Equations of motion
2.1What the syllabus asks of you
- define and use distance, displacement, speed, velocity and acceleration
- use graphical methods to represent distance, displacement, speed, velocity and acceleration
- determine displacement from the area under a velocity–time graph
- determine velocity using the gradient of a displacement–time graph
- determine acceleration using the gradient of a velocity–time graph
- derive, from the definitions of velocity and acceleration, equations that represent uniformly accelerated motion in a straight line
- solve problems using equations that represent uniformly accelerated motion in a straight line, including the motion of bodies falling in a uniform gravitational field without air resistance
- describe an experiment to determine the acceleration of free fall using a falling object
- describe and explain motion due to a uniform velocity in one direction and a uniform acceleration in a perpendicular direction
Read the notes first
Chapter 2: Equations of motion
Drill this topic
Every question on Equations of motion, untimed, marked as you go.
Make a worksheet
Choose the years and how many questions, then print it with the answers on separate pages.
167 questions
1May / June 2026 · Paper 11 · Q5Show mark scheme →
AThe ball lands with the same velocity at which it is thrown.
BThe horizontal component of acceleration is constant throughout the motion.
CThe horizontal and vertical components of acceleration are both zero at the highest point of the motion.
DThe horizontal and vertical components of velocity are both zero at the highest point of the motion.
Mark scheme
2May / June 2026 · Paper 11 · Q6Show mark scheme →
Adistance travelled per unit time
Bproduct of acceleration and change in time
Cproduct of frequency and wavelength
Drate of change of displacement
Mark scheme
3May / June 2026 · Paper 11 · Q7Show mark scheme →
A10 m s–1
B29 m s–1
C31 m s–1
D39 m s–1
Mark scheme
4May / June 2026 · Paper 11 · Q8Show mark scheme →
Aaverage kinetic energy
Baverage velocity
Cgravitational potential energy gained
Dpower generated
Mark scheme
5May / June 2026 · Paper 11 · Q18Show mark scheme →
A1.0
B1.8
C2.5
D3.1
Mark scheme
6May / June 2026 · Paper 12 · Q5Show mark scheme →
AThey have the same average speed.
BThey have the same average velocity.
CThey have the same displacement.
DThey travel the same distance.
Mark scheme
7May / June 2026 · Paper 12 · Q6Show mark scheme →
AGraph A
BGraph B
CGraph C
DGraph D
Mark scheme
8May / June 2026 · Paper 12 · Q7Show mark scheme →
A200 m
B400 m
C500 m
D600 m
Mark scheme
9May / June 2026 · Paper 12 · Q8Show mark scheme →
Athe distance fallen by the body
Bthe initial velocity of the body
Cthe mass of the body
Dthe time taken for the body to fall
Mark scheme
10May / June 2026 · Paper 13 · Q6Show mark scheme →
AX and Y are at rest at their maximum heights.
BX and Y are fired with the same speed.
CX and Y take the same time to return to the ground.
DX and Y travel the same distance.
Mark scheme
11May / June 2026 · Paper 13 · Q7Show mark scheme →
A15 m
B75 m
C80 m
D2000 m
Mark scheme
12May / June 2026 · Paper 14 · Q5Show mark scheme →
Avelocity-time graph: straight line with negative gradient starting from positive velocity, crossing time axis
Bvelocity-time graph: triangular shape, velocity increases from zero then decreases back to zero
Cvelocity-time graph: horizontal line at constant positive velocity
Dvelocity-time graph: horizontal line at constant positive velocity then instantaneously drops to constant negative velocity
Mark scheme
13May / June 2026 · Paper 14 · Q6Show mark scheme →
A3.5 m s–1
B6.6 m s–1
C13 m s–1
D20 m s–1
Mark scheme
14May / June 2026 · Paper 14 · Q7Show mark scheme →
AIt is equal to half the value of the gradient of the graph.
BIt is equal to the value of the gradient of the graph.
CIt is equal to twice the value of the gradient of the graph.
DIt is equal to the reciprocal of the gradient of the graph.
Mark scheme
15Oct / Nov 2025 · Paper 11 · Q5Show mark scheme →
A6 s
B10 s
C20 s
D40 s
Mark scheme
16Oct / Nov 2025 · Paper 11 · Q6Show mark scheme →
AThe area under line PQ represents the initial height of the ball.
BThe collisions of the ball with the ground are inelastic.
CThe gradient of the line RS represents the acceleration due to free fall.
DThe maximum upwards velocity of the ball is reached at point P.
Mark scheme
17Oct / Nov 2025 · Paper 11 · Q7Show mark scheme →
Ax
B2x
C4x
D16x
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18Oct / Nov 2025 · Paper 12 · Q5Show mark scheme →
Av − ad
B√(v² − 2ad)
Cv + ad
Dv − ½ad²
Mark scheme
19Oct / Nov 2025 · Paper 12 · Q6Show mark scheme →
A7 m
B9 m
C16 m
D21 m
Mark scheme
20Oct / Nov 2025 · Paper 12 · Q7Show mark scheme →
Aarea 1 – area 2
B(area 1 + area 2) / 2
Carea 1 + area 2
Darea 2 – area 1
Mark scheme
21Oct / Nov 2025 · Paper 13 · Q5Show mark scheme →
A6 s
B10 s
C20 s
D40 s
Mark scheme
22Oct / Nov 2025 · Paper 13 · Q6Show mark scheme →
AThe area under line PQ represents the initial height of the ball.
