Kinematics HL Paper 1&2

A.1 Kinematics

Multiple-choice questions, Extended responses and a Lab-style investigation

This original question bank develops Paper 1A multiple-choice skills, Paper 2 structured problem-solving and Paper 1B-style data analysis. It covers displacement and velocity vectors, motion graphs, constant acceleration, free fall and projectile motion. These are practice questions, not reproduced IB examination questions.

Instructions and data

Select one answer for each multiple-choice question before opening its explanation.
For extended responses, show the equation, substitution and answer with units.
Use three significant figures unless an exact value or another precision is requested. Retain extra digits during calculations.
Use degree mode for angles. State a positive direction before using signed quantities.
Take $g=9.81\,\mathrm{m\,s^{-2}}$ . Neglect air resistance unless a question explicitly includes it.
Use constant-acceleration equations only over intervals where acceleration is constant.
A.1 is shared SL/HL content. “Challenge” identifies a demanding application, not a separate HL-only syllabus requirement.
Suggested marks are learning guides. Equivalent correct methods are acceptable.

Equation reminders

$\bar v=\frac{\Delta x}{\Delta t},\qquad\text{average speed}=\frac{\text{total distance}}{\text{total time}}$

$\bar a=\frac{\Delta v}{\Delta t}$

$v=u+at,\qquad s=ut+\frac12at^2$

$v^2=u^2+2as,\qquad s=\frac{u+v}{2}t$

$u_x=u\cos\theta,\qquad u_y=u\sin\theta$

$x=u_xt,\qquad y=u_yt-\frac12gt^2,\qquad v_y=u_y-gt$

Here s and y are displacements, not automatically distances travelled. The projectile equations use an origin at the launch point, upward positive and no horizontal acceleration. Add an initial height when using ground-level coordinates.

Common misconceptions

Negative acceleration does not always mean decreasing speed: compare the signs of velocity and acceleration.
Displacement is the signed area under a velocity–time graph; distance uses the magnitudes of all areas.
At a projectile’s highest point, vertical velocity is zero but acceleration is still downward.
Horizontal and vertical projectile motion share the same flight time.
Doubling ascent time gives total flight time only when launch and landing heights are equal and drag is negligible.

Section A — Paper 1A-style multiple choice

15 questions × 1 mark = 15 marks.

Multiple choice 1 — Distance and displacement [1 mark]

A student walks 60.0 m east and then 40.0 m west in 20.0 s. Which pair gives the average speed and average velocity?

1.00 m/s; 5.00 m/s east.
5.00 m/s; 1.00 m/s east.
5.00 m/s; 1.00 m/s west.
1.00 m/s; 1.00 m/s east.

Show answer and explanation

Correct option: B.

Distance is 100 m, whereas displacement is +20.0 m if east is positive.

$\text{average speed}=100/20.0=5.00\,\mathrm{m\,s^{-1}}$

$\bar v=20.0/20.0=1.00\,\mathrm{m\,s^{-1}}$

Adding the path lengths gives speed; subtracting opposing displacements gives velocity.

Multiple choice 2 — Velocity components [1 mark]

A particle has velocity components +12.0 m/s east and −5.00 m/s north. What are its speed and direction?

7.00 m/s; 22.6° south of east.
13.0 m/s; 22.6° north of east.
17.0 m/s; 67.4° south of east.
13.0 m/s; 22.6° south of east.

Show answer and explanation

Correct option: D.

$v=\sqrt{12.0^2+(-5.00)^2}=13.0\,\mathrm{m\,s^{-1}}$

$\theta=\tan^{-1}(5.00/12.0)=22.6^\circ$

The negative northward component means motion toward the south.

Multiple choice 3 — Position–time graph [1 mark]

A position–time graph is a horizontal line over a time interval. What does this show?

The object is at rest throughout the interval.
The object has constant non-zero speed.
The object has constant non-zero acceleration.
The object returns to its starting position during the interval.

Show answer and explanation

Correct option: A.

The position does not change. The gradient, and therefore the velocity, is zero. A horizontal line can occur at any fixed position; it need not be at the origin.

Multiple choice 4 — Signs of velocity and acceleration [1 mark]

At an instant, a particle has velocity −6.00 m/s and acceleration +2.00 m/s². Which statement describes its motion at that instant?

It moves in the positive direction and speeds up.
It moves in the positive direction and slows down.
It moves in the negative direction and slows down.
It moves in the negative direction and speeds up.

Show answer and explanation

Correct option: C.

Velocity determines direction. Acceleration opposes the negative velocity, so its magnitude is decreasing at this instant.

Multiple choice 5 — Signed graph areas [1 mark]

Velocity decreases linearly from +4.00 m/s at t = 0 to −2.00 m/s at t = 3.00 s. What are the displacement and distance travelled?

