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Kinematics
Chapter 2 pg. 42
MQ: ​2-9
Chapter 3 pg. 67-68
MQ: 1-12
Newton’s Laws
Chapter 4 pg. 99
MQ: 1-13
Energy, Momentum, Impulse, Collisions, and Work
Chapter 6 pg. 163
MQ: 1-14
Chapter 7 pg.191
MQ: 1-12
Circular Motion and Simple Harmonic Motion
Chapter 5 pg. 131-132
MQ: 1-12
Chapter 11 pg. 321
MQ: 1-4, 6-8
Kepler’s Laws and Gravity
Chapter 5 pg.131-132
MQ: 7-11
Rotational Motion
Chapter 8 pg. 221
MQ: 1-13
Waves and Sound
Chapter pg. 322
MQ: 9, 10, 12-15
Chapter pg. 353
MQ: 1, 2, 6-8, 10, 12, 13
Electricity and Circuits
Chapter 16 pg. 468
MQ: 1-3,13
Chapter 18 pg.520
MQ: 1, 2, 4, 5
Chapter 19 pg.550
MQ: 1-3
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Chapter 2 pg. 42
MQ: ​2-9
2. D b/c if you imagine a car going in the negative direction and if you are slowing down then the
acceleration is in the direction that your velocity is changing. If you are slowing down then you
are accelerating the opposite direction.
3. D b/c if you apply the kinematic equations then you see that the car comes to a stop in 5
seconds, but you realize that the acceleration is still present so you will now accelerate in the
negative direction.
4. C b/c the gravitational acceleration is on a object on the surface of the earth is always
downwards and at the highest point of any vertical path velocity is 0 m/s.
5. A b/c the gravitational acceleration is always the value of 9.8 m/s downwards so both objects
have the same acceleration aka change in velocity.
6. B b/c 4 km out 8 km you drove at 30 km/h and the other 4 km you drove 50 km/h, so half of
the trip you drove at 30 km/h and the other half 50 km/h so the average is 40 km/h.
7. C b/c the gravitational acceleration is always the value of 9.8 m/s downwards so both objects
have the same acceleration aka change in velocity.
8. B and C b/c in the other equations the fact that the velocity is downwards (negative) plays a
factor in the calculations. In B the equations is set up properly and in C the velocity not being
negative plays no role. It can’t be A because the initial velocity must be -20 in the equation not
20.
9. A b/c the car is accelerating in a direction and and it is also traveling in the same direction, so
you conclude the graph will get steeper as time increases b/c the slope represents the velocity
and none of the graph and none of the graphs match other than A.
Chapter 3 pg. 67-68
MQ: 1-12
1) C, b/c A, B, D, E, are impossible because the maximum is when the vectors are in
the same direction which would be 20+40=60. The minimum would be in opposite
directions which would be 20 and the only one in the range 20-60 is C
2) D, b/c if you have a component that is broken into parts y and x it is always less than
because it is the original vector times sin or cos theta which is always less than 1. If the
component vector is in the same direction as the original than it can be equal to the
original vector. It can also be greater than; for example, net force is a vector but a
component could be FA in one direction that increases it granted a FB in the other
direction that decreases it.
3) B, b/c you have to create an isosceles triangle for each leg to equal 100m and thus
angle is 120
4) A, the force of gravity acts as soon as it leaves the barrel making it answer A
5) B, Acceleration is always down towards the earth (b) because gravity goes in that
direction
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6) B, both balls have the same height so their vertical velocity will be the same at the
ground; however, the thrown ball also has a horizontal velocity making the overall
velocity (once you combine components) larger than that of just the vertical
7) C, when you throw a baseball up, it continues to have the same horizontal velocity of
the train and you at 90 km/hr and thus when it comes down it will stay with you in the air
and land in your hand
8) D, Recall that time in the air is only contingent on the height that it reaches granted
that the force acting on all of them is the same because it is the same gravitational field
on all of them. Thus, if they all reach the same height, using kinematics, time is
(2d/a)^0.5.