BThe collisions of the ball with the ground are inelastic.
CThe gradient of the line RS represents the acceleration due to free fall.
DThe maximum upwards velocity of the ball is reached at point P.
Mark scheme
23Oct / Nov 2025 · Paper 13 · Q7Show mark scheme →
Ax
B2x
C4x
D16x
Mark scheme
24Oct / Nov 2025 · Paper 14 · Q5Show mark scheme →
Aconstant velocity / constant acceleration
Bconstant velocity / varying acceleration
Cvarying velocity / constant acceleration
Dvarying velocity / varying acceleration
Mark scheme
25Oct / Nov 2025 · Paper 14 · Q6Show mark scheme →
A300 m
B540 m
C590 m
D800 m
Mark scheme
26Oct / Nov 2025 · Paper 14 · Q7Show mark scheme →
Afrom the area under a displacement–time graph
Bfrom the area under a velocity–time graph
Cfrom the gradient of a displacement–time graph
Dfrom the gradient of a velocity–time graph
Mark scheme
27May / June 2025 · Paper 11 · Q5Show mark scheme →
A11 m
B19 m
C37 m
D41 m
Mark scheme
28May / June 2025 · Paper 11 · Q7Show mark scheme →
A$s = \frac{1}{2}(u+v)t$
B$s = ut + \frac{1}{2}at^2$
C$v = u + at$
D$v^2 = u^2 + 2as$
Mark scheme
29May / June 2025 · Paper 11 · Q8Show mark scheme →
Ahorizontal Q, vertical P
Bhorizontal Q, vertical S
Chorizontal R, vertical P
Dhorizontal R, vertical S
Mark scheme
30May / June 2025 · Paper 12 · Q6Show mark scheme →
AThe car accelerates for 2 s, then stops for 4 s and then reverses.
BThe car accelerates at 12 m s–2 for 2 s.
CThe car travels a distance of 36 m in the first 4 s.
DThe car travels a distance of 48 m in the last 4 s.
Mark scheme
31May / June 2025 · Paper 12 · Q8Show mark scheme →
AThe stone has a constant horizontal acceleration and constant vertical velocity.
BThe stone has a constant horizontal velocity and constant vertical acceleration.
CThe stone has a constant upward acceleration followed by a constant downward acceleration.
DThe stone has a constant upward velocity followed by a constant downward velocity.
Mark scheme
32May / June 2025 · Paper 13 · Q5Show mark scheme →
A–3 m
B21 m
C24 m
D27 m
Mark scheme
33May / June 2025 · Paper 13 · Q6Show mark scheme →
A1.2 m
B2.5 m
C4.9 m
D9.8 m
Mark scheme
34May / June 2025 · Paper 13 · Q7Show mark scheme →
ABoth coins hit the ground with the same speed.
BBoth coins travel the same vertical distance.
CCoin Y hits the ground before coin X.
DCoin Y has a smaller vertical acceleration.
Mark scheme
35May / June 2025 · Paper 14 · Q5Show mark scheme →
AGraph A
BGraph B
CGraph C
DGraph D
Mark scheme
36May / June 2025 · Paper 14 · Q6Show mark scheme →
A1.1 s
B1.2 s
C1.6 s
D2.4 s
Mark scheme
37Feb / March 2025 · Paper 12 · Q4Show mark scheme →
A0.23 m s⁻²
B0.46 m s⁻²
C3.0 m s⁻²
D6.0 m s⁻²
Mark scheme
38Feb / March 2025 · Paper 12 · Q5Show mark scheme →
AIts speed decreases to a value greater than zero, then increases to 20 m s⁻¹.
BIts speed decreases to a value greater than zero, then increases to a value greater than 20 m s⁻¹.
CIts speed decreases to zero, then increases to 20 m s⁻¹.
DIts speed decreases to zero, then increases to a value less than 20 m s⁻¹.
Mark scheme
39Feb / March 2025 · Paper 12 · Q18Show mark scheme →
A0.50E
B0.71E
C0.87E
DE
Mark scheme
40Oct / Nov 2024 · Paper 11 · Q5Show mark scheme →
AThe horizontal component of the object's velocity is constant and has the value ucosθ.
BThe horizontal distance travelled by the object is tucosθ.
CThe time taken for the object to reach maximum height is t/2.
DThe vertical component of the object's velocity is constant and has the value usinθ.