+5.00 m; 5.00 m.
+3.00 m; 5.00 m.
+3.00 m; 3.00 m.
−3.00 m; 5.00 m.

Show answer and explanation

Correct option: B.

$a=\frac{-2.00-4.00}{3.00}=-2.00\,\mathrm{m\,s^{-2}}$

Velocity reaches zero at 2.00 s. The positive triangular area is 4.00 m; the negative area has magnitude 1.00 m.

$s=4.00-1.00=3.00\,\mathrm{m},\qquad d=4.00+1.00=5.00\,\mathrm{m}$

The trapezium gives signed displacement, not distance when direction changes.

Multiple choice 6 — Acceleration–time graph [1 mark]

Acceleration is +2.00 m/s² for 3.00 s, then −1.00 m/s² for 2.00 s. What is the total change in velocity?

+4.00 m/s.
+8.00 m/s.
−4.00 m/s.
+1.00 m/s.

Show answer and explanation

Correct option: A.

$\Delta v=(2.00)(3.00)+(-1.00)(2.00)=4.00\,\mathrm{m\,s^{-1}}$

The signed area under an acceleration–time graph gives change in velocity.

Multiple choice 7 — Constant acceleration [1 mark]

A trolley has initial velocity +5.00 m/s and constant acceleration +3.00 m/s² for 4.00 s. What are its final velocity and displacement?

17.0 m/s; 68.0 m.
12.0 m/s; 44.0 m.
17.0 m/s; 24.0 m.
17.0 m/s; 44.0 m.

Show answer and explanation

Correct option: D.

$v=5.00+(3.00)(4.00)=17.0\,\mathrm{m\,s^{-1}}$

$s=(5.00)(4.00)+\frac12(3.00)(4.00)^2=44.0\,\mathrm{m}$

Using final speed multiplied by time would overestimate displacement.

Multiple choice 8 — Stopping distance [1 mark]

A vehicle slows uniformly from 18.0 m/s to rest with acceleration −3.00 m/s². What is its braking distance?

6.00 m.
108 m.
54.0 m.
27.0 m.

Show answer and explanation

Correct option: C.

$s=\frac{v^2-u^2}{2a}=\frac{0-(18.0)^2}{2(-3.00)}=54.0\,\mathrm{m}$

Both numerator and denominator are negative, giving positive forward displacement.

Multiple choice 9 — Highest point [1 mark]

A ball is thrown vertically upward with negligible air resistance. What are its velocity and acceleration at its highest point?

Zero velocity; zero acceleration.
Zero velocity; 9.81 m/s² downward.
9.81 m/s downward; zero acceleration.
Zero velocity; 9.81 m/s² upward.

Show answer and explanation

Correct option: B.

The ball momentarily stops rising, so velocity is zero. Gravity continues to produce downward acceleration.

Multiple choice 10 — Horizontal launch [1 mark]

Two balls are projected horizontally from the same height. Ball B has twice the launch speed of ball A. Neglect air resistance. How do B’s flight time and range compare with A’s?

Same flight time; twice the range.
Twice the flight time; twice the range.
Same flight time; four times the range.
Half the flight time; same range.

Show answer and explanation

Correct option: A.

$t=\sqrt{2h/g},\qquad R=u_xt$

Flight time depends on vertical motion and is unchanged. Range is proportional to horizontal launch speed.

Multiple choice 11 — Projectile at the apex [1 mark]

A projectile is launched at an angle strictly between 0° and 90° above the horizontal. At the highest point, which statement is correct when drag is negligible?

Its speed is zero and acceleration is zero.
Its horizontal velocity is zero and acceleration is downward.
Its vertical velocity is zero and horizontal velocity is unchanged.
Its vertical velocity is unchanged and acceleration is zero.

Show answer and explanation

Correct option: C.

Gravity changes the vertical component. There is no horizontal acceleration, so the horizontal component remains non-zero and constant.

Multiple choice 12 — Complementary launch angles [1 mark]

Projectiles are launched at the same speed from level ground at 30° and 60°, landing at the launch height. Neglect drag. Which comparison is correct?

The 30° projectile has greater range and longer flight time.
The 60° projectile has greater range and the same flight time.
The ranges and flight times are both equal.
The ranges are equal; the 60° projectile has longer flight time.

Show answer and explanation

Correct option: D.

$R=\frac{u^2\sin2\theta}{g},\qquad T=\frac{2u\sin\theta}{g}$

The range factors sin 60° and sin 120° are equal. The 60° launch has a larger vertical component and hence longer flight time.

Multiple choice 13 — Signed vertical velocity [1 mark]

A ball is thrown upward at 10.0 m/s. Taking upward as positive, what is its velocity 2.00 s later? Neglect drag.

−9.62 m/s.
+9.62 m/s.
−19.6 m/s.
+29.6 m/s.

Show answer and explanation

Correct option: A.

