chabay ch 1 solutions

1

1.X.1

(a) This smooth sailing ship’s motion constitutes motion with a constant velocity.

(b) Orbital motion does not constitute motion with a constant velocity; Moon’s direction in space continually changes.

(c) Tennis balls, and other projectiles, do not move with constant velocity.

(d) A velocity can have a magnitude of zero, so this is indeed motion with a constant velocity. The important thing is thatthe velocity isn’t changing.

(e) A person on a Ferris wheel experiences a continual change in direction regardless of the Ferris wheel’s speed, andregardless of whether the wheel is turning at a constant rate or a variable rate.

1.X.2

(a) The ball’s direction changes, so that’s evidence of a significant interaction.

(b) The baseball’s direction changed, so that’s evidence of a significant interaction.

(c) The satellite’s direction in space continually changes, so that’s evidence of a significant interaction.

(d) There is no evidence of significant interactions in this case.

(e) The particle’s changing direction is evidence of a significant interaction.

1.X.3

(a) Changing velocity (in this case, the velocity’s magnitude) indicates a net interaction.

(b) Changing velocity (in this case, the velocity’s magnitude) indicates a net interaction.

(c) Changing velocity (in this case, the velocity’s direction) indicates a net interaction.

(d) There is no sign of a net interaction.

(e) There is no sign of a net interaction.

1.X.4

(a) This statement is correct.

(b) This statement is incorrect.

(c) This statement is incorrect.

(d) This statement is correct.

(e) This statement is incorrect.

(f) This statement is incorrect.

2

1.X.5

(a) This statement is incorrect.

(b) This statement is incorrect.


(d) This statement is correct.

(e) This statement is correct.

1.X.6 Three numbers (signed) are needed to specify a 3D position vector.

1.X.7 One number (signed) is needed to specify a scalar.

1.X.8

|~r| =

√(r

x

)2

+(

ry

)2

+(

rz

)2

|~r| =√

(−3)2

+ (−4)2

+ (1)2m

|~r| ≈√

26 m ≈ 5.10 m

1.X.9 A vector’s magnitude corresponds to the length of the arrow used to represent the vector. It’s impossible for an arrowto have a negative length. Therefore, a vector cannot have a negative magnitude.

1.X.10 Within the framework of vector algebra, adding a vector and a scalar is not defined, and therefore has no meaning.Therefore, the correct answer choice is (d). (NOTE: There is a mathematical language, called geometric algebra, used bysome physicists in which adding vectors and scalars is not only defined, but also useful. However, this text does not usegeometric algebra.)

1.X.11 Within the framework of vector algebra, dividing a scalar (or anything else for that matter) by a vector is not defined,and therefore has no meaning. (NOTE: There is a mathematical language, called geometric algebra, used by some physicistsin which dividing by vectors is not only defined, but also useful. However, this text does not use geometric algebra.)

1.X.12 Simply multiply each component of ~a by f .

3

f~a = (−3) 〈0.03,−1.4, 26.0〉f~a = 〈−0.09, 4.2,−78.0〉

1.X.13 Simply divide each component of ~r by 2. Note that this is equivalent to multiplying ~r by 12 .

~r/2 =〈2,−3, 5〉m

2~r/2 = 〈1,−1.5, 2.5〉m

1.X.14 The magnitude of 3~v will simply be three times the magnitude of ~v.

∣∣∣ ~3v∣∣∣ = |3| |~v| = 3

√(v

x

)2

+(

vy

)2

+(

vz

)2

∣∣∣ ~3v∣∣∣ = (3)

√(2)

2+ (−3)

2+ (5)

2m/s∣∣∣ ~3v

∣∣∣ = (3)√

38 m/s ≈ 18.5 m/s

1.X.15 Any vector ~a and its opposite −~a will always have the same magnitude.

1.X.16

〈0, 6, 0〉|〈0, 6, 0〉|

=〈0, 6, 0〉√

(0)2

+ (6)2

+ (0)2

=〈0, 6, 0〉

6= 〈0, 1, 0〉

1.X.17 ~a = 〈400, 200,−100〉m/s2. First, we need the magnitude of ~a.

|~a| =

√(a

x

)2

+(

ay

)2

+(

az

)2

|~a| =√

(400)2

+ (200)2

+ (−100)2m/s2

|~a| = 458 m/s2

Now we need to get the direction a.