9) C, At every point the horizontal component is the same of the ball and only the
vertical component changes due to the force of gravity. Thus, B can’t be true and gravity
is always there so A can’t be true. Thus, we look to minimize the vertical component
which is at a minimum at the highest point and thus velocity is at the minimum C.
10) B, Even if the monkey falls, the hunter was aiming horizontally meaning the vertical
component would be the same as the monkey because the only force acting is gravity
and thus it would hit it.
11) B, E, Acceleration is always constant in free fall B is not true. Velocity is still positive
because the horizontal component never goes away so E can’t be true.
12) A, We have to remember that we cannot add directly because both cars are not
moving in the same direction. Thus, using pythagorean theorem we find that the relative
motion is less than 20 m/s
Chapter 4 pg. 99
MQ: 1-13
1) A, Net forces only occur when there is something acting on it that is unbalanced. We
know there is no friction in the problem and the only forces acting is Normal Force and
Gravity both of which are balanced and thus net force is 0.
2) A, B, D, if you apply 400 N to the car, Newton’s third law tells us the force the car
applies on you is also 400N. If no movement occurs, the friction forces on both the car
and yourself must be equal and 400 N.
3) D, Looking above, if there is movement, the friction forces must be unequal as those
are the only forces providing for horizontal movement. Thus that on Matt must be more
than that on the Truck.
4) D, We need to remeber components in this problem, and as the rope becomes more
horizontal, the vertical component to hold up the gravity force on the sack decreases
and thus overall tension and the Force must increase. (thus, can’t be A or B). The rope
can never be fully straight as there needs to be some angle/sag to keep the sack up
and thus it is D not C
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5) C, Newton's third law is in application here because where you apply a force on the
water through the paddle that must be applied back by the water on the paddle which is
what causes it to move. Free body diagrams would help here.
6) C, Remember your Fnet equation where Fn is his ‘apparent mass.’ Fn is equal to
m(a+g) and thus when accel is positive (upward) the apparent mass is more.
Conceptually, when you accelerate upward, it is as if the Elevator is pushing into you
increasing the Fn.
7) C, When skiing down a hill, the Fn is equal to mg cos angle of the hill, but the weight
is still mg so it has to be more.
8) B, The only force when the ball is in motion is that of gravity as it is in free fall
9) C, Again, if Net Force is 0, the object will either stay at rest OR continue in motion at
that speed per Newton’s first law. Thus the answer is C as the forces are now balanced.
10) D, This is an action reaction pair per Newton’s Third Law. The force you apply on
the box is the force the box applies on you.
11) B, Remember first that maximum static friction is 50 times 0.5 which is 25 so it won’t
move because 20<25. Then know that static friction opposes the direction of the force
applied so we can narrow to B and D and then we know that static friction isn’t the max
friction but rather equal to the force applied. If it was 25, it would mean that if I push a
box it would magically start moving in the opposite direction because there would be a
net force of 5N.
12) D, If the slope is 90 and only 90, the component of gravity is mgcos90 which would
be 0.
13) A, If you are not in the same direction you cannot add forces and thus when you do
Pythagorean because the forces are components of the overall, you would get less than
950.
Chapter 6 pg. 163
MQ: 1-14
1) B, Remember Work=Force times Distance so even if you apply a force, only when it
starts moving do you actually do work.
2) C, Again remember that Work= Force times Distance, given that the distance and
force are in the same direction. If they are perpendicular there is not a component of the
distance that is in the same direction and thus it is 0 Work done.
3) C, Remember KE= ½ m v^2. And doubling v quadruples KE
4) D, Remember kinematics where d equals v^2 and thus doubling quadruples braking
distance
5) C, Remember that energy is conserved and there is no input of energy in this
situation. Thus, GPE of the ball is changed to elastic potential on the trampoline and
back to GPE but some is probably lost to friction and heat so a little less makes sense.
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6) E, Again remember the conservation of energy equation where energy is not created
or destroyed merely changed forms. Thus, at all points KE + PE will be equal and E is
correct.