Mark scheme
41Oct / Nov 2024 · Paper 11 · Q6Show mark scheme →
A110 km h⁻¹
B120 km h⁻¹
C122 km h⁻¹
D126 km h⁻¹
Mark scheme
42Oct / Nov 2024 · Paper 11 · Q13Show mark scheme →
A1.8
B2.5
C3.1
D3.5
Mark scheme
43Oct / Nov 2024 · Paper 11 · Q14Show mark scheme →
Ajumper at rest: Q; jumper with zero acceleration: P
Bjumper at rest: Q; jumper with zero acceleration: R
Cjumper at rest: R; jumper with zero acceleration: Q
Djumper at rest: R; jumper with zero acceleration: R
Mark scheme
44Oct / Nov 2024 · Paper 12 · Q4Show mark scheme →
Avelocity-time graph: velocity starts at 0, increases with increasing gradient then decreasing gradient, approaching a constant positive value
Bvelocity-time graph: velocity starts at 0, rises to a peak then decreases back to 0
Cvelocity-time graph: velocity starts at 0, rises to a positive peak, decreases through 0 to a negative minimum, then returns to 0
Dvelocity-time graph: velocity starts at 0 and increases with increasing gradient (concave up)
Mark scheme
45Oct / Nov 2024 · Paper 12 · Q5Show mark scheme →
A2.0 m s–1
B2.8 m s–1
C3.9 m s–1
D7.8 m s–1
Mark scheme
46Oct / Nov 2024 · Paper 12 · Q6Show mark scheme →
AThe acceleration of the ball is always at right angles to the path of the ball.
BThe ball follows a circular path until it hits the ground.
CThe ball has constant acceleration.
DThe ball's time in the air is proportional to the velocity at which the ball is thrown.
Mark scheme
47Oct / Nov 2024 · Paper 13 · Q4Show mark scheme →
A0.61 s
B0.80 s
C1.3 s
D1.6 s
Mark scheme
48May / June 2024 · Paper 11 · Q4Show mark scheme →
Au = v - 2at
Bu = v - at
Cv = u + at^2
Dv = u + 2at^2
Mark scheme
49May / June 2024 · Paper 11 · Q7Show mark scheme →
Ahorizontal: constant deceleration; vertical: constant acceleration downwards
Bhorizontal: constant deceleration; vertical: acceleration decreases upwards then increases downwards
Chorizontal: constant velocity; vertical: constant acceleration downwards
Dhorizontal: constant velocity; vertical: acceleration decreases upwards then increases downwards
Mark scheme
50May / June 2024 · Paper 12 · Q4Show mark scheme →
AAcceleration can be determined from the gradient of a velocity–time graph.
BAcceleration is the rate of change of velocity.
CDisplacement can be determined from the area under a velocity–time graph.
DVelocity is the rate of change of distance.
Mark scheme
51May / June 2024 · Paper 12 · Q5Show mark scheme →
A0.13L
B0.25L
C0.38L
D0.50L
Mark scheme
52May / June 2024 · Paper 12 · Q6Show mark scheme →
A21m
B45m
C67m
D83m
Mark scheme
53May / June 2024 · Paper 13 · Q5Show mark scheme →
A6 m
B12 m
C18 m
D24 m
Mark scheme
54Feb / March 2024 · Paper 12 · Q6Show mark scheme →
A1.25 m
B3.75 m
C5.00 m
D6.25 m
Mark scheme
55Feb / March 2024 · Paper 12 · Q7Show mark scheme →
Achange in velocity per unit time
Brate of change of speed per unit time
Crate of change of velocity per unit time
Dresultant force per unit mass
Mark scheme
56Feb / March 2024 · Paper 12 · Q9Show mark scheme →
A4.2 m s⁻¹
B4.8 m s⁻¹
C18 m s⁻¹
D23 m s⁻¹
Mark scheme
57Oct / Nov 2023 · Paper 11 · Q6Show mark scheme →
AR = VT cos θ
BR = VT sin θ
CR = VT cos θ – 1/2 gT²
DR = VT sin θ – 1/2 gT²
Mark scheme
58Oct / Nov 2023 · Paper 12 · Q5Show mark scheme →
Athe rate of change of displacement
Bthe rate of change of kinetic energy
Cthe rate of change of momentum
Dthe rate of change of velocity
Mark scheme
59Oct / Nov 2023 · Paper 12 · Q6Show mark scheme →
A6.58 m s–1
B7.70 m s–1
C10.6 m s–1
D14.6 m s–1
Mark scheme
60Oct / Nov 2023 · Paper 13 · Q5Show mark scheme →
AGraph A: distance rises to 50 m at 40 s, stays flat to 50 s, then falls to 0 at 80 s
BGraph B: distance rises to 50 m at 40 s, stays flat to 50 s, then rises to 100 m at 80 s
CGraph C: distance rises to 50 m at 40 s, stays flat to 50 s, then falls to 0 at 80 s
DGraph D: distance rises to 50 m at 40 s, stays flat to 50 s, then rises to 100 m at 80 s
Mark scheme
61Oct / Nov 2023 · Paper 13 · Q6Show mark scheme →
A0.378
B0.615
C1.63
D2.64
Mark scheme
62May / June 2023 · Paper 11 · Q5Show mark scheme →
AThey have the same average speed.
BThey have the same average velocity.
CThey have the same displacement.
DThey travel the same distance.
