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
(a)
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

(b)
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

(a)
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

(b)
(i)
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%)

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



Question 3
(a)
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.

(b)
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.
(i)
Names and directions of other 2 forces acting on box:
The weight: vertically down
The normal / reaction / contact (force): perpendicular / normal to the slope

(ii)
Variation with time t of velocity v of box as it moves down slope is shown.
1.
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

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

3.
Resistive force R on box:
Component of weight along slope = mgsinθ (= 218N)
218 – R = 169
Resistive force, R = 49N



Question 4
(a)
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

(b)
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.
(i)
1.
Initial kinetic energy of ball:
Initial Kinetic energy = ½ mv2 = ½ (0.4)(30)2 = 180J

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

3.
Gravitational potential energy of ball at height H:
Gravitational potential energy, GPE = mgh = 0.4(9.81)(22.88) = 89.8 (90) J

(ii)
1.
Kinetic energy of ball at maximum height:
KE at maximum height = initial KE – GPE = 180 – 90 = 90J

2.
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
(a)
Young modulus is defined as the ratio of stress to strain

(b)
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.
(i)
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)

(ii)
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.
(a)
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.

(b)
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).
OR
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.

(c)
For R = 6.0 Ω,
(i)
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
OR
Total resistance = (24V / 2.4A =)10 Ω
(parallel resistance = (10 – 6 =) 4Ω), p.d. = 2.4 x (4/10) = 9.6V  

(ii)
Resistance of X:
Resistance through AB, R = 9.6 / 2.4 = 4.0Ω
1/6 + 1/X = 1/4
So, resistance of X = 12Ω
OR
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

(iii)
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

(d)
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 - http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-137.html}

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