Physics
For the IB Diploma
Paper 1 Multiple Choice
Worked Solutions
Complete for the 2016-2022 syllabus
PLUS a complete IA guide
and level 7 IA/EE samples
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,Physics for the IB Diploma
Paper 1 Multiple Choice
Worked Solutions
1
,ISBN: 9798589009293
Imprint: Independently published
All rights reserved. No part of this publication may be reproduced, distributed,
or transmitted in any form or by any means, including photocopying, recording,
or other electronic or mechanical methods, without the prior written permission
of the publisher, except in the case of brief quotations embodied in critical
reviews and certain other non-commercial uses permitted by copyright law.
2
, Table of Contents
Introduction……………………………………….…………………………………………………. 4
Worked Solutions:
May 2016……………………………………………………………….……….............. 5
November 2016……………………….…………………………………………..…….. 19
May 2017 TZ1………………………………….………………………………………….. 33
May 2017 TZ2………………………………….………………………………………….. 46
November 2017……………………………….……………………………………..….. 59
May 2018 TZ1………………………………….………………………………………….. 67
May 2018 TZ2………………………………….………………………………………….. 78
November 2018……………………………….……………………………………….... 94
May 2019 TZ1………………………………….………………………………………….. 109
May 2019 TZ2………………………………….………………………………………….. 118
November 2019……………………………….……………………………………….... 131
Paper 1 Guide……………………………………….………………………………………………. 142
Physics IA Guide…………………………………….……………………………………………… 145
Sample Physics IA...................................................................................... 155
Author’s Note......………………………………….……………………………………………… 170
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, Introduction
What is paper 1?
Paper 1 for IB Physics HL is a set of 40 multiple-choice questions, to be completed in 1 hour.
Each question has 4 options, of which only one is correct. All questions are equally weighted,
with one mark for a correct answer and no negative marking for incorrect attempts. From
2016-2019, paper 1 constituted 20% of total marks for HL. In 2021, as there will be no paper
3 (as of 12/2020), paper 1 has an even higher weightage.
Why is paper 1 preparation important?
Paper 1 is widely considered very challenging, even by top candidates. That’s why it’s crucial
for all students to dedicate specific time to tackle paper 1 questions. In addition, there are
specific techniques needed to answer paper 1 (see the tips section on page 142 for a
complete compilation) that are separate to understanding the syllabus.
What’s in this book?
This book is divided into 4 sections:
1. Detailed solutions for every HL paper 1 question (every season and time-zone) from
2016 to 2019 (all released papers of the current syllabus)
2. General paper 1 tips and strategies from the perspective of an examiner
3. IA success strategies (including a thorough checklist) and a sample level-7 IA
4. [Bonus] A sample A-grade EE written by a 45-point student (will be sent by email if
you request for it; you can find our email ID on page 170)
Who is this book for?
We expect most candidates who use our book to be familiar with most IB Physics theory (or
at least familiar with the theory for the questions you’re answering). However, we will always
provide you with the equations you need in the answer itself, and a brief description of what
it does to jog your memory. This book can’t be used in isolation to teach you material from
scratch (remember, you shouldn’t really be attempting past papers till you have covered the
syllabus!).
How to use the multiple-choice worked solutions in this book?
1. Complete the past paper questions you want (this could be a complete paper in
examination conditions, part of a paper, or specific questions to tackle a certain topic)
2. Find the answer section for each question in section 1
3. Check the letter-answer on the left column of the worked solutions
a. If your answer was correct, still skim through our sample solution; we often put
shortcuts and other efficient ways to answer the question that you might have
missed
b. If your answer was incorrect, begin to read through our solution. Try to minimise
the number of steps you read – ideally, you should only read through the answer
until you become ‘unstuck’ and then go back and try the question again.
4
, May 2016
1. B Recall that the volume of a sphere, 𝑉, given radius 𝑟 is:
4
𝑉 = π𝑟 !
3
Similarly, the volume of a cube is 𝑠 ! .
The general formula for calculating percentage uncertainty for an equation 𝑦 = 𝑎" is:
∆𝑦 ∆𝑎
= +𝑛 +
𝑦 𝑎
(In simple terms, you take the uncertainty in 𝑎, multiply by the power and if the result is
negative, make it positive)
From here, we need to know the absolute uncertainty of the radius of the sphere and the
side length of a cube. These are:
𝑢# = 5 ± 0.1cm
𝑢$ = 10 ± 0.2cm
We can now calculate the percentage uncertainty for the sphere:
0.1
3 6 7 × 100 = 6%
5
And the cube:
0.2
3 6 7 × 100 = 6%
10
From here we know that the ratio is
6
=1
6
Which corresponds to B.
2. A As the person jumps, her velocity increases because of the acceleration due to gravity. As
air resistance keeps increasing with speed, and weight remains constant, her velocity begins
to increase at a slower rate, until velocity stops increasing (reaching terminal velocity). After
she opens the parachute, there is a much greater value of air resistance when compared to
the downward weight, and so her velocity begins to decrease until a new terminal velocity
is reached. Therefore, this corresponds to diagram A.