Mark scheme
63May / June 2023 · Paper 11 · Q6Show mark scheme →
A0.47 m
B3.0 m
C4.3 m
D7.3 m
Mark scheme
64May / June 2023 · Paper 12 · Q5Show mark scheme →
A2.0 m
B2.0 m due south
C3.1 m
D3.1 m due south
Mark scheme
65May / June 2023 · Paper 12 · Q6Show mark scheme →
A12 m
B15 m
C19 m
D24 m
Mark scheme
66May / June 2023 · Paper 12 · Q17Show mark scheme →
A\sqrt{\frac{gx \sin\theta}{2}}
B\sqrt{2gx \sin\theta}
C\sqrt{2gx}
D2gx \sin\theta
Mark scheme
67May / June 2023 · Paper 13 · Q5Show mark scheme →
As = ut + \frac{1}{2}at^2
Bs = \frac{1}{2}(u + v)t
Cv = u + at
Dv^2 = u^2 + 2as
Mark scheme
68May / June 2023 · Paper 13 · Q6Show mark scheme →
AGraph A: horizontal displacement against time, curve with decreasing gradient
BGraph B: vertical displacement against time, curve with decreasing gradient
CGraph C: horizontal displacement against time, curve with increasing gradient
DGraph D: vertical displacement against time, curve with increasing gradient
Mark scheme
69May / June 2023 · Paper 13 · Q17Show mark scheme →
A0 J
B2.4 J
C11 J
D29 J
Mark scheme
70Feb / March 2023 · Paper 12 · Q5Show mark scheme →
Athe horizontal component of the ball's acceleration
Bthe horizontal component of the ball's velocity
Cthe vertical component of the ball's momentum
Dthe vertical component of the ball's velocity
Mark scheme
71Feb / March 2023 · Paper 12 · Q6Show mark scheme →
A[graph of a against t: triangular wave starting at positive maximum, crossing zero]
B[graph of a against t: square wave, positive then negative then positive]
C[graph of a against t: square wave, positive then negative then positive, offset]
D[graph of a against t: impulses/spikes at intervals]
Mark scheme
72Oct / Nov 2022 · Paper 11 · Q5Show mark scheme →
Agraph: v increases with a, curve concave down starting at origin
Bgraph: v increases with a, curve concave up starting at origin
Cgraph: v constant (horizontal line) as a increases
Dgraph: v vertical line at a = 0
Mark scheme
73Oct / Nov 2022 · Paper 11 · Q6Show mark scheme →
A390 m
B570 m
C710 m
D770 m
Mark scheme
74Oct / Nov 2022 · Paper 12 · Q5Show mark scheme →
Av₂₀ = 0.32, v₄₀ = 0
Bv₂₀ = 0.32, v₄₀ = 0.32
Cv₂₀ = 0.50, v₄₀ = 0
Dv₂₀ = 0.50, v₄₀ = 0.50
Mark scheme
75Oct / Nov 2022 · Paper 12 · Q6Show mark scheme →
AGraph A
BGraph B
CGraph C
DGraph D
Mark scheme
76Oct / Nov 2022 · Paper 12 · Q7Show mark scheme →
A6 s
B10 s
C20 s
D40 s
Mark scheme
77Oct / Nov 2022 · Paper 13 · Q5Show mark scheme →
AGraph A: v_H constant horizontal line, v_V increasing linearly from origin
BGraph B: v_V constant horizontal line, v_H increasing linearly from origin
CGraph C: v_H constant horizontal line, v_V increasing with increasing gradient
DGraph D: v_H constant horizontal line, v_V decreasing with increasing gradient
Mark scheme
78Oct / Nov 2022 · Paper 13 · Q6Show mark scheme →
A1/6
B1/√6
C√6
D6
Mark scheme
79May / June 2022 · Paper 11 · Q5Show mark scheme →
Athe area below the curve PQ
Bthe area of the triangle PQS
Cthe gradient of the straight line PQ
Dthe gradient of the tangent at Q
Mark scheme
80May / June 2022 · Paper 11 · Q6Show mark scheme →
A10 m s⁻¹
B29 m s⁻¹
C31 m s⁻¹
D39 m s⁻¹
Mark scheme
81May / June 2022 · Paper 12 · Q5Show mark scheme →
A4.2 m s–1
B4.7 m s–1
C6.0 m s–1
D14 m s–1
Mark scheme
82May / June 2022 · Paper 12 · Q6Show mark scheme →
A3.75 s
B3.99 s
C4.19 s
D4.22 s
Mark scheme
83May / June 2022 · Paper 13 · Q6Show mark scheme →
AX and Y are at rest at their maximum heights.
BX and Y are fired with the same speed.
CX and Y take the same time to return to the ground.
DX and Y travel the same distance.
