Physics (AS & A Level) Paper 23

scalars: energy, power and time vectors: momentum and weight [2]

triangle with right angles between 120 m and 80 m, arrows in correct direction and result displacement from start to finish arrow in correct direction and labelled R [1]

1. average speed (= 200 / 27) = 7.4 m s–1 [1]

2. resultant displacement (= [1202 + 802]1/2) = 144 (m) average velocity (= 144 / 27) = 5.3(3) m s–1 direction (= tan–1 80 / 120) = 34° (33.7) [3]

systematic: the reading is larger or smaller than (or varying from) the true reading by a constant amount random: scatter in readings about the true reading [2]

precision: the size of the smallest division (on the measuring instrument) or 0.01 mm for the micrometer accuracy: how close (diameter) value is to the true (diameter) value [2]

(gravitational potential energy is) the energy/ability to do work of a mass that it has or is stored due to its position/height in a gravitational field kinetic energy is energy/ability to do work a object/body/mass has due to its speed/velocity/motion/movement [2]

s = [(u + v) t] / 2 or acceleration = 9.8/9.75 (using gradient) = [(7.8 + 3.9) × 0.4] / 2 or s = 3.9 × 0.4 + 1 2 × 9.75 × (0.4) 2 s = 2.3(4) m [3]

a = (v – u) / t or gradient of line = (7.8 – 3.9) / 0.4 = 9.8 (9.75) m s–2 (allow ± 1 2 small square in readings) [2]

KE = 1 2 mv2 change in kinetic energy = 1 2 mv2 – 1 2 mu2 = 1 2 × 1.5 × (7.82 – 3.92) = 34 (34.22) J [3]

work done = force × distance (moved) or Fd or Fx or mgh or mgd or mgx = 1.5 × 9.8 × 2.3 = 34 (33.8) J (equals the change in KE) [2]

(resultant force = 0) (equilibrium) therefore: weight – upthrust = force from thin wire (allow tension in wire) or 5.3 (N) – upthrust = 4.8 (N) [1]

difference in weight = upthrust or upthrust = 0.5 (N) 0.5 = ρghA or m = 0.5 / 9.81 and V = 5.0 × 13 × 10–6 (m3) ρ = 0.5 / (9.81 × 5.0 × 13 × 10–6) = 780 (784) kg m–3 [3]

the total momentum of a system (of colliding particles) remains constant provided there is no resultant external force acting on the system/ isolated or closed system [2]

the total kinetic energy before (the collision) is equal to the total kinetic energy after (the collision) [1]

p (= mv = 1.67 × 10–27 × 500) = 8.4 (8.35) × 10–25 N s [1]

1. mvA cos 60° + mvB cos 30° or m(vA 2 + vB 2) 1/2 2. mvA sin 60° + mvB sin 30° [2]

8.35 × 10–25 or 500m = mvA cos 60° + mvB cos 30° and 0 = mvA sin 60° + mvB sin 30° or using a vector triangle vA = 250 m s–1 vB = 430 (433) m s–1 [3]

ohm is volt per ampere or volt / ampere [1]

R = ρl / A RP = 4ρ(2l) / πd2 or 8ρl / πd2 or RQ = ρl / πd2 or ratio idea e.g. length is halved hence R halved and diameter is halved hence R is 1/4 RQ (= 4ρl / π4d2) = ρl / πd2 = RP / 8 (= 12 / 8) = 1.5 Ω [3]

power = I 2R or V 2 / R or VI = (1.25) 2 × 12 + (10) 2 × 1.5 or (15) 2/12 + (15) 2/1.5 or 15 × 11.25 = (18.75 + 150 =) 170 (168.75) W [3]

IP = (15 / 12 =) 1.25 (A) and IQ = (15 / 1.5 =) 10 (A) vP / vQ = IPnAQe / IQnAPe or (1.25 × πd 2) / (10 × πd 2/4) = 0.5 [3]

alter distance from vibrator to pulley alter frequency of generator (change tension in string by) changing value of the masses any two [2]

points on string have amplitudes varying from maximum to zero/minimum [1]

60° or π / 3 rad [1]

ratio = [3.4 / 2.2]2 = 2.4 (2.39) [2]

α-particle is 2 protons and 2 neutrons;

β+-particle is positive electron/positron α-particle has charge +2e;

β+-particle has +e charge α-particle has mass 4u;

β-particle has mass (1/2000)u α-particle made up of hadrons;

β+-particle a lepton any three [3]

ν β 0 0 0 1 1 0 1 1 n p + + → all terms correct all numerical values correct (ignore missing values on ν) [2]

1. proton: up, up, down / uud 2. neutron: up, down, down / udd [2]

up quark has charge +2 / 3 (e) and down quark has charge –1 / 3 (e) total is +1 (e) [1]