$v=10.0-(9.81)(2.00)=-9.62\,\mathrm{m\,s^{-1}}$

The negative sign indicates downward motion.

Multiple choice 14 — Resultant displacement [1 mark]

A walker travels 3.00 m east and then 4.00 m north. What is the resultant displacement?

7.00 m at 53.1° north of east.
5.00 m at 53.1° north of east.
5.00 m at 36.9° north of east.
1.00 m due north.

Show answer and explanation

Correct option: B.

$s=\sqrt{3.00^2+4.00^2}=5.00\,\mathrm{m}$

$\theta=\tan^{-1}(4.00/3.00)=53.1^\circ$

The 7.00 m path length is the distance travelled, not the displacement magnitude.

Multiple choice 15 — Negative acceleration [1 mark]

A particle has initial velocity −3.00 m/s and constant acceleration −2.00 m/s² for 4.00 s. Which statement is correct?

It slows down and moves 28.0 m in the positive direction.
It speeds up and has displacement −16.0 m.
It speeds up and has displacement −28.0 m.
It slows down and has displacement −28.0 m.

Show answer and explanation

Correct option: C.

$v=-3.00+(-2.00)(4.00)=-11.0\,\mathrm{m\,s^{-1}}$

$s=(-3.00)(4.00)+\frac12(-2.00)(4.00)^2=-28.0\,\mathrm{m}$

Velocity and acceleration have the same sign, so speed increases. The distance is 28.0 m because there is no reversal.

Section B — Paper 2-style extended responses

The 15 questions below total 102 suggested marks. Complete them in smaller sets if using this bank for revision.

Extended response 1 — Resolving displacement vectors [6 marks]

A particle undergoes a displacement of 12.0 m due east followed by a displacement of 8.00 m at 60.0° north of east.

Resolve the second displacement into eastward and northward components. [2]
Determine the magnitude of the resultant displacement. [2]
Determine its direction relative to east. [2]

Show worked answer

(a) Take east and north as positive. Resolve the second displacement:

$s_{2x}=8.00\cos60.0^\circ=4.00\,\mathrm{m}$

$s_{2y}=8.00\sin60.0^\circ=6.9282\,\mathrm{m}$

One mark for each component. [2]

(b) Add corresponding components, then apply Pythagoras:

$s_x=12.0+4.00=16.0\,\mathrm{m},\quad s_y=6.9282\,\mathrm{m}$

$s=\sqrt{16.0^2+6.9282^2}=17.4\,\mathrm{m}$

Component addition/method [1]; resultant magnitude [1].

(c)

$\theta=\tan^{-1}\left(\frac{6.9282}{16.0}\right)=23.4^\circ$

Angle calculation [1]; correct direction [1].

Answer: 17.4 m at 23.4° north of east.

Check: The displacement magnitude is smaller than the 20.0 m distance travelled.

Extended response 2 — Velocity–time interpretation [6 marks]

A car’s velocity increases uniformly from +5.00 m/s to +25.0 m/s in 8.00 s.

Calculate the acceleration. [2]
Calculate the displacement. [2]
Explain the meanings of the gradient and area of a velocity–time graph. [2]

Show worked answer

(a)

$a=\frac{v-u}{t}=\frac{25.0-5.00}{8.00}=2.50\,\mathrm{m\,s^{-2}}$

Equation [1]; answer [1].

(b) Constant acceleration permits average velocity (u + v)/2:

$s=\frac{5.00+25.0}{2}(8.00)=120\,\mathrm{m}$

Equation or trapezium method [1]; answer [1].

(c) Gradient gives acceleration [1]. Signed area gives displacement [1].

Answer: Acceleration +2.50 m/s²; displacement +120 m.

Check: Velocity stays positive, so distance and displacement magnitude are both 120 m.

Extended response 3 — Vertical launch [7 marks]

A ball is thrown vertically upward at 18.0 m/s and returns to its launch height. Neglect air resistance.

Determine the maximum height above the launch point. [2]
Determine the time to the highest point. [2]
Determine total flight time. [2]
State its acceleration at the highest point. [1]

Show worked answer

Take upward as positive: u = +18.0 m/s and a = −9.81 m/s².

(a) At the top, v = 0:

$0=u^2-2gH,\qquad H=\frac{18.0^2}{2(9.81)}=16.5\,\mathrm{m}$

Equation [1]; height [1].

(b)

$0=18.0-9.81t,\qquad t_{\mathrm{up}}=1.83\,\mathrm{s}$

Equation [1]; time [1].

(c) Equal launch and landing heights give symmetric flight:

$T=\frac{2(18.0)}{9.81}=3.67\,\mathrm{s}$

Symmetry or non-zero root [1]; time [1].

(d) Acceleration remains 9.81 m/s² downward. [1]

Answer: H = 16.5 m; ascent time 1.83 s; total time 3.67 s.