Mark scheme
84May / June 2022 · Paper 13 · Q7Show mark scheme →
AGraph A: velocity v against time t, square wave with decreasing amplitude, alternating positive and negative, starting positive
BGraph B: velocity v against time t, square wave with decreasing amplitude, alternating positive and negative, starting positive
CGraph C: velocity v against time t, square wave of constant amplitude, alternating positive and negative, with decreasing period
DGraph D: velocity v against time t, square wave of constant amplitude, alternating positive and negative, with constant period
Mark scheme
85Feb / March 2022 · Paper 12 · Q5Show mark scheme →
Afrom the area under a displacement–time graph
Bfrom the area under a velocity–time graph
Cfrom the gradient of a displacement–time graph
Dfrom the gradient of a velocity–time graph
Mark scheme
86Feb / March 2022 · Paper 12 · Q6Show mark scheme →
A0.5 m s–2
B0.91 m s–2
C1.7 m s–2
D5.0 m s–2
Mark scheme
87Oct / Nov 2021 · Paper 11 · Q6Show mark scheme →
Ahorizontal component of displacement: p; vertical component of displacement: q
Bhorizontal component of displacement: q; vertical component of displacement: r
Chorizontal component of displacement: r; vertical component of displacement: p
Dhorizontal component of displacement: r; vertical component of displacement: q
Mark scheme
88Oct / Nov 2021 · Paper 11 · Q7Show mark scheme →
A1.0 s
B1.7 s
C2.5 s
D5.0 s
Mark scheme
89Oct / Nov 2021 · Paper 12 · Q6Show mark scheme →
A$0.67\,\mathrm{m\,s^{-2}}$
B$1.0\,\mathrm{m\,s^{-2}}$
C$1.3\,\mathrm{m\,s^{-2}}$
D$2.0\,\mathrm{m\,s^{-2}}$
Mark scheme
90Oct / Nov 2021 · Paper 12 · Q7Show mark scheme →
A$0.76\,\mathrm{m}$
B$11\,\mathrm{m}$
C$23\,\mathrm{m}$
D$110\,\mathrm{m}$
Mark scheme
91Oct / Nov 2021 · Paper 13 · Q6Show mark scheme →
Athe average velocity of the car
Bthe change in velocity of the car
Cthe final velocity of the car
Dthe initial velocity of the car
Mark scheme
92Oct / Nov 2021 · Paper 13 · Q7Show mark scheme →
AThe final displacement of the stone depends upon its initial horizontal velocity.
BThe stone travels with a constant horizontal component of velocity until it hits the water.
CThe stone travels with an increasing vertical component of velocity.
DThe time taken for the stone to hit the surface of the water depends on its initial horizontal velocity.
Mark scheme
93May / June 2021 · Paper 11 · Q6Show mark scheme →
AGraph A: speed constant (horizontal line)
BGraph B: speed decreases linearly to zero then increases linearly
CGraph C: speed decreases to a minimum then increases (curved)
DGraph D: speed decreases to zero then increases (curved)
Mark scheme
94May / June 2021 · Paper 11 · Q7Show mark scheme →
A75 s
B150 s
C230 s
D300 s
Mark scheme
95May / June 2021 · Paper 12 · Q6Show mark scheme →
AGraph A: v decreasing linearly from a positive value to zero as t increases
BGraph B: v decreasing exponentially from a positive value towards zero as t increases
CGraph C: v increasing from zero with decreasing gradient (curve) as t increases
DGraph D: v increasing linearly from zero as t increases
Mark scheme
96May / June 2021 · Paper 12 · Q7Show mark scheme →
A$R = \dfrac{V^2 \sin\theta \cos\theta}{g}$
B$R = \dfrac{2V^2 \sin\theta \cos\theta}{g}$
C$R = \dfrac{V^2 \sin\theta \cos\theta}{2g}$
D$R = \dfrac{V^2 g \sin\theta \cos\theta}{2}$
Mark scheme
97May / June 2021 · Paper 13 · Q6Show mark scheme →
A2.5 m
B3.0 m
C20 m
D28 m
Mark scheme
98May / June 2021 · Paper 13 · Q7Show mark scheme →
AG/2
BG
C2G
DG^2
Mark scheme
99Feb / March 2021 · Paper 12 · Q6Show mark scheme →
AA
BB
CC
DD
Mark scheme
100Feb / March 2021 · Paper 12 · Q7Show mark scheme →
A$H = \dfrac{v\sin\alpha}{2g}$
B$H = \dfrac{g\sin\alpha}{2v}$
C$H = \dfrac{(v\sin\alpha)^2}{2g}$
D$H = \dfrac{g^2\sin\alpha}{2v}$
Mark scheme
101Oct / Nov 2020 · Paper 11 · Q6Show mark scheme →
Agraph A
Bgraph B
Cgraph C
Dgraph D
Mark scheme
102Oct / Nov 2020 · Paper 11 · Q7Show mark scheme →
A40 m
B60 m
C180 m
D300 m
Mark scheme
103Oct / Nov 2020 · Paper 12 · Q6Show mark scheme →
A15 m above X
B15 m below X
C25 m above X
D25 m below X
Mark scheme
104Oct / Nov 2020 · Paper 12 · Q7Show mark scheme →
A1.6 m
B2.0 m
C2.7 m
D3.1 m
Mark scheme
105Oct / Nov 2020 · Paper 13 · Q6Show mark scheme →
A1.6 m
B1.8 m
C2.0 m
D3.2 m
Mark scheme
106Oct / Nov 2020 · Paper 13 · Q7Show mark scheme →
A3.0 m s⁻¹
B5.0 m s⁻¹
C6.0 m s⁻¹
D15 m s⁻¹
Mark scheme
107May / June 2020 · Paper 11 · Q6Show mark scheme →
A210 m s^-2
B58 m s^-2
C26 m s^-2
D7.3 m s^-2
Mark scheme
108May / June 2020 · Paper 11 · Q7Show mark scheme →
AR = VT cos θ
BR = VT sin θ
CR = VT cos θ − 1/2 gT^2
DR = VT sin θ − 1/2 gT^2
Mark scheme
109May / June 2020 · Paper 11 · Q8Show mark scheme →
AGraph A: p against s, curve with decreasing gradient starting from origin
BGraph B: p against s, straight line through origin
CGraph C: p against s, curve with increasing gradient starting from origin
DGraph D: p against s, horizontal line
Mark scheme
110May / June 2020 · Paper 11 · Q18Show mark scheme →
AGraph A: height above ground decreasing linearly with speed from a positive value to zero
BGraph B: height above ground increasing linearly with speed from zero
CGraph C: height above ground decreasing with speed, curve concave downward
DGraph D: height above ground increasing with speed, curve concave upward
Mark scheme
111May / June 2020 · Paper 12 · Q5Show mark scheme →
AIt is at a distance of 10 m above the surface of the water.