4

a =~a|~a|

a =〈400, 200,−100〉��m/s2

458 ��m/s2= 〈0.873, 0.437,−0.218〉

1.X.18 If ~b and ~a are equal, then they must have equal components. So bymust be 7.

1.X.19 If ~r1and ~r

2are equal, then they must have the same magnitude and the same direction. However, these two vectors

have different directions so they are not equal even if they have the same magnitude. Note that the familiar expression equaland opposite often seen referring to vectors in introductory physics textbooks is inherently oxymoronic; it contradicts itself.Two vectors cannot be equal if they have opposite directions. Two vectors can indeed have equal magnitudes and oppositedirections though, and this is the correct way of articulating the relationship.

1.X.20 ~F1= 〈300, 0,−200〉 and ~F

2= 〈150,−300, 0〉

∣∣∣~F1

∣∣∣ =

√(F

1,x

)2

+(

F1,y

)2

+(

F1,z

)2

=√

(300)2

+ (0)2

+ (−200)2

= 361 N∣∣∣~F2

∣∣∣ =

√(F

2,x

)2

+(

F2,y

)2

+(

F2,z

)2

=√

(150)2

+ (−300)2

+ (0)2

= 335 N∣∣∣~F1

+ ~F2

∣∣∣ =∣∣∣⟨F

1,x+ F

2,x,F

1,y+ F

2,y,F

1,z+ F

2,z

⟩∣∣∣ =√

(450)2

+ (−300)2

+ (−200)2

= 577 N∣∣∣~F1

∣∣∣+∣∣∣~F

2

∣∣∣ = 361 N + 335 N = 696 N

In general, it it not the case that∣∣∣~F

1+ ~F

2

∣∣∣ and ∣∣∣~F1

∣∣∣+∣∣∣~F

2

∣∣∣ are equal. The only time it is true is when ~F1and ~F

2have the

same direction.

1.X.21 ~A =⟨

3× 103,−4× 10

3,−5× 10

3⟩and ~B =

⟨−3× 10

3, 4× 10

3, 5× 10

3⟩so this is again very straightfoward.

(a)

~A + ~B =⟨

3× 103,−4× 10

3,−5× 10

3⟩

+⟨−3× 10

3, 4× 10

3, 5× 10

3⟩

= 〈0, 0, 0〉

Note that technically, writing ~A + ~B = 0 is incorrect since the right hand side is a scalar and the left hand side is avector.

(b) ∣∣∣~A + ~B∣∣∣ = 0

5

(c)

∣∣∣~A∣∣∣ =

√(3× 103

)2+(−4× 103

)2+(−5× 103

)2≈ 7071

(d)

∣∣∣~B∣∣∣ =

√(−3× 103

)2+(4× 103

)2+(5× 103

)2≈ 7071

(e) ∣∣∣~A∣∣∣+∣∣∣~B∣∣∣ ≈ 14142

So here we have two vectors, neither of which has zero magnitude but when they’re added together, the resultanthas zero magnitude. The magnitude of the sum of two vectors is not in general equal to the sum of their individualmagnitudes.

1.X.22 ~F1= 〈300, 0,−200〉 Nand ~F

2= 〈150,−300, 0〉 Nso this is again very straightfoward.

~F1

+ ~F2

= 〈300, 0,−200〉 N + 〈150,−300, 0〉 N= 〈450,−300,−200〉 N

~F1− ~F

2= 〈300, 0,−200〉 N− 〈150,−300, 0〉 N= 〈150, 300,−200〉 N

~F2− ~F

1= −

(~F

1− ~F

2

)= 〈−150,−300, 200〉 N

1.X.23

(a) This statement is correct.

(b) This statement is correct.


(d) This statement is incorrect.

(e) This statement is correct.

1.X.24 ~ri= 〈−3, 2, 5〉 m at t

iand ~r

f= 〈6, 4, 25〉 m at t

f.

6

(a)

∆~r = ~rf−~r

i= 〈6, 4, 25〉 m− 〈−3, 2, 5〉 m

= 〈9, 2, 20〉 m

(b) ∆t = tf− t

i= 10 : 02− 10 : 00 = 00 : 02 = 120 s

1.X.25 ~ri= 〈3, 0,−7〉 m at t

iand ~r

f= 〈2, 0,−8〉 m at t

f.