Check: Calculate total time from unrounded values.

Extended response 4 — Horizontal projectile [7 marks]

A ball is projected horizontally at 12.0 m/s from a cliff 45.0 m above level ground.

Find the time to reach the ground. [2]
Find the horizontal range. [2]
Find the magnitude and direction of impact velocity. [3]

Show worked answer

Use downward as positive for vertical calculations. The initial vertical velocity is zero.

(a)

$45.0=\frac12(9.81)t^2,\qquad t=\sqrt{\frac{90.0}{9.81}}=3.03\,\mathrm{s}$

Equation [1]; time [1].

(b) Use the unrounded flight time:

$R=12.0\sqrt{\frac{90.0}{9.81}}=36.3\,\mathrm{m}$

Horizontal method [1]; range [1].

(c)

$v_y=gt=29.7136\,\mathrm{m\,s^{-1}},\qquad v_x=12.0\,\mathrm{m\,s^{-1}}$

$v=\sqrt{12.0^2+29.7136^2}=32.0\,\mathrm{m\,s^{-1}}$

$\phi=\tan^{-1}(29.7136/12.0)=68.0^\circ$

Vertical component [1]; resultant speed [1]; angle below horizontal [1].

Answer: Flight time 3.03 s; range 36.3 m; impact velocity 32.0 m/s at 68.0° below horizontal.

Check: Horizontal velocity is unchanged; the vertical component is downward.

Extended response 5 — Angled projectile on level ground [7 marks]

A projectile is launched at 25.0 m/s at 35.0° above horizontal and lands at the launch height.

Determine time of flight. [3]
Determine maximum height above launch. [2]
Determine range. [2]

Show worked answer

First resolve the launch velocity, retaining extra digits:

$u_x=25.0\cos35.0^\circ=20.4788\,\mathrm{m\,s^{-1}},\quad u_y=25.0\sin35.0^\circ=14.3394\,\mathrm{m\,s^{-1}}$

(a) At landing y = 0:

$0=u_yT-\frac12gT^2,\qquad T=\frac{2u_y}{g}=2.92\,\mathrm{s}$

Resolution [1]; non-zero-root method [1]; time [1].

(b) At the highest point, vertical velocity is zero:

$H=\frac{u_y^2}{2g}=10.5\,\mathrm{m}$

Equation [1]; height [1].

(c)

$R=u_xT=59.9\,\mathrm{m}$

Equation [1]; range [1].

Answer: T = 2.92 s; H = 10.5 m; R = 59.9 m.

Check: The equal-height range expression also gives 59.9 m when intermediate values remain unrounded.

Extended response 6 — Challenge: unequal launch and landing heights [7 marks]

A projectile is launched from a platform 20.0 m above ground at 18.0 m/s, directed 30.0° above horizontal.

Calculate its horizontal and vertical launch components. [2]
Determine its flight time to the ground. [3]
Determine its horizontal range. [2]

Show worked answer

(a)

$u_x=18.0\cos30.0^\circ=15.5885\,\mathrm{m\,s^{-1}},\qquad u_y=9.00\,\mathrm{m\,s^{-1}}$

One mark per component. [2]

(b) Use the launch point as y = 0 and take upward positive. Landing occurs at y = −20.0 m:

$-20.0=9.00t-4.905t^2$

$4.905t^2-9.00t-20.0=0$

$t=\frac{9.00+\sqrt{9.00^2+4(4.905)(20.0)}}{9.81}=3.14\,\mathrm{s}$

Signed displacement/equation [1]; quadratic method [1]; positive root [1]. The negative root does not describe motion after launch.

(c)

$R=(18.0\cos30.0^\circ)(3.135347\ldots)=48.9\,\mathrm{m}$

Equation [1]; range [1].

Answer: Flight time 3.14 s; range 48.9 m.

Check: Twice the ascent time would return the projectile to platform level, not ground level.

Extended response 7 — Piecewise velocity–time graph [8 marks]

A particle follows the velocity–time graph defined by the coordinates below. Positive and negative velocities describe opposite directions.

Velocity–time coordinates; join adjacent points with straight lines
t / s	v / m s⁻¹
0	−4.00
2.00	0
5.00	+6.00
7.00	+6.00
9.00	0

Determine acceleration on each of the four intervals. [2]
Calculate total displacement. [2]
Calculate total distance travelled. [2]
Calculate average velocity and average speed over 9.00 s. [2]

Show worked answer

(a) Gradients are +2.00, +2.00, 0 and −3.00 m/s², in interval order. [2]

(b) Signed areas:

$s=-\frac12(2.00)(4.00)+\frac12(3.00)(6.00)+(2.00)(6.00)+\frac12(2.00)(6.00)$

$s=-4.00+9.00+12.0+6.00=23.0\,\mathrm{m}$

Signed-area method [1]; displacement [1].