BIt is at a distance of 10 m below the surface of the water.
CIt is at a distance of 20 m below the surface of the water.
DIt is at a distance of 30 m below the surface of the water.
Mark scheme
112May / June 2020 · Paper 12 · Q6Show mark scheme →
AGraph A
BGraph B
CGraph C
DGraph D
Mark scheme
113May / June 2020 · Paper 13 · Q6Show mark scheme →
A15 m
B25 m
C30 m
D45 m
Mark scheme
114May / June 2020 · Paper 13 · Q7Show mark scheme →
Asame / same
Bsame / different
Cdifferent / same
Ddifferent / different
Mark scheme
115Feb / March 2020 · Paper 12 · Q6Show mark scheme →
Ahorizontally to the right
Bparallel to its velocity
Cperpendicular to its velocity
Dvertically downwards
Mark scheme
116Feb / March 2020 · Paper 12 · Q7Show mark scheme →
A0.75 m s–2
B1.5 m s–2
C3.0 m s–2
D6.0 m s–2
Mark scheme
117Oct / Nov 2019 · Paper 12 · Q6Show mark scheme →
A1.61 s
B3.22 s
C3.88 s
D5.18 s
Mark scheme
118Oct / Nov 2019 · Paper 13 · Q6Show mark scheme →
A15 m
B75 m
C80 m
D2000 m
Mark scheme
119May / June 2019 · Paper 11 · Q6Show mark scheme →
A21 m
B45 m
C67 m
D83 m
Mark scheme
120May / June 2019 · Paper 11 · Q7Show mark scheme →
AA
BB
CC
DD
Mark scheme
121May / June 2019 · Paper 13 · Q7Show mark scheme →
Adisplacement 60 m, acceleration 4.0 m s^-2
Bdisplacement 70 m, acceleration 4.0 m s^-2
Cdisplacement 60 m, acceleration 4.8 m s^-2
Ddisplacement 70 m, acceleration 4.8 m s^-2
Mark scheme
122Feb / March 2019 · Paper 12 · Q7Show mark scheme →
A3.5 m s–1
B6.6 m s–1
C13 m s–1
D20 m s–1
Mark scheme
123Oct / Nov 2018 · Paper 12 · Q6Show mark scheme →
A8.9 s
B10 s
C12 s
D14 s
Mark scheme
124Oct / Nov 2018 · Paper 13 · Q6Show mark scheme →
AGraph A
BGraph B
CGraph C
DGraph D
Mark scheme
125May / June 2018 · Paper 12 · Q5Show mark scheme →
A[graph: s against t, curve rising from origin to a maximum then falling back down, like an inverted parabola] s t 0 0
B[graph: s against t, straight line rising from origin then a less steep straight line continuing to rise] s t 0 0
C[graph: s against t, curve rising from origin with decreasing gradient, levelling off] s t 0 0
D[graph: s against t, straight line rising from origin to a peak then straight line falling back to zero] s t 0 0
Mark scheme
126May / June 2018 · Paper 12 · Q6Show mark scheme →
Aspeed against t: horizontal straight line at constant speed
Bspeed against t: straight line decreasing to zero then straight line increasing
Cspeed against t: curve decreasing to a minimum then increasing, minimum above zero
Dspeed against t: curve decreasing to zero then increasing
Mark scheme
127May / June 2018 · Paper 12 · Q7Show mark scheme →
A[graph of d against t: curve starting at origin with increasing gradient, concave upwards]
B[graph of d against t: curve starting at origin, gradient decreasing to zero, becoming horizontal]
C[graph of d against t: curve starting at origin, initially flat then increasing gradient, concave upwards]
D[graph of d against t: curve starting at origin with decreasing gradient, concave downwards]
Mark scheme
128May / June 2018 · Paper 12 · Q11Show mark scheme →
Anon-zero and constant; constant
Bnon-zero and constant; decreasing
Czero; constant
Dzero; decreasing
Mark scheme
129May / June 2018 · Paper 12 · Q18Show mark scheme →
A1/2²
B1/2
C1/√2
D1 − 1/√2
Mark scheme
130May / June 2018 · Paper 13 · Q6Show mark scheme →
A0.39 m s^{-2}
B3.7 m s^{-2}
C7.4 m s^{-2}
D9.8 m s^{-2}
Mark scheme
131Feb / March 2018 · Paper 12 · Q8Show mark scheme →
Aarea 1 – area 2
B(area 1 + area 2) / 2
Carea 1 + area 2
Darea 2 – area 1
Mark scheme
132Feb / March 2018 · Paper 12 · Q9Show mark scheme →
A3.07 m s–1
B7.85 m s–1
C15.7 m s–1
D31.4 m s–1
Mark scheme
133Oct / Nov 2017 · Paper 11 · Q7Show mark scheme →
Ahorizontal component of displacement: p; vertical component of displacement: q
Bhorizontal component of displacement: q; vertical component of displacement: r