7) B, Energy is change in Kinetic Energy. If you do ½ mv^2 at both 0, 30, and 60, the
difference between the two will be greater for 30-60 because you end up squaring the
velocities.
8) D, HorsePower is Work/Time as is the definition of Power. Thus, the answer is D.
9) E, Instead of a kinematics approach, use conservation of energy where both balls are
given the same PE and KE at the beginning and thus energy in the entire system will be
the same. Thus, speed at the end has to be equal as at the ground, PE for both is the
same (0) and time must also be equal because they have the same velocity.
10) E, Use conservation of energy where PE= KE and v=(2gh)^0.5. Thus the speed is
only contingent on height of the initial where C and D will be the same.
11) C, If there is friction we want to minimize distance so that the friction force acts for
less distance. thus, C has less distance and will lose less to friction and have the
highest velocity at the bottom
12) C, PE increases because you are going up so mgh increases and PE increases. KE
remains the same because the man is applying a force in order to keep the speed
constant and thus ½ m v^2.
13) A, Friction force always opposes direction of motion and thus is pointing up the
ramp. All forces act for the same distance so Work is now only contingent on which
force is the largest. Friction is the largest force because it has to equal Fa of the man
and Gravity and thus it is larger than the other forces. Fn does no work because no
distance is moved vertically.
14) A, PE= mgh and thus it cannot decrease as h increases.
Chapter 7 pg.191
MQ: 1-12
1) D, Remember in all collisions in Physics 1 total initial momentum is equal to total final
momentum, meaning that if one object loses 100 kg*m/s of momentum than the other
will gain 100 kg*m/s.
2) B, the speed will decrease because this can be thought of as an inelastic collision
where the objects stick together after colliding so total mass goes up and momentum
stays the same velocity has to go down to compensate.
3) C, no difference because the entire equation is multiplied by a factor of 2 and that
doesn't change the velocity
4) A, away because the consider the 3rd law that when you throw the wrench away you
experience a force in the opposite direction and start moving at a constant speed
toward the space station.
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5) A, Only momentum is conserved because kinetic energy is only conserved in specific
elastic collisions and this is inelastic collision so kinetic energy is not conserved.
Momentum is conserved in all collisions.
6) A, because the ball has a vertical velocity after the collision and that means that the
change in velocity is vi + vf and the produces a higher change in momentum than just vi
by itself.
7) A, again as the stated by the previous problem the bouncy ball has a higher impulse
and that results in more force on the door.
8) C, because b is less than c because
9) D, We both know that F=ma and thus the heavy truck will have a lower acceleration
and if v=at, and the force is applied for the same time on both, the truck will end up at a
lower velocity while the car will have a greater one. The crucial thing to note here
however is that momentum is equal to mv and thus although the velocity is lower it will
have a larger mass and thus the same momentum (p) as a greater velocity but lighter
mass. This is also evident where if you replace the variable v in momentum in terms of
F, t, and m, would end up getting p=Ft for both which makes it equal. The difference
with KE is that it is v^2 and thus the lighter and faster car would have more.
10) E, If you push with the same force for the same distance, final velocity will be the
same due to distance being proportional to v^2. This would indicate that Kinetic Energy
is also equal for both as it is equal to ½ mv^2. However is v is equal, mass is more for
the truck and thus momentum is more.
11) C, This is the tricky question where because the car is coasting at a constant speed
Newton’s First Law tells us an object will continue in motion unless acted upon by an
external force. Thus, milk leaking out does not mean that there is magically a force
acting on the car and it will remain at the same speed.
12) C, So here we are using the Impulse Momentum theorem. We set F*t equal to
m(vf-vi). It is important to note that the direction of velocity matters so that in situation
one (inelastic) the velocity goes from 5 m/s to 0 and the change is 5. In situation two
(elastic), the velocity goes from 5 to -5 and the change is actually 10 not 0. Thus, on
face value the momentum change is more for scenario 2. Then we have to divide by
delta t to get F where t is larger for scenario one because the ball sticks to the bowling
ball which would further reduce F. If we know F is larger in scenario two we can
conclude it will swing further. Conceptually, in scenario one it is inelastic and thus only
momentum is conserved but in two, kinetic energy is also conserved and thus would
have more.