∆~r = ~rf−~r

i= 〈2, 0,−8〉 −〈3, 0,−7〉 m

= 〈−1, 0,−1〉 m

1.X.26 Let’s arbitrarily call the unknown unit vector a.

a = 〈cos 160◦, cos 20◦, 0〉= 〈−0.940, 0.342, 0〉

1.X.27 Let’s arbitrarily call the unknown unit vector s.

s = 〈cos 130◦, cos 40◦, 0〉= 〈−0.643, 0.766, 0〉

1.X.28

(80 ��cm5 ��min

)(1 m

100 ��cm

)(1 ��min60 s

)= 0.00267 m/s

1.X.29

(a) The average velocity of the tennis ball is

7

~vavg

=∆~r∆t

=~r

f−~r

i

tf− t

i

=〈9, 2, 8〉 m− 〈5, 7, 2〉 m

0.7 s− 0.2 s

=〈4,−5, 6〉 m

0.5 s= 〈8,−10, 12〉 m/s

(b) Average speed is defined as:

average speed =distance traveled

∆t

In general, this is not the same as |~vavg|. However, if the direction of motion of the object does NOT change

during the time interval, then

average speed = |~vavg| (if direction of ~v stays the same)

In this case, we assume that during a small time interval, the ball’s velocity is approximately constant and therefore,

average speed = |~vavg|

=(√

(8)2 + (−10)2 + (12)2)

m/s

= 17.5 m/s

(c) The direction of the average velocity is given by its unit vector

vavg

=~v

avg∣∣∣~vavg

∣∣∣=〈8,−10, 12〉 m/s

17.5 m/s= 〈0.46,−0.57, 0.68〉

1.X.30

(a) The average velocity of the spacecraft is

8

~vavg

=∆~r∆t

=~r

f−~r

i

∆t

=〈325, 25,−550〉 m− 〈200, 300,−400〉 m

5 s

=〈125,−275,−150〉 m

5 s= 〈25,−55,−30〉 m/s

(b) Average speed is defined as:

average speed =distance traveled

∆t

In general, this is not the same as |~vavg|. However, since the spacecraft’s direction of motion is presumably constant

during this time interval

average speed = |~vavg| (if direction of ~v stays the same)

=(√

(25)2 + (−55)2 + (−30)2)

m/s

= 67 m/s

(c) The direction of the average velocity is given by its unit vector

vavg

=~v

avg∣∣∣~vavg

∣∣∣=〈25,−55,−30〉 m/s

67 m/s= 〈0.37,−0.82,−0.44〉

1.X.31

~rf

= ~ri

+ ~vavg

∆t

= < 0, 0, 0 > +(⟨3× 105, 2× 105,−4× 105

⟩m/s)(9.7 s− 9.0 s)

= (⟨3× 105, 2× 105,−4× 105

⟩m/s)(0.7 s)

=⟨2.1× 105, 1.4× 105,−2.8× 105

⟩m

1.X.32

The velocity of the baseball is

9

~v =∆~r∆t

Note that you cannot simply solve for ∆t because you can’t divide a vector by a vector. Thus, to solve for ∆t, write thevelocity in component form.

vx

=∆x∆t

vy

=∆y∆t

vz

=∆z∆t

Use any of the components of the velocity to solve for the time interval. For example, using the x-component of the baseball’smotion gives

vx

=∆x∆t

∆t =∆xv

x

=x

2− x

1

vx

=18 m− 3 m

30 m/s

=15 m

30 m/s= 0.5 s

Check your answer by using the other components. You should get the same result, for example

vy

=∆y∆t

∆t =∆yv

y

=y

2− y

1

vy

=17 m− 7 m

30 m/s

=10 m

20 m/s= 0.5 s

1.X.33

10

The average velocity of a particle is its velocity during a finite time interval and is calculated as the particle’s displacementdivided by the time interval. It is in the direction of the displacement of the particle during the time interval. It does not tellyou anything about whether the particle’s velocity was constant during the time interval or whether it changed in magnitudeor direction during the time interval. A car that makes one lap around a track has zero average velocity during one lapbecause it’s displacement is zero. However, it certainly was moving and was perhaps speeding up or slowing down during thelap.

Instantaneous velocity of a particle is its velocity at an instant of time. It is calculated as the average velocity of a particleduring a time interval in the limit that the time interval approaches zero. In other words, it is the derivative of the particle’sposition as a function of time. The instantaneous velocity of a particle is tangent to the particle’s path at a single location(i.e. point), thus it defines the direction of motion of the particle at an instant. On a graph of x-position vs. time, theinstantaneous x-velocity of the particle at an instant t is the slope of a line tangent to the graph, at that instant.