Chorizontal component of displacement: r; vertical component of displacement: p
Dhorizontal component of displacement: r; vertical component of displacement: q
Mark scheme
134Oct / Nov 2017 · Paper 11 · Q18Show mark scheme →
A0.76 m
B11 m
C23 m
D110 m
Mark scheme
135Oct / Nov 2017 · Paper 12 · Q6Show mark scheme →
A29 m
B108 m
C123 m
D138 m
Mark scheme
136Oct / Nov 2017 · Paper 12 · Q7Show mark scheme →
A[velocity-time graph A: steeper rise than original, same maximum speed, same deceleration slope, shorter total time]
B[velocity-time graph B: steeper rise than original, same maximum speed, same deceleration slope, shorter total time]
C[velocity-time graph C: steeper rise than original, same maximum speed, same deceleration slope, shorter total time]
D[velocity-time graph D: steeper rise than original, same maximum speed, same deceleration slope, shorter total time]
Mark scheme
137Oct / Nov 2017 · Paper 13 · Q6Show mark scheme →
Ax-t graph: curve starting at origin, increasing with decreasing gradient, approaching a horizontal asymptote
Bx-t graph: straight line through the origin with constant positive gradient
Cx-t graph: curve starting at origin, increasing with increasing gradient
Dx-t graph: curve starting at origin, increasing with increasing gradient (parabolic)
Mark scheme
138Oct / Nov 2017 · Paper 13 · Q7Show mark scheme →
Av-t graph: two straight lines with negative gradient; Q starts higher and reaches zero later than P
Bv-t graph: two curves decreasing to zero; Q starts higher than P
Cv-t graph: two straight lines with negative gradient; Q starts higher and reaches zero at the same time as P
Dv-t graph: two curves decreasing to zero; Q starts higher and reaches zero at the same time as P
Mark scheme
139Oct / Nov 2017 · Paper 13 · Q11Show mark scheme →
AXV and XH
BXV only
CXH only
DXT only
Mark scheme
140May / June 2017 · Paper 11 · Q4Show mark scheme →
Aarea under a velocity-time graph
Bgradient of a displacement-time graph
Cgradient of a velocity-time graph
Dgradient of an acceleration-time graph
Mark scheme
141May / June 2017 · Paper 11 · Q5Show mark scheme →
A
B
C
D
Mark scheme
142May / June 2017 · Paper 11 · Q6Show mark scheme →
Athe horizontal component of the ball's acceleration
Bthe horizontal component of the ball's velocity
Cthe vertical component of the ball's momentum
Dthe vertical component of the ball's velocity
Mark scheme
143May / June 2017 · Paper 12 · Q6Show mark scheme →
AGraph A: v decreases linearly from a maximum, then is constant, then increases linearly
BGraph B: v decreases, then decreases less steeply, then increases
CGraph C: v decreases linearly, then is constant, then increases steeply
DGraph D: v increases from zero, then increases more steeply
Mark scheme
144May / June 2017 · Paper 13 · Q5Show mark scheme →
A2.2 m s–2
B8.7 m s–2
C27 m s–2
D54 m s–2
Mark scheme
145May / June 2017 · Paper 13 · Q6Show mark scheme →
Agraph A
Bgraph B
Cgraph C
Dgraph D
Mark scheme
146Feb / March 2017 · Paper 12 · Q6Show mark scheme →
A0.23 m s⁻²
B0.46 m s⁻²
C3.0 m s⁻²
D6.0 m s⁻²
Mark scheme
147Feb / March 2017 · Paper 12 · Q7Show mark scheme →
A–1.5 m s⁻¹
B0.0 m s⁻¹
C4.0 m s⁻¹
D4.5 m s⁻¹
Mark scheme
148Oct / Nov 2016 · Paper 11 · Q6Show mark scheme →
A6.3 m s–1
B9.4 m s–1
C9.9 m s–1
D15 m s–1
Mark scheme
149Oct / Nov 2016 · Paper 11 · Q7Show mark scheme →
A6.6 m s–1
B7.7 m s–1
C10.6 m s–1
D14.6 m s–1
Mark scheme
150Oct / Nov 2016 · Paper 11 · Q8Show mark scheme →
A16 m
B42 m
C58 m
D84 m
Mark scheme
151Oct / Nov 2016 · Paper 12 · Q7Show mark scheme →
Adistance-time graph: curve starting at origin with increasing gradient, then straightening to constant gradient
Bdistance-time graph: straight line from origin then curve with decreasing gradient levelling off
Cdistance-time graph: starting at non-zero distance, flat then decreasing to zero