(c) Sum area magnitudes:

$d=4.00+9.00+12.0+6.00=31.0\,\mathrm{m}$

Magnitude-area method [1]; distance [1].

(d)

$\bar v=23.0/9.00=2.56\,\mathrm{m\,s^{-1}},\qquad\text{average speed}=31.0/9.00=3.44\,\mathrm{m\,s^{-1}}$

One mark per average. [2]

Answer: Displacement +23.0 m; distance 31.0 m; average velocity +2.56 m/s; average speed 3.44 m/s.

Check: Direction reverses at t = 2.00 s.

Extended response 8 — Piecewise position–time graph [7 marks]

The idealized position–time graph of an object consists of straight segments between the following points.

Position–time coordinates; join adjacent points with straight lines
t / s	x / m
0	0
2.00	6.00
5.00	6.00
7.00	2.00

Determine velocity on each interval. [3]
Determine displacement and distance over the full 7.00 s. [2]
Determine average velocity and average speed. [2]

Show worked answer

(a) Velocity is the gradient on each segment:

$v_1=6.00/2.00=3.00\,\mathrm{m\,s^{-1}}$

$v_2=0,\qquad v_3=\frac{2.00-6.00}{7.00-5.00}=-2.00\,\mathrm{m\,s^{-1}}$

One mark per velocity. [3]

(b) Displacement = 2.00 − 0 = +2.00 m. [1] Distance = 6.00 + 0 + 4.00 = 10.0 m. [1]

(c)

$\bar v=2.00/7.00=0.286\,\mathrm{m\,s^{-1}},\qquad\text{average speed}=10.0/7.00=1.43\,\mathrm{m\,s^{-1}}$

One mark per average. [2]

Answer: Velocities +3.00, 0 and −2.00 m/s; displacement +2.00 m; distance 10.0 m.

Check: The horizontal segment shows rest, not motion at constant non-zero speed.

Extended response 9 — Reaction and braking distance [7 marks]

A car travels at 24.0 m/s. The driver has reaction time 0.750 s, during which speed is constant. Braking then produces constant acceleration −6.00 m/s² until rest.

Determine distance travelled during reaction. [1]
Determine braking distance. [2]
Determine total stopping distance and total stopping time. [2]
Determine total stopping distance at an initial speed of 32.0 m/s, with unchanged reaction time and braking acceleration. [2]

Show worked answer

(a)

$d_{\mathrm{reaction}}=24.0(0.750)=18.0\,\mathrm{m}$

Correct value. [1]

(b)

$d_{\mathrm{braking}}=\frac{0-(24.0)^2}{2(-6.00)}=48.0\,\mathrm{m}$

Equation [1]; distance [1].

(c)

$d_{\mathrm{stop}}=18.0+48.0=66.0\,\mathrm{m}$

$t_{\mathrm{stop}}=0.750+\frac{24.0}{6.00}=4.75\,\mathrm{s}$

One mark per total. [2]

(d)

$d_{\mathrm{stop,new}}=(32.0)(0.750)+\frac{(32.0)^2}{2(6.00)}=109\,\mathrm{m}$

Both contributions included [1]; answer [1].

Answer: Original stopping distance 66.0 m and time 4.75 s; at 32.0 m/s, stopping distance 109 m.

Check: Reaction distance scales with speed, but braking distance scales with speed squared.

Extended response 10 — Reversal under constant acceleration [7 marks]

A particle has initial velocity +12.0 m/s and constant acceleration −3.00 m/s² for 6.00 s.

Determine when its direction reverses. [1]
Determine final velocity. [1]
Determine displacement after 6.00 s. [2]
Determine total distance travelled. [2]
Explain its change in speed before and after reversal. [1]

Show worked answer

(a)

$0=12.0-3.00t,\qquad t=4.00\,\mathrm{s}$

Correct time. [1]

(b)

$v=12.0-3.00(6.00)=-6.00\,\mathrm{m\,s^{-1}}$

Correct signed velocity. [1]

(c)

$s=(12.0)(6.00)+\frac12(-3.00)(6.00)^2=18.0\,\mathrm{m}$

Equation [1]; displacement [1].

(d) Before reversal the particle moves +24.0 m. In the final 2.00 s it moves −6.00 m:

$d=24.0+6.00=30.0\,\mathrm{m}$

Separate motion at reversal [1]; total distance [1].

(e) Speed decreases while velocity and acceleration have opposite signs, then increases when both are negative. [1]

Answer: Reversal at 4.00 s; final velocity −6.00 m/s; displacement +18.0 m; distance 30.0 m.

Extended response 11 — Challenge: a delayed pursuit [6 marks]

Runner A passes a marker at t = 0 with constant velocity +3.00 m/s. Runner B starts from rest at the same marker at t = 2.00 s and accelerates at +2.00 m/s². Treat both runners as particles; the idealized acceleration continues until B catches A.

