Quiz Samples for Chapter 4 General Physics I Name: Department:...

General Physics IQuiz Samples for Chapter 4

Motion in Two and Three DimensionsMarch 16, 2020

Name: Department: Student ID #:

Notice

� +2 (−1) points per correct (incorrect) answer.

� No penalty for an unanswered question.

� Fill the blank ( ) with � (8) if the statement iscorrect (incorrect).

� Textbook: Walker, Halliday, Resnick, Principlesof Physics, Tenth Edition, John Wiley & Sons(2014).

4-1 Position and Displacement

1. (�) The position of a particle in three dimensionscan be represented by a linear combination(sum of elements after multiplied by numbers) ofthree basis vectors i, j, k as

r = ix+ j y + k z.

Here, the coefficients x, y, z of the basis vectors i,j, k are the components (coordinates) along thex, y, z axes, respectively.

2. (�)

The displacement of a particle from a position r1

to another position r2 in three dimensions can berepresented by a linear combination (sum of

elements after multiplied by numbers) of threebasis vectors i, j, k as

∆r = r2 − r1

= i (x2 − x1) + j (y2 − y1) + k (z2 − z1).

Here, the coefficients x, y, z of the basis vectors i,j, k are the components (coordinates) along thex, y, z axes, respectively.

4-2 Average Velocity and InstantaneousVelocity

1. (�) The average velocity during the timeinterval [t1, t2 = t1 + ∆t] and the instantaneousvelocity at time t of a particle in three dimensionscan be represented by

vaverage =∆r

∆t

= i∆x

∆t+ j

∆y

∆t+ k

∆z

∆t,

v = i vx + j vy + k vz

= lim∆t→0

∆r

∆t

=dr

dt

= idx

dt+ j

dy

dt+ k

dz

dt= r

= i x+ j y + k z.

The components of the instantaneous velocity are

vx =dx

dt= x,

vy =dy

dt= y,

vz =dz

dt= z.

©2020 KPOPEEE All rights reserved. Korea University of 8



4-3 Average Acceleration and InstantaneousAcceleration

1. (�) The average acceleration during the timeinterval [t1, t2 = t1 + ∆t] and the instantaneousacceleration (acceleration) at time t of aparticle in three dimensions can be represented by

aaverage =∆v

∆t

= i∆vx∆t

+ j∆vy∆t

+ k∆vz∆t

,

a = i ax + j ay + k az

= lim∆t→0

∆v

∆t

=dv

dt

= idvxdt

+ jdvydt

+ kdvzdt

= v

= i vx + j vy + k vz.

The acceleration can also be expressed as thesecond-order time derivative of the position vectoras

a =d2x

dt2

= id2x

dt2+ j

d2y

dt2+ k

d2z

dt2

= x

= i x+ j y + k z.

The components of the instantaneous accelerationare

ax =dvxdt

= vx =d2x

dt2= x,

ay =dvydt

= vy =d2y

dt2= y,

az =dvzdt

= vz =d2z

dt2= z.

4-4 Projectile Motion

1. (�) The projectile is a particle in motion in threedimensions under the gravitational acceleration gneglecting any other external forces except for theEarth’s constant gravitational force near thesurface.

2. (�)

A projectile motion can always be expressed ina two dimensional plane whose basis vectors can bechosen as the vertical one and a horizontal one onthe plane.

3. (�) The horizontal motion of a projectile isindependent of the vertical motion. The horizontalmotion is free of any acceleration and, therefore,the horizontal component of the velocity isinvariant. The vertical motion is the same as thefree-fall acceleration that has constant accelerationg = −gk, −k is the unit vertical vector headed forthe center of mass of the Earth.

4. Consider a projectile motion of a particle on the xyplane with the initial position and velocity,

r0 = ix0 + j y0,

v0 = i v0x + j v0y.

Here, i and j are, respectively, the unit vectorsalong the horizontal and vertical directions. Thegravitational acceleration is given by

g = −gj.

(a) (�) If the angle between the horizontal axisand the initial velocity is θ0, where0 < θ0 <

π2 . Then

v0x = v0 cos θ0,

v0y = v0 sin θ0.




(b) (�) The horizontal component of thedisplacement of the particle is

x− x0 = v0xt = v0t cos θ0.

Therefore,

t =x− x0

v0x=

x− x0

v0 cos θ0. (1)

(c) (�) The displacement of the vertical motion is

y − y0 = v0yt−1

2gt2

= v0t sin θ0 −1

2gt2.

(d) (�) Time t can be eliminated by making useof Eq.(1) as

y − y0 = v0yt−1

2gt2

= v0y

(x− x0

v0x

)− 1

2g

(x− x0

v0x

)2

.

