Physics (AS & A Level) Paper 21
May / June 2016·9702·7 questions
Mock exam: 75 min, official timing
Question paper
Question 1
Question 2
Question 3
Question 4
Question 5
Question 6
Question 7
Mark scheme
Question 1
1(a)(i)[1]
(50 to 200) × 10–3 kg or (0.05 to 0.2) kg [1]
1(a)(ii)[1]
(50 to 300) cm3 [1]
1(b)[5]
density = mass / volume or ρ = M / V V = [π(0.38 × 10–3) 2 × 25.0 × 10–2] / 4 (= 2.835 × 10–8 m3) ρ = (0.225 × 10–3) / 2.835 × 10–8 = 7940 (kg m–3) ∆ρ / ρ = 2(0.01/0.38) + (0.1/25.0) + (0.001/0.225) [= 0.061] or %ρ = 5.3% + 0.40% + 0.44% (= 6.1%) ∆ρ = 0.061 × 7940 = 480 (kg m–3) density = (7.9 ± 0.5) × 103 kg m–3 or (7900 ± 500) kg m–3 [5]
Question 2
2(a)(i)[1]
horizontal component (= 12 cos 50°) = 7.7 m s–1 [1]
2(a)(ii)[1]
vertical component (= 12 sin 50° or 7.7 tan 50°) = 9.2 m s–1 [1]
2(b)[2]
v2 = u2 + 2as and v = 0 or mgh = ½mv2 or s = v2 sin2θ / 2g 9.22 = 2 × 9.81 × h hence h = 4.3 (4.31) m [2]
alternative methods using time to maximum height of 0.94 s: s = ut + ½at2 and t = 0.94 (s) (C1) s = 9.2 × 0.94 – ½ × 9.81 × 0.942 hence s = 4.3 m (A1) or s = vt – ½at2 and t = 0.94 (s) (C1) s = ½ × 9.81 × 0.942 hence s = 4.3 m (A1) or s = ½(u + v)t and t = 0.94 (s) (C1) s = ½ × 9.2 × 0.94 hence s = 4.3 m (A1)
2(c)[4]
t (= 9.2 / 9.81) = 0.94 (0.938) s horizontal distance = 0.938 × 7.7 (= 7.23 m) displacement = [4.32 + 7.232]1/2 = 8.4 m [4]
Question 3
3(a)(i)[1]
force (= mg = 0.15 × 9.81) = 1.5 (1.47) N [1]
3(a)(ii)[1]
resultant force (on ball) is zero so normal contact force = weight or the forces are in opposite directions so normal contact force = weight or normal contact force up = weight down [1]
3(b)(i)[1]
(resultant) force proportional/equal to rate of change of momentum [1]
3(b)(ii)[3]
change in momentum = 0.15 × (6.2 + 2.5) (= 1.305 N s) magnitude of force = 1.305 / 0.12 = 11 (10.9) N or (average) acceleration = (6.2 + 2.5) / 0.12 (= 72.5 m s–2) (C1) magnitude of force = 0.15 × 72.5 = 11 (10.9) N (A1) (direction of force is) upwards/up [3]
3(b)(iii)[2]
there is a change/gain in momentum of the floor this is equal (and opposite) to the change/loss in momentum of the ball so momentum is conserved [2]
or change of (total) momentum of ball and floor is zero (M1) so momentum is conserved (A1) or (total) momentum of ball and floor before is equal to the (total) momentum of ball and floor after (M1) so momentum is conserved (A1)
Question 4
4(a)[1]
the energy (stored) in a body due to its extension/compression/deformation/ change in shape/size [1]
4(b)(i)[2]
two values of F/x are calculated which are the same e.g. 10.4 / 40 = 0.26 and 6.5 / 25 = 0.26 or ratio of two forces and the ratio of the corresponding two extensions are calculated which are the same e.g. 5.2 / 10.4 = 0.5 and 20 / 40 = 0.5 (B1) or gradient of graph line calculated and coordinates of one point on the line used with straight line equation y = mx + c to show c = 0 (B1) (so) force is proportional to extension (and so Hooke’s law obeyed) [2]
4(b)(ii)[5]
1. k = F / x or k = gradient gradient or values from a single point used e.g. k = 10.4 / (40 × 10–2) k = 26 N m–1 [2]
2. work done = area under graph or ½Fx or ½(F2 + F1)(x2 – x1) or ½kx2 or ½k(x2 2 – x1 2) = ½ × 10.4 × 0.4 – ½ × 5.2 × 0.2 or ½ × (5.2 + 10.4) × 20 × 10–2 or ½ × 26 × (0.42 − 0.22) = 1.6 J [3]
4(c)[1]
remove the force and the spring goes back to its original length [1]
Question 5
5(a)[2]
T = 4 (ms) or 4 × 10–3
5(b)[3]
intensity ∝ (amplitude) 2 and amplitude = 2.8 (2.83) (cm) curve with same period and with amplitude 2.8 cm curve shifted 1.0 ms to left or to right of wave X [3]
5(c)(i)[4]
gradient = (4.5 – 2.4) × 10–3 / (3.25 – 1.75) [= 1.4 × 10–3] wavelength = 0.45 × 10–3 × 1.4 × 10–3 = 6.30 × 10–7
5(c)(ii)[2]
(Gradient is equal to λ/a, therefore) gradient of line is reduced. (1)
Value of x will be reduced for all values of D, or new line is completely below old line, or intercept is less. (1)
Question 6
6(a)[1]
(coulomb is) ampere second [1]
6(b)[3]
(total) charge or Q = nAle I = Q / t and l / t = v I = nAle / t = nAve therefore v = I / nAe [3]
6(c)(i)[3]
ratio = (I / nAYe) / (I / nAZe) = AZ / AY or 4A / A or πd2 / (πd2 / 4) = 4 [3]
6(c)(ii)[2]
R = ρl / A or R = 4ρl / πd2 RY = ρl / A and RZ = ρ(2l) / 4A so RY / RZ = 2 or RY = 4ρl /πd2 and RZ = 4ρ(2l) /π4d2 or 2ρl / πd2 so RY / RZ = 2 [2]
6(c)(iii)[2]
V = 12RY / (RY + RZ) or I = 12 / (RY + RZ) and V = IRY V = 12 × 2/3 = 8(.0) V [2]
6(c)(iv)[1]
ratio = I2RY / I2RZ or (VY 2 / RY) / (VZ 2 / RZ) or (VYI) / (VZI) = 2 [1]
Question 7
7(a)[1]
hadron: neutron/proton and lepton: electron/(electron) neutrino [1]
(allow other correct particles)
7(b)(i)[1]
proton: up up down or uud [1]
7(b)(ii)[1]
neutron: up down down or udd [1]
7(c)(i)[1]
neutron → proton + electron + (electron) antineutrino [1]
7(c)(ii)[1]
up down down (quarks) change to up up down (quarks) or down (quark) changes to up (quark) [1]