Write each runner’s position using τ, the time since B starts. [2]
Determine when B catches A, measured from t = 0. [2]
Determine the catch position and B’s velocity at that instant. [2]

Show worked answer

(a) A has been moving for τ + 2.00 s:

$x_A=3.00(\tau+2.00),\qquad x_B=\frac12(2.00)\tau^2=\tau^2$

One mark per position expression. [2]

(b) Equate positions:

$\tau^2=3.00\tau+6.00$

$\tau=\frac{3.00+\sqrt{33.0}}{2}=4.37228\ldots\,\mathrm{s}$

$t=\tau+2.00=6.37\,\mathrm{s}$

Equation/root method [1]; time from the stated origin [1].

(c)

$x=3.00(6.37228\ldots)=19.1\,\mathrm{m}$

$v_B=2.00(4.37228\ldots)=8.74\,\mathrm{m\,s^{-1}}$

One mark per result. [2]

Answer: Catch time 6.37 s after A passes the marker; position 19.1 m; B velocity +8.74 m/s.

Check: Use B’s own elapsed time when calculating B’s velocity.

Extended response 12 — An object released from a rising balloon [7 marks]

A balloon rises vertically at constant speed 5.00 m/s. It releases a small object at height 50.0 m above ground. Neglect drag after release.

State the object’s initial velocity relative to the ground. [1]
Determine its time to reach the ground. [2]
Determine its impact velocity. [1]
Determine its maximum height above ground. [1]
Determine its total distance travelled after release. [1]
State why constant-acceleration equations would become inaccurate if drag were significant. [1]

Show worked answer

(a) The object initially shares the balloon’s upward velocity: +5.00 m/s. [1]

(b) Take ground level as y = 0 and upward as positive:

$0=50.0+5.00t-4.905t^2$

$t=\frac{5.00+\sqrt{5.00^2+4(4.905)(50.0)}}{9.81}=3.74\,\mathrm{s}$

Equation [1]; positive root [1].

(c)

$v=5.00-9.81t=-31.7\,\mathrm{m\,s^{-1}}$

Correct signed velocity. [1]

(d)

$H=50.0+\frac{5.00^2}{2(9.81)}=51.3\,\mathrm{m}$

Correct maximum height. [1]

(e) The object first rises 1.27421 m, then falls 51.27421 m:

$d=1.27421+51.27421=52.5\,\mathrm{m}$

Correct total distance. [1]

(f) Drag changes as velocity changes, so acceleration would not remain constant. [1]

Answer: Initial velocity +5.00 m/s; flight time 3.74 s; impact velocity −31.7 m/s; maximum height 51.3 m; distance 52.5 m.

Extended response 13 — Challenge: clearing a wall [7 marks]

A projectile is launched from ground level at 20.0 m/s at 40.0° above horizontal. A thin wall is 25.0 m horizontally from the launch point and 7.00 m high. Treat the projectile as a particle.

Determine the launch velocity components. [2]
Determine the time at which the projectile reaches the wall. [1]
Determine its height at the wall. [2]
State whether it clears the wall and by how much. [1]
Determine whether it is rising or falling there, using vertical velocity. [1]

Show worked answer

(a)

$u_x=20.0\cos40.0^\circ=15.3209\,\mathrm{m\,s^{-1}}$

$u_y=20.0\sin40.0^\circ=12.8558\,\mathrm{m\,s^{-1}}$

One mark per component. [2]

(b)

$t=25.0/u_x=1.63\,\mathrm{s}$

Correct time. [1]

(c) Use unrounded components and time:

$y=u_yt-\frac12gt^2=7.92\,\mathrm{m}$

Equation [1]; height [1].

(d) It clears the wall by approximately 0.917 m. [1]

(e)

$v_y=u_y-gt=-3.15\,\mathrm{m\,s^{-1}}$

It is falling, as shown by the negative vertical velocity. [1]

Answer: Height at wall 7.92 m; clearance 0.917 m; falling with vertical velocity −3.15 m/s.

Check: Clearing the wall depends on height at the wall, not merely maximum height.

Extended response 14 — Estimating motion with varying acceleration [7 marks]

A sensor records the velocities below. Between readings, assume the true velocity–time curve is smooth and bends upward, so straight chords lie above it.

Measured velocities at equally spaced times
t / s	v / m s⁻¹
0	0
1.00	2.00
2.00	6.00
3.00	12.0
4.00	20.0

Calculate average acceleration on each one-second interval. [2]
Estimate displacement using trapezia between adjacent readings. [2]
Estimate average velocity over the full interval. [1]
Explain why one constant-acceleration calculation using only initial and final velocities is inappropriate. [1]
State whether the trapezium estimate is above or below the true displacement, and explain. [1]

Show worked answer

(a) Successive velocity changes divided by 1.00 s give +2.00, +4.00, +6.00 and +8.00 m/s². [2]

(b) Add the areas of four trapezia:

$s_{\mathrm{interval}}=\frac{v_i+v_f}{2}\Delta t$

$s_{\mathrm{estimate}}=1.00+4.00+9.00+16.0=30.0\,\mathrm{m}$

Trapezium method [1]; displacement estimate [1].