(e) (�) We set x0 = y0 = 0 and make use of theidentity,

v0y

v0x=

sin θ0

cos θ0= tan θ0,

We find that

y = x tan θ0 −gx2

2v20 cos2 θ0

.

(f) The horizontal range R is the totaltranslation along the horizontal directionduring the whole flight of the projectile tocome back to the ground, y = 0. Then we findthat

R =v2

0

gsin 2θ0,

where we have made use of the identity

2 sin θ0 cos θ0 = sin 2θ0.

4-5 Uniform Circular Motion

1. (�)

In two dimensions, the position of a point on acircle of radius r whose center is the origin can beparametrized by

r = r(i cos θ + j sin θ),

where θ is the angle rotated from the x axiscounterclockwise. Suppose that the angle is afunction of time as

θ = ωt,

where the angular frequency ω is constant:

θ =dθ

dt= ω.

We say that this particle is making a uniformcircular motion.

(a) (�) The chain rule of differentiation can beemployed to evaluate the following derivatives:

d

dtcos θ =

dθ

dt

d cos θ

dθ= −ω sin θ,

d

dtsin θ =

dθ

dt

d sin θ

dθ= ω cos θ.




(b) (�)

We define

r = i cos θ + j sin θ.

� = −i sin θ + j cos θ.

The vectors r and � are unit vectors:

r2 = r · r = 1,

�2 = � · � = 1.

The vectors r and � are perpendicular to eachother:

r · � = 0.

We call r the unit radial vector. We call �the unit tangent vector.

(c) (�) The unit tangent vector is tangent to thecircle at a point on the circle.

(d) (�) The unit radial vector is normal to thecircle at a point on the circle.

(e) (�) The velocity of the particle at t is given by

v = r

= rω (−i sin θ + j cos θ)

= rω �.

(f) (�) The acceleration of the particle at t isgiven by

a = v

= −rω2(i cos θ + j sin θ)

= −rω2r.

2. The position, velocity, and acceleration of aparticle under a uniform circular motion can beexpressed as

r = r r,

v = rω �,

a = −rω2 r.

(a) (�) Let T be the minimum time to return tothe same position. It is called the period ofthe uniform circular motion. The period andthe angular frequency ω are related as

ωT = 2π → ω =2π

T, T =

2π

ω.

(b) (�) The average speed of the particle in anytime interval is the same as the magnitude ofthe instantaneous velocity. This value isconstant and

saverage = |v| = rω =2πr

T.

(c) (�) The instantaneous velocity is alwaystangent to the circle and perpendicular to theunit radial vector r.

(d) (�) The instantaneous acceleration is alwaystoward the center that is antiparallel to theunit radial vector r.




(e) (�) The position, velocity, and acceleration ofa particle under a uniform circular motion canbe expressed as

r =v

ωr,

v = v �,

a = −v2

rr.

3. The latitude of Seoul is approximately θ = 37◦. Weassume that the Earth is a sphere with radius Rand rotates about its axis along the straight lineconnecting the north and south poles with theangular frequency ω:

R = 6.371× 106 m,

ω = 1 revolution/day.

(a) (�) The perpendicular distance from therotation axis to Seoul is about

r = R cos θ ≈ 5.1 × 106 m.

(b) (�) The angular frequency is

ω =2π

revolution× revolution

day× day

24× 60× 60s

=π

43200rad/s.

(c) (�) The magnitude of velocity of Seoul is

v = rω = 3.7 × 102 m/s.

(d) (�) The magnitude of centrifugal accelerationof Seoul is

a = rω2 = 2.7× 10−2 m/s2.

(e) (�) The fraction of the magnitude of thecentrifugal acceleration with respect to taht ofthe gravitational acceleration g is

a

g= 2.8× 10−3.

4-7 Relative Motion

1. (�)

Suppose that there are two observers A and Bwhose positions are given, respectively, by rA andrB in a frame of reference S. The position vectorof a particle at point P is rP in the frame S. LetSA and SB be the reference frames in which A andB are at rest, respectively. Figure shows thepositions of the particle, r and rP/A in the framesof S and SA, respectively.

(a) (�) The position, velocity, and acceleration ofthe particle measured in the frame S are givenby

r = ix+ j y + k z,

v = r,

a = v = x.

(b) (�) The position, velocity, and acceleration ofthe particle measured in the frame SA are




given by

rP/A = rP − rA,

vP/A = vP − vA

= rP − rA,

aP/A = aP − aA

= vP − vA

= rP − rA.