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Chapter 5 pg. 131-132
MQ: 1-12
1) B, We know that centrifugal force always points inside and thus it cannot be A
however we know that their is a normal force back in when you press against something
thus B is correct.
2) E, If the circular motion is uniform there is a centripetal acceleration and tangential
velocity where we also know that Net Force is ALWAYS in the same direction as the
acceleration thus it is E
3) C, In circular motion, which would be the ping pong ball in the tube, the velocity is
always tangential to motion; thus it would exit tangentially and leave in the direction of
C. Because we also have to remember that once it leaves there are no forces acting on
it and there is no reason for it to curve in any direction aka not A, B, D, or E.
4) D, If it is a circular track we know there is centripetal acceleration and thus net force
also points in as established in 1. Thus is it D.
5) B, Here we have to set up a Fnet equation where Fc equals Ft minus Fg. Thus, Ft
equals Fc plus Fg only at the bottom and this is the max as you are adding both.
6) A, We know there is a Centripetal Force towards the center, which means it can’t be
E. We know there is a Fg downwards and we know that Static Friction opposes this
motion which would be straight upwards which is A.
7) D, Remember what Fabo says: satellites are always in free fall, crashing towards the
planet but just missing. This would imply D right away. However, upon deeper analysis
we know it is not A because it has Net Force inwards, we know it can’t be B because
there is a gravitational force mutually between the two and we know is can’t be C
because it is irrelevant.
8) F, Obviously we know that per Newton’s third law, the force of the Moon on the Earth
is equal to the force of the Earth on the Moon. Then, we know that because F=ma and
the F is equal for both, the a must be the most the Moon because it has a smaller mass.
9) C, We know it can’t be D because the Earth still exerts a force on the spacemen as
per the Fg equation equals Gm1m2/r^2. We know it can’t be A by the same logic, there
still is a gravitational force. We know it can’t be E because it is irrelevant once again,
and we know that it can’t be B because the Centrifugal Force DOESN’T EXIST. Thus, it
is C but even still, refer to question 7 where we establish that satellites are in free fall.
10) B, Here we must set up equations where Fc is equal to Fg. Fc is equal to ma which
is equal to mv^2/r. Fg is equal to Gm1m2/r^2 and when we set them equal we are able
to observe m1 and m cancel out, thus leaving us with the relationship that v is equal to
sqrt(Gm2/r). Thus, here we can see that v is NOT contingent on the mass of the satellite
and speed is the same for both.
11) C, Remember all the way back to Kinematics where releasing an object on a train
will keep the same horizontal velocity. Likewise, releasing a payload doesn’t magically
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disappear the velocity it had while on the spacecraft, and with the same velocity and
radius of orbit it would continue next to the spacecraft.
12) C, If the penny is still spinning there is only centripetal acceleration inwards which
would be C.
Chapter 11 pg. 321
MQ: 1-4,6-8
1) E, Acceleration is contingent on force on the spring as F=ma and F also equals -kx
thus acceleration is 0 only at the equilibrium position where x equals 0. However, we
know that velocity is at the maximum at the equilibrium position and 0 at the edges thus
it has to be E.
2) A, C, D; We know C can be true as velocity is max at equilibrium and acceleration is
0 here. We know B is false as proven in question 1. We know A is true as this occurs at
the ends of oscillation (the amplitude) where velocity is 0 (because it stops and starts
going back to the equilibrium) whereas force is the maximum as it equals -kx and thus
acceleration is a maximum. We know D is true as it occurs anywhere between
situations A and C.
3) C, We know that T=2*pi*(sqrt(m/k)) and thus if we quadruple the mass we double the
period.
4) B, Again refer to the above equation where decreasing mass decreases period.