(c) Estimated average velocity = 30.0/4.00 = 7.50 m/s. [1]

(d) Acceleration varies, so the constant-acceleration average (u + v)/2 cannot describe the full interval. [1]

(e) The straight chords lie above the specified curve, so trapezia overestimate the true area and displacement. [1]

Answer: Trapezium displacement estimate 30.0 m; estimated average velocity 7.50 m/s.

Check: The initial-to-final trapezium would give 40.0 m, a less accurate estimate for this curve.

Extended response 15 — Projectile motion with air resistance [6 marks]

A projectile is launched upward and to the right. There is no wind, and drag always acts opposite its instantaneous velocity. Consider its motion while it continues travelling right.

Describe how its horizontal velocity changes. [1]
Explain why its vertical acceleration generally differs from the constant value −g away from the highest point. [2]
At the highest point, state its vertical velocity, whether its horizontal velocity is zero, and the direction and magnitude of its vertical acceleration. [2]
Explain why the trajectory is not the ideal symmetric parabola. [1]

Show worked answer

(a) Horizontal velocity decreases because drag has a leftward component while the projectile moves right. [1]

(b) Drag’s vertical component opposes vertical motion. On ascent it acts downward, adding to gravity; on descent it acts upward, reducing the magnitude of downward acceleration. [1] Its magnitude depends on velocity, so vertical acceleration generally varies. [1]

(c) At the highest point, vertical velocity is zero while horizontal velocity is still positive. [1] Drag is horizontal at that instant, so vertical acceleration is −g, or 9.81 m/s² downward. [1] Total acceleration also has a horizontal component.

(d) Horizontal velocity changes and vertical acceleration is generally variable, so the constant-horizontal-velocity, constant-vertical-acceleration conditions for the symmetric parabolic path do not hold. [1]

Answer: Drag changes both components of motion; at the highest point vᵧ = 0 but aᵧ = −g.

Check: Zero vertical velocity does not imply zero vertical acceleration.

Section C — Lab-style investigation: measuring g

Research question and method [20 suggested marks]

How does fall height affect the fall time of a ball released from rest, and what value of gravitational acceleration can be inferred?

Use a small, dense ball released without a push, with an electronic timer triggered by release and stopped at a contact sensor. Measure the distance travelled by the ball’s centre between release and the stop event. Use a catch tray, keep the drop zone clear and secure the apparatus.

The following are illustrative practice data, not a claim of measurements collected in a particular laboratory. Each height has absolute uncertainty ±0.002 m. Use ±0.003 s as the absolute uncertainty of each mean time for this exercise; repetition does not remove this supplied timing limit.

Raw data and processing columns
h / m	t₁ / s	t₂ / s	t₃ / s	Mean time / s	(Mean time)² / s²
0.200	0.202	0.201	0.203	________	________
0.400	0.286	0.284	0.285	________	________
0.600	0.350	0.349	0.351	________	________
0.800	0.403	0.405	0.404	________	________
1.000	0.450	0.452	0.451	________	________

Identify the independent and dependent variables. [2]
State three controls or procedure choices that improve validity. [3]
Calculate all mean times and their squares. [3]
Derive a linear plotting relationship, state the graph axes and use the first and last processed rows to estimate g. [4]
Use the supplied endpoint uncertainties to estimate a range for g and a conservative uncertainty. [3]
Evaluate one systematic limitation and one random limitation, giving an improvement for each. [3]
State a conclusion and compare the estimated g with 9.81 m/s². [2]

Show investigation answers and processed table

(a) Independent variable: fall height h. [1] Dependent variable: measured fall time. [1]

(b) Accept three valid points: use the same ball; release from rest without a push; use the same timer and stopping event; keep the drop vertical; measure centre displacement consistently; repeat readings at each height. Award one mark per distinct relevant point. [3]

(c) Calculate the arithmetic mean of the three times at each height, then square the mean. One mark for a correct mean-time method, one for all means and one for all squares. [3]

Processed practice data
h / m	Mean time / s	(Mean time)² / s²
0.200	0.202	0.040804
0.400	0.285	0.081225
0.600	0.350	0.122500
0.800	0.404	0.163216
1.000	0.451	0.203401

(d) With zero initial vertical speed and negligible drag:

$h=\frac12gt^2$

Plot h vertically against squared mean time horizontally. The gradient b has units m/s² and equals g/2. Relationship [1]; axes and gradient meaning [1].