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,3. D The weight of the object is acting downwards, there is friction between the block and
the slope and there is a normal reaction force acting at 90 degrees. The below
diagram shows the scenario:
The component of 𝑚𝑔 that is perpendicular to the plane of the slope is the value of
normal reaction:
𝑚𝑔𝑐𝑜𝑠𝜃
The component of 𝑚𝑔 that is parallel to the plane of the slope is equal to the value
of frictional force which is
𝜇$ 𝑚𝑔𝑠𝑖𝑛𝜃
Equating the two we have
𝑚𝑔𝑐𝑜𝑠𝜃 = 𝜇$ 𝑚𝑔𝑠𝑖𝑛𝜃
Simplifying this, we get
𝑠𝑖𝑛𝜃
𝜇$ =
𝑐𝑜𝑠𝜃
So, the coefficient of static friction is 𝑡𝑎𝑛𝜃 and the normal reaction force is 𝑚𝑔𝑐𝑜𝑠𝜃.
Therefore, the correct answer is option D.
4.A The difference between kinetic energy before and after the collision gives you the loss
in KE. So where 𝑢 is initial velocity and 𝑣 is final velocity, we have:
1
𝑚(𝑢% − 𝑣 % )
2
∆' )∆*
The formula for average force is ∆( which is also ∆( .
Since 𝑢 and 𝑣 are acting in different directions, we get the average force to be
𝑚(𝑣 − (−𝑢))
𝑇
𝑢 is negative of 𝑣 because they act in different directions and we know that the change
in time is T. Therefore, the correct answer is option A.
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,5.D We know that 𝑃 = 𝐹𝑣. From the question, we know that velocity is increasing and,
because acceleration is constant while resistive force is increasing, we know that thrust
force also increases to overcome the friction. Therefore, the correct answer is D.
6.C The formula for work done is 𝐹𝑠𝑐𝑜𝑠𝜃 and 𝐹 = 𝑚𝑎. Therefore, we know that
𝑊 = 𝑚𝑎𝑠
We can assume that the direction of force and distance needs to be the same for
work to be done so 𝜃 = 0. The area under the graph gives the 𝑎 ∙ 𝑠 part of the above
equation which is just adding the area of a rectangle and a triangle which is 400.
Multiplying by the mass with the area gives us 1200J. Therefore, the correct answer is
option C.
7.B From the question we know that mass is 0.6𝑘𝑔, the latent heat of fusion if 200𝐾𝐽/𝑘𝑔
and the time acting is 40𝑠. The energy supplied to the substance is obtained by
𝑄 = 𝑚𝐿
+,+-./
And now to get power, which is 012+ , we have
𝑄 𝑚𝐿
=
∆𝑡 ∆𝑇
So simply Power is 3 KJ if we substitute the appropriate values into the above
equation. Therefore, this is simply 3000J every second which is just 3000W.
Therefore, the correct answer is option B.
8.A An ideal gas has a high temperature, very low pressure, and a very low density. This is
because the model of the ideal gas states that the gas has negligible intermolecular
forces of attraction which only happens at the conditions mentioned previously.
Therefore, the correct answer is option A.
9.B The light passes through two polarizers which are at different angles. When initial
intensity, 𝐼3 , passes through P, we can find the final intensity using
𝐼 = 𝐼4 cos % 𝜃
This is 0 for the polarizer at 0 degrees for polarizer A.
For a value of 𝜃 = 45, the intensity coming from P is half of initial velocity and using
that in the equation
𝐼 = 𝐼4 cos % 𝜃
5
we get 6 (𝐼4 ), thus obtaining the answer. Therefore, the correct answer is option B.
7
, 10.D In a pipe that is open on both sides, there is half of a wave and when there is a pipe
that is open on only one side we see one-fourth of a wave. We know the formula
𝑐 = 𝑓𝜆
For the first pipe, this is
𝑐 = 𝑓(2𝐿)
And the second pipe,
𝑐 = 𝑓(4𝐿′)
These formulae were obtained by comparing the wavelength of the wave to the length
of the pipe by seeing how much of the wave fit into the pipe. We can now see that the
ratio
𝐿 1
=
𝐿7 2
Therefore, the correct answer is option D.
11.D When light moves from an optically rarer medium to optically denser medium we
know that the light slows down and so the ray bends towards the normal. In essence,
the angle with the normal decreases. The speed of the wave increases, and the
frequency remains the same, and for the frequency to remain same we need to
wavelength to decrease as 𝑣 = 𝑓𝜆. Therefore, the correct answer is option D.
12.C The resistance in parallel can be calculated by
1 1 1
= +
𝑅 𝑅1 𝑅2
The resistance is now 2 ohms when we substitute values. Since the ammeter has a
resistance of 1 ohm, the total resistance is 3 ohms. We know that
8 8
𝑉 = 𝐼𝑅 and so 𝑅 = 9
and 𝐼 = :
;
and so, we know that the current through the ammeter is ! = 2. The voltage across
the voltmeter can be calculated by finding the current flowing through one resistor
(1𝐴) and its resistance (4 ohms). Hence, the voltage is 4𝑉. Thus, the answer is C.
13.A For this question, we must use Fleming’s left-hand rule and use the formula
𝐹 = 𝐵𝐼𝐿𝑠𝑖𝑛𝜃
By applying the left-hand rule, we know that the direction of force is the direction
pointing to the one shown in option A and B. Since the angle is 90 degrees, we get
𝐹 = 𝐵𝐼𝐿
Therefore, the correct answer is option A.
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