6) E, Remember that the equation for Period of a pendulum is T=2*pi*(sqrt(l/g)) and
thus is not affected by the Amplitude that you pull it. Moreover, conceptually speaking,
even if you increase the distance you pull it and you increase distance the pendulum
has to travel, you also increase the restoring force acting on it (the component of
gravity) and thus it can travel the distance in the same time as previously.
7) C, Refer to the period of pendulum equation where T=2*pi*(sqrt(l/g)) and mass does
not factor in thus C is correct. However, conceptually again, more mass might mean
that it’s harder to pull but it also means that mg is more and thus can pull it in the same
time.
8) A, Only A works here because C and D are irrelevant to the period as shown above.
We know it can’t be B as lengthening increases period and thus A works to decrease it.
Chapter 8 pg. 221
MQ: 1-13
1) C, 2 times the radius means 2 times the distance (using the circumference), and
double the distance means double the velocity.
2) B, because when you use Θ=.5𝜶t^2 the factor by which Θ(t) and Θ(.5t) is .25.
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3) B, because the car will travel a larger distance than it would if it had normal tires
because Θ=d*r and if r is larger Θ is larger. So if the car travels more than the
estimation than the speed will be read too slow.
4) C, it has the largest lever arm out of the other forces.
5) C, E, F, when you do a force diagram and analyze the torques C, E, F will be
produced.
6) B, if you reference the equations for each shape than it will yield that that the hollow
one will have an higher moment of inertia.
7) B, Look at the Moment of Inertia’s where if it is concentrated at the edge it is MR^2
versus uniform being ½ MR^2. Therefore, WITH ALL ELSE EQUAL, Energy will be
double.
8) B, Energy is fixed as per conservation of energy law. Thus, if you use 1000J, if it is
spinning some is changed to both rotational and kinetic whereas without spinning it is
just kinetic and will go faster.
9) B, The hoop will go further because we are looking for what has the larger moment of
inertia. If I have two objects with the same angular speed, the one with the larger
moment will be harder to stop as the only force acting on them on the ramp is gravity
and thus the hoop goes further. (⅖ MR^2 versus MR^2)
10) A, this question relies on Conservation of Angular Momentum where because the
problem states there is no external torque we can apply it. Thus, because Iw equals Iw
and by bringing in the mass we reduce the radius and thus moment of inertia, the
angular momentum has to increase.
11) A, although on first thought we are tempted to think angular momentum times radius
is tangential velocity and thus conclude it goes down because radius decreases, we
need to take a closer look at the problem. I is equal to MR^2, and after plugging in
w=v/r, we see that we end up with MVR on both sides of the conservation of angular
momentum equation. Thus, reducing R increases V again, albeit by a lesser amount
than it increases w (one is r squared the other is just r).
12) A, Remember that L=L and if we move everyone to the radius of the earth, we
would increase the moment of inertia and thus have to reduce angular speed.
13) C, This is a very important question to note as there is NOT AN INCREASE in
angular speed as many people think. We think conservation of angular momentum but
we forget that when we release the bricks they continue on with the same velocity they
had when they were in our hands just tangentially so instead of the energy being given
to us and ‘conserved’ it still is conserved just linearly as the bricks would fly out of our
hands and land a certain distance away. This happens because we forget that we have
to investigate what happens to the bricks when we release them, not just assume the
energy goes to us.
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Chapter 11pg. 322
MQ: 9,10,12-15
* A note on waves: A useful metaphor for thinking about them conceptually is
thinking about them as ghosts of different shapes (frequencies/speeds, etc.) and
sizes (amplitudes). So when they come near each other they react in weird ways
but after they pass through each other they go back to normal. Wave interactions
don’t change any of fundamental properties of either wave assuming no loss of
energy.
9) E, b/c when you have a transverse wave motion on a cord each infinitely small piece
on the cord that is experiencing transverse motion is put into simple harmonic motion,
and just by playing out the scenario in you head you can deduce that both B and C must
be true b/c the for the small piece of the cord to remain attached to the rest of the cord (
and not fly off) it has to have the same frequency and amplitude. A is not true b/c it is
saying that horizontal speed has to be equal to vertical and that makes no sense.