$b=\frac{1.000-0.200}{0.203401-0.040804}=4.92014\,\mathrm{m\,s^{-2}}$

$g_{\mathrm{estimate}}=2b=9.84\,\mathrm{m\,s^{-2}}$

Gradient calculation [1]; g [1]. A full investigation should fit all points with error bars rather than rely only on endpoints. Do not force a fit through the origin without checking that the data and uncertainties support it.

(e) A conservative endpoint estimate uses the largest numerator and smallest denominator for the maximum gradient, and vice versa for the minimum:

$g_{\max}=\frac{2[(1.000+0.002)-(0.200-0.002)]}{(0.451-0.003)^2-(0.202+0.003)^2}=10.1\,\mathrm{m\,s^{-2}}$

$g_{\min}=\frac{2[(1.000-0.002)-(0.200+0.002)]}{(0.451+0.003)^2-(0.202-0.003)^2}=9.56\,\mathrm{m\,s^{-2}}$

Before rounding, the range is approximately 9.56–10.13 m/s². Use the larger deviation from the central estimate, approximately 0.29 m/s². Maximum/minimum method [1]; range [1]; uncertainty statement [1].

Endpoint estimate: g = (9.84 ± 0.29) m/s², using a conservative endpoint uncertainty of about 3.0%.

This estimates the effect of the stated endpoint limits; it is not a statistical confidence interval. An error-bar analysis using all readings is preferable.

(f) A systematic limitation is an offset between the release trigger and actual release, or inconsistent height reference points. Calibrate the trigger and measure the ball’s centre displacement between the physical start and stop events. A random limitation is variation in release or detection. Repeat trials, improve alignment and use a consistent release mechanism. One relevant systematic issue [1], one random issue [1] and appropriate improvements [1].

(g) The data support height proportional to time squared, consistent with approximately constant acceleration from rest. [1] The accepted value 9.81 m/s² lies within the endpoint uncertainty range, so these data are consistent with it; they do not establish exact equality. [1]

Mastery checklist

Distinguish distance from displacement and speed from velocity.
Resolve vectors and state directions clearly.
Use motion-graph gradients and signed areas correctly.
Split motion at reversals or changes in acceleration.
Use independent horizontal and vertical equations with a common time.
Retain unrounded values and reject inappropriate quadratic roots.
Explain the limitations of constant-acceleration models when drag is significant.
Process repeated measurements, identify uncertainties and evaluate investigations.

IB Physics HL Book 2025; Papers 1 & 2 Practice 5 Sections65 Lessons54 Weeks

Curriculum

Kinematics HL Paper 1&2

Instructions and data

Equation reminders

Common misconceptions

Section A — Paper 1A-style multiple choice

Multiple choice 1 — Distance and displacement [1 mark]

Multiple choice 2 — Velocity components [1 mark]

Multiple choice 3 — Position–time graph [1 mark]

Multiple choice 4 — Signs of velocity and acceleration [1 mark]

Multiple choice 5 — Signed graph areas [1 mark]

Multiple choice 6 — Acceleration–time graph [1 mark]

Multiple choice 7 — Constant acceleration [1 mark]

Multiple choice 8 — Stopping distance [1 mark]

Multiple choice 9 — Highest point [1 mark]

Multiple choice 10 — Horizontal launch [1 mark]

Multiple choice 11 — Projectile at the apex [1 mark]

Multiple choice 12 — Complementary launch angles [1 mark]

Multiple choice 13 — Signed vertical velocity [1 mark]

Multiple choice 14 — Resultant displacement [1 mark]

Multiple choice 15 — Negative acceleration [1 mark]

Section B — Paper 2-style extended responses

Extended response 1 — Resolving displacement vectors [6 marks]

Extended response 2 — Velocity–time interpretation [6 marks]

Extended response 3 — Vertical launch [7 marks]

Extended response 4 — Horizontal projectile [7 marks]

Extended response 5 — Angled projectile on level ground [7 marks]

Extended response 6 — Challenge: unequal launch and landing heights [7 marks]

Extended response 7 — Piecewise velocity–time graph [8 marks]

Extended response 8 — Piecewise position–time graph [7 marks]

Extended response 9 — Reaction and braking distance [7 marks]

Extended response 10 — Reversal under constant acceleration [7 marks]

Extended response 11 — Challenge: a delayed pursuit [6 marks]

Extended response 12 — An object released from a rising balloon [7 marks]

Extended response 13 — Challenge: clearing a wall [7 marks]

Extended response 14 — Estimating motion with varying acceleration [7 marks]

Extended response 15 — Projectile motion with air resistance [6 marks]

Section C — Lab-style investigation: measuring g

Research question and method [20 suggested marks]

Mastery checklist

Related

Leave A Reply Cancel reply

Company

Links

Support

Recommend

alphy school publishers

Customise Consent Preferences

Log in with your account

Modal title