(c) (�) If vA and vB are constant, then theacceleration is invariant:

aP = aP/A = aP/B.

2. Particle A is moving along the x axis and B isalong the y axis as

rA = i vx0t, (2)

rB = j

(vy0t−

1

2gt2), (3)

where vx0 and vy0 are constants.

(a) (�) The position, velocity, and acceleration ofA relative to B are

rA/B = i vx0t+ j

(1

2gt2 − vy0t

),

vA/B = i vx0 + j (gt− vy0) ,

aA/B = j g.

(b) (�) The position, velocity, and acceleration ofB relative to A are

rB/A = −rA/B,vB/A = −vA/B,aB/A = −aA/B.

These are the consequences of the relativity.The magnitude is the same and the directionis opposite.

(c) (�)

The motion of A relative to an observer at theorigin is in one dimension along the x axis.

(d) (�)

The motion of B relative to an observer at theorigin is in one dimension along the y axis.

(e) (�)




The motion of A relative to B is on the xyplane and the trajectory is a parabola.

(f) (�)

The motion of B relative to A is on the xyplane and the trajectory is a parabola.

3. [optional: This problem does not appear inthe quiz. Just for advanced students.] Thefollowing is the basic strategy for theMichelson-Morley experiment to measure the speedof light and confirm the existence of the medium oflight called aether. This experiment actuallydisproved the existence of the aether. Considerthe following assumptions:

� The light travels with the speed c in the restframe S? of aether (the reference frame inwhich aether is at rest).

� Time t is invariant under the transformationS? ↔ SE .

� The speed of light depends on the frame ofreference as any other particles.

� In SE that is fixed on the Earth surface, theaether is flowing along the x axis with theconstant velocity

V = V i.

� In SE we let a beam of light À travel alongthe x axis from the origin O to A (x = L)and, then, bounce back to O as O → A→ O.

� In SE we let another beam of light `B travelalong the y axis from the origin O to B

(y = L) and, then, bounce back to O asO → B → O.

Note that these assumptions were proved to bewrong.

(a) (�) Any velocity v? measured in S? can beconverted to the corresponding value vmeasured in SE as

v = v? − (−V i).

(b) (�) The velocity of À in each part of thetravel measured in S? is

v?O→A = c i,

v?A→O = −c i.

(c) (�) The velocity of À in SE in each part ofthe travel is given by

vO→A = (c+ V )i,

vA→O = (−c+ V )i.

The time elapsed for À in each part of thetravel measured in SE is

∆tO→A =L

c+ V,

∆tA→O =L

c− V.

Thus the total time elapsed for the wholetravel O → A→ O is

∆tO→A→O =L

c+ V+

L

c− V=

2L

c

(1− V 2

c2

) .(d) (�) In S?, `B travels from O → B with the

distance c∆tO→B. This distance is equal tothe hypotenuse of a right triangle whosehorizontal side is V∆tO→B and the verticalside is L. According to the Pythagorastheorem, we find that

(V∆tO→B)2 + L2 = (c∆tO→B)2,

→ ∆tO→B =L√

c2 − V 2.

In a similar manner,

(V∆tB→O)2 + L2 = (c∆tB→O)2,

→ ∆tB→O =L√

c2 − V 2= ∆tO→B.




(e) (�) The velocity of `B in each part of thetravel measured in S? is

v?O→B = −V i +√c2 − V 2j,

v?B→O = −V i−√c2 − V 2j.

(f) (�) The velocity of `B in SE in each part ofthe travel is given by

vO→B = +√c2 − V 2j,

vB→O = −√c2 − V 2j.

The time elapsed for `B in each part of thetravel measured in SE is

∆tO→B =L√

c2 − V 2,

∆tB→O =L√

c2 − V 2.

Thus the total time elapsed for the wholetravel O → B → O is

∆tO→B→O =2L√

c2 − V 2=

2L

c

√1− V 2

c2

.

(g) (�) Note that c� V . Thus `A returns to Olater than `B.

∆tO→A→O −∆tO→B→O

=2L

c

√1− V 2

c2

1√1− V 2

c2

− 1

> 0.

(h) (�) However, Michelson and Morley havefailed to detect the time difference. It wasfinally proved that their assumptions werewrong. The Nobel Prize in Physics 1907 wasawarded to Albert Abraham Michelson “forhis optical precision instruments and thespectroscopic and metrological investigationscarried out with their aid.” This failure isactually the confirmation of Einstein’s specialtheory of relativity. For more details of theMichelson-Morley experiment, refer toWikipedia.


https://en.wikipedia.org/wiki/Michelson-Morley__experiment

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