Ddistance-time graph: starting at non-zero distance, decreasing curve to zero
Mark scheme
152Oct / Nov 2016 · Paper 12 · Q8Show mark scheme →
A390 m
B570 m
C710 m
D770 m
Mark scheme
153Oct / Nov 2016 · Paper 12 · Q9Show mark scheme →
Agraph A
Bgraph B
Cgraph C
Dgraph D
Mark scheme
154Oct / Nov 2016 · Paper 12 · Q10Show mark scheme →
A1.20 x
B1.25 x
C1.44 x
D1.56 x
Mark scheme
155Oct / Nov 2016 · Paper 13 · Q6Show mark scheme →
A6.3 m s–1
B9.4 m s–1
C9.9 m s–1
D15 m s–1
Mark scheme
156Oct / Nov 2016 · Paper 13 · Q7Show mark scheme →
A6.6 m s–1
B7.7 m s–1
C10.6 m s–1
D14.6 m s–1
Mark scheme
157Oct / Nov 2016 · Paper 13 · Q8Show mark scheme →
A16 m
B42 m
C58 m
D84 m
Mark scheme
158May / June 2016 · Paper 11 · Q6Show mark scheme →
ANPTU + PQST
BNPWV + VRSU
CNPWV + WRST
DPST + PQS
Mark scheme
159May / June 2016 · Paper 11 · Q7Show mark scheme →
AGraph A: r against h, curve starting at origin with decreasing gradient (concave down)
BGraph B: r against h, curve starting at origin with increasing gradient (concave up)
CGraph C: r against h, curve starting at non-zero r with increasing gradient
DGraph D: r against h, straight line through origin
Mark scheme
160May / June 2016 · Paper 11 · Q8Show mark scheme →
A1.0 s
B1.7 s
C2.5 s
D5.0 s
Mark scheme
161May / June 2016 · Paper 12 · Q6Show mark scheme →
AGraph A: straight line from origin (0,0) to (1, 1.6)
BGraph B: curve starting at origin with increasing gradient, reaching (1, 1.6)
CGraph C: curve starting at origin with decreasing gradient, reaching (1, 1.6)
DGraph D: horizontal line at speed 1.6 from time 0 to 1
Mark scheme
162May / June 2016 · Paper 12 · Q8Show mark scheme →
AGraph A: s against t, starting at 0, rising to a single maximum and returning to 0 at t = T
BGraph B: s against t, sinusoidal starting at 0, rising to a maximum, crossing zero and reaching a minimum before returning to 0 at t = T
CGraph C: s against t, two positive humps with a cusp at zero in the middle, returning to 0 at t = T
DGraph D: s against t, two positive humps separated by a flat zero region, returning to 0 at t = T
Mark scheme
163May / June 2016 · Paper 13 · Q6Show mark scheme →
AGraph A: velocity against time, straight line with negative gradient starting from positive velocity, crossing time axis
BGraph B: velocity against time, triangular shape starting at zero, rising to a peak then falling back to zero
CGraph C: velocity against time, rising straight line from zero to a peak then dropping vertically to negative and rising back
DGraph D: velocity against time, straight line with negative gradient from positive velocity to zero, then rising with positive gradient
Mark scheme
164May / June 2016 · Paper 13 · Q7Show mark scheme →
Ax = Vt cos θ; y = Vt sin θ + 1/2 gt^2
Bx = Vt cos θ; y = Vt sin θ – 1/2 gt^2
Cx = Vt sin θ; y = Vt cos θ + 1/2 gt^2
Dx = Vt sin θ; y = Vt cos θ – 1/2 gt^2
Mark scheme
165Feb / March 2016 · Paper 12 · Q6Show mark scheme →
AThe direction of a is constant and the direction of a is the same as the direction of s.
BThe direction of a is constant and the direction of a is the same as the direction of u.
CThe magnitude of a is constant and the direction of a is constant.
DThe magnitude of a is constant and the direction of a is the same as the direction of v.
Mark scheme
166Feb / March 2016 · Paper 12 · Q7Show mark scheme →
AGraph A
BGraph B
CGraph C
DGraph D
Mark scheme
167Feb / March 2016 · Paper 12 · Q8Show mark scheme →
AThe ball lands with the same velocity at which it is thrown.
BThe horizontal component of acceleration is constant throughout the motion.
CThe horizontal and vertical components of acceleration are both zero at the highest point of the motion.
DThe horizontal and vertical components of velocity are both zero at the highest point of the motion.
Mark scheme