10) A, remember when waves come near each other they interfere (meaning they
cancel or add) but then they keep going past each other. ​AFTER​ COMPLETELY
DESTRUCTIVE INTERFERENCE, THEY DON’T DISAPPEAR. (use ghost metaphor
from above to make sure you fully understand that)
12) D b/c imagine you are the particle at point B, the wave that is about to pass through
you is at a higher amplitude and you to get to that higher amplitude for the wave to keep
traveling.
13) D, b/c go back to the ghost metaphor above and number 10 for clarification.
14) C, b/c if you imagine the creation of the one wavelength of the wave is done when
the student has moved the slinky up, down, down, and then back up so basically gone
through all of the altitudes of that one wave length. So the faster the creation the smaller
the wavelength.
15) A, for the purposes of AP Physics 1 waves only transport energy. Though all of
modern physics relies on waves transporting both matter and energy (you do not not
need to know this). Whenever you are asked about waves transporting things it is only
energy.
Chapter 12 pg. 353
MQ: 1,2,6-8,10,12,13
1) A, Speed of sound is a function of temperature (v = 331 m/s + (0.6 m/s℃)T), and as
temperature increases, so does speed.
2) D, Sound waves are characterized either as the longitudinal displacement of air
molecules or the pressure differences that cause the displacement.
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6) E, For a string or open tube the lowest vibration mode is half of a wavelength.. For a
closed tube the lowest vibration mode is a quarter of a wavelength.
7) E, The frequency of the sound remains the same (emitted frequency, dependent on
the source, does not change). However, the speed of sound is greater in water than in
air, and given that frequency remains the same, if speed changes, wavelength also
changes.
8) E, We know that the frequency stays constant when we switch mediums of a wave
however going from more dense (metal string) to less dense (air) would decrease our
speed and thus our wavelength because speed = wavelength time frequency and
frequency stays constant.
10) A, b/c now the wave has to end in the middle so it doesn't have to travel to the
bottom, therefore wavelength gets cut in half.
12) C, b/c you are moving at the same velocity there is no doppler effect as the equation
suggest there is n​o change in frequency.
13) E, b/c on a guitar string of length L, fundamental frequency: wavelength equals 2L
and not L. This is because it is closed on both ends so L=½ lambda.
Chapter 16 pg. 468
MQ: 1-3,13
1) A, Because the Q​1​ and Q​2​ have opposite charges, they are attracted to each other.
Q​1​ is pulled towards Q​2​, whose position is x = 1.0 m. So, Q​1​ (which is at the origin) must
travel in a positive direction.
2) D, If Q​1 is
instead located at 1.0 m, it is still attracted to Q​2​, which would be at the
​
origin. So, Q​1​ must travel in the negative direction.
3) E, Their attraction is equal. They share a common electrostatic force of attraction,
regardless of which one has a greater charge or mass.
13) E, The ball’s charge can be negative because opposites attract, so the positively
charged rod would be attracted to the negative ball. But, it could also be neutral
because induction would allow the positively charged rod to attract the negative charges
of the ball, then attract the ball.
Chapter 18 pg.520
MQ: 1,2,4, 5
1) C, Bulb brightness is based on power. Reversing a battery does not affect current,
resistance, voltage, or power.
2) C, Current is constant through a series circuit. Batteries don’t create charge, nor do
lightbulbs absorb charge.
4) E, Charge is conserved throughout a circuit. The charge from the electrons is not
used when it goes through the bulb.
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5) E, Again, in a series circuit, current is always the same at all points because all
current has to flow through the circuit.
Chapter 19 pg.550
MQ: 1-3
1) A, C, In a series circuit, the current has only one path and it has to go through all
resistors. This is the case in A and C - in B and D, the current branches off in two
different paths.
2) D, The current must travel through R​2​, but then it branches off into R​1​ and R​3​. So, R​1
and R​2​ are in parallel because current travels through either one or the other.
3) C, In a series current, the current flowing through any given point of a circuit is the
same as the current flowing through any other given point. Thus, even if the resistors
are swapped, current stays the same.