10) = 57.6W

Explain qualitatively how power provided by battery changes as resistance R is increased:
The power provided by the battery decreases since the e.m.f. is constant or power = 24 x current, and the current decreases OR the e.m.f. is constant or power = 242 / resistance, and the resistance increases.

Question 7
{Detailed explanations for this question is available as Solution 679 at Physics 9702 Doubts | Help Page 137 -}


  1. Question 4 b ii kinetic energy should be zero at maximum height!!! Plz explain it

    1. It is not zero because the KE is due to both the horizontal velocity and the vertical velocity.

      At maximum height, the vertical velocity is zero, but NOT the horizontal velocity since it still continues to move. Thus, the KE is not zero


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