Physics (AS & A Level) Paper 22
Feb / March 2016·9702·6 questions
Mock exam: 75 min, official timing
Question paper
Question 1
Question 2
Question 3
Question 4
Question 5
Question 6
Mark scheme
Question 1
1(a)[1]
metre rule / tape measure
1(b)(i)[2]
v = [(1.8 × 126 × 10–2) / 5.1 × 10–3]1 / 2 = 21.1 (m s–1)
1(b)(ii)[3]
percentage uncertainty = 4% or fractional uncertainty = 0.04 ∆v = 0.04 × 21.1 = 0.84 v = 21.1 ± 0.8 (m s–1)
Question 2
2(a)[1]
change in velocity / time (taken) or rate of change of velocity
2(b)(i)[1]
vX = (24 / 1.5) = 16 (m s–1)
2(b)(ii)[2]
tan 28° = vY / vX or vX = v cos 28° and vY = v sin 28° vY = 16 tan 28° or vY = 16 × (sin 28° / cos 28°) so vY = 8.5 (m s–1)
2(b)(iii)[2]
v = u + at t = (0 – 8.5) / (–9.81) = 0.87 (s)
2(b)(iv)[2]
straight line from positive vY at t = 0 to negative vY at t =1.5 s line starts at (0, 8.5) and crosses t-axis at (0.87, 0) and does not go beyond t = 1.5 s. A1
2(c)(i)[2]
(v 2 = u 2 + 2as) 0 = 8.52 + 2(–9.81)s or (s = ut + ½at 2) s = 8.5×0.87 + ½ × (–9.81) × 0.872 or (s = vt – ½at 2) s = 0 – ½×(–9.81)×0.872 or (s = ½(u + v)t or area under graph) s = 0.5 × 8.5 × 0.87 s = 3.7 (m)
2(c)(ii)[2]
∆EP = mg∆h (allow E = mgh) m = 22 / (9.81 × 3.7) = 0.61 (kg)
2(d)[1]
acceleration (of freefall) is unchanged / not dependent on mass, and so no effect (on maximum height) or explanation in terms of energy: (initial) KE ∝ mass, (∆)KE = (∆)PE, (max) PE ∝ mass, and so no effect (on maximum height)
Question 3
3(a)(i)[1]
(work =) force × distance moved in the direction of the force.
3(a)(ii)[1]
the energy stored (in an object) due to extension / compression / change of shape
3(b)(i)[2]
EK = ½mv 2 = 0.5 × 0.40 × 0.302 = 1.8 × 10–2 (J)
3(b)(ii)[3]
(change in) kinetic energy = work done on spring / (change in) elastic potential energy C1 1.8 × 10–2 = ½ × F × 0.080 FMAX = 0.45 (N)
3(b)(iii)[1]
a = F / m = 0.45 / 0.40 = 1.1 (m s–2)
3(b)(iv)[2]
1. constant velocity / resultant force is zero, so in equilibrium 2. decelerating / resultant force is not zero, so not in equilibrium
3(c)[2]
curved line from the origin with decreasing gradient
Question 4
4(a)(i)[1]
Displacement of particles perpendicular to direction of energy propagation
4(a)(ii)[2]
waves meet / overlap (at a point) (resultant) displacement is sum of the individual displacements
4(b)(i)[2]
λ = vT or λ = v / f and f = 1 / T λ = 4.0 × 1.5 λ = 6.0 (cm)
4(b)(ii)[3]
path difference [= (44 cm – 29 cm) / 6 cm] = 2.5λ either waves have path difference = (n + ½)λ or waves have phase difference = 180° so destructive interference
4(c)(i)[2]
intensity ∝ (amplitude) 2 ratio = (0.602 / 0.902) = 0.44
4(c)(ii)[1]
phase difference = 90°
Question 5
5(a)(i)[1]
movement / flow of charge carriers
5(a)(ii)[1]
work (done) or energy (transformed)(from electrical to other forms) charge
5(b)(i)[2]
p.d. across one lamp = 2.5 V resistance = [(8.7 – 7.5) / 0.3] / 2 = 2.0 (Ω)
5(b)(ii)[2]
straight line through the origin with gradient of 0.5
5(b)(iii)[2]
P = I 2R or P = VI and V= IR or P = V 2 / R and V= IR = 0.302 × 2.0 = 0.60 × 0.30 = 0.602 / 2.0 = 0.18 (W)
5(b)(iv)[4]
1 R = ρl / A l = (2.0 × 0.40 × 10–6) / 1.7 × 10–8 = 47 (m) 2 I = Anvq v = 0.30 / (0.40 × 10–6 × 8.5 × 1028 × 1.6 × 10–19) = 5.5 × 10–5 (m s–1)
Question 6
6(a)[2]
p 1 1 ν β− − 0 0 0 1 and
6(b)[1]
an (electron) antineutrino
6(c)[1]
lepton(s)
6(d)(i)[1]
down, down, up / ddu
6(d)(ii)[1]
a down / d (quark) changes to an up / u (quark) or ddu → uud