1.X.34

The velocity of the comet is tangent to its path, in the direction of motion. Thus, the correct arrows are:

location 1 : a

location 2 : h

location 3 : g

location 4 : f

location 5 : e

1.X.35

Acceleration is defined as

~a =∆~v∆t

The direction of velocity is not given in the problem; therefore, assume that the motion is in the +x-direction.

~a =〈25, 0, 0〉 m/s− < 0, 0, 0 >

5 s= 〈5, 0, 0〉 m/s2

Thus, the magnitude of the acceleration of the car is

|~a| = 5 m/s2

The magnitude of the acceleration of a falling rock is approximately 10 m/s2. The magnitude of the car’s acceleration is halfthe magnitude of the acceleration of a falling rock.

11

1.X.36

~r(t) = < 3 + 5t, 4t2, 2t− 6t3 >

~v(t) =d~vdt

= < 5, 8t, 2− 18t2 >

~a(t) =d~vdt

= < 0, 8,−36t >

To calculate ~v at t = 0, substitute t = 0 into the expression for ~v(t).

~v(0) = < 5, 0, 2 >

To calculate ~a at t = 0, substitute t = 0 into the expression for ~a(t).

~a(0) = < 0, 8, 0 >

The constants in the function for ~r should have units so that if t is in seconds, for example, ~r might be calculated in metersor kilometers, or some other unit for position. Then, we would know the units for velocity and acceleration as well.

1.X.37

For |~v| << c,

~p = m~v|~p| = m |~v|

= (65 kg)(10 m/s)= 650 kg · m/s

1.X.38

The electron’s momentum is

12

~p =

1√1− |~v|

2

c2

m~v

=(9.1× 10

−31kg)(

⟨0, 0,−2× 108

⟩m/s)√

1− (−2×108m/s)2

(3×108m/s)2

=(9.1× 10

−31kg)(

⟨0, 0,−2× 108

⟩m/s)√

1−(

23

)2=

⟨0, 0,−2.4× 10−22

⟩kg · m/s

The magnitude of its momentum is

|~p| = 2.4× 10−22 kg · m/s

1.X.39

~p and ~v have the same direction, so the comet’s velocity is also in the direction of arrow b.

1.X.40

Since |~v| << c, use

~p = m~v

~v =~pm

=〈0.9, 0, 22.5〉 kg · m/s

4.5 kg= 〈0.2, 0, 5〉 m/s

~rf

= ~ri

+ ~v∆t= < 0, 0, 0 > +(〈0.2, 0, 5〉 m/s)(2 s)= 〈0.4, 0, 10〉 m

1.X.41

Since |~v| << c, use

13

~p = m~v

~v =~pm

=〈4500, 0,−3000〉 kg · m/s

1300 kg= 〈3.46, 0,−2.31〉 m/s

~rf

= ~ri

+ ~v∆t= 〈94, 0, 30〉 m + (〈3.46, 0,−2.31〉 m/s)(17 s− 12 s)= 〈94, 0, 30〉 m + (〈3.46, 0,−2.31〉 m/s)(5 s)= 〈94, 0, 30〉 m + 〈17.3, 0,−11.6〉 m= 〈111, 0, 18〉 m

1.X.42

(a) Sketch a picture showing the ball before the collision with the wall and after the collision with the wall, like those shownin Figure ?? and Figure ??.

pi

Figure 1: The initial momentum ~pi

(b) Sketch the change in momentum vector by drawing the initial momentum and final momentum vectors tail to tail andthen sketching the change in momentum vector from the head of the initial momentum vector to the head of the finalmomentum vector, as shown in Figure ??.

∆~p = ~pf− ~p

i

= m~vf−m~v

i

= m(~vf− ~v

i)

= (0.057 kg)(〈−48, 0, 0〉 m/s− 〈50, 0, 0〉 m/s)= (0.057 kg)(〈−98, 0, 0〉 m/s)= 〈−5.6, 0, 0〉 kg · m/s

14

pf

Figure 2: The final momentum ~pf

pf pi

pf

Figure 3: The change in momentum ∆~p.

Note that the change in momentum points to the left, which is consistent with the sketch in Figure ??.

(c)

∆ |~p| =∣∣∣~p

f

∣∣∣− ∣∣~pi

∣∣= m

∣∣∣~vf

∣∣∣−m ∣∣~vi

∣∣= m(

∣∣∣~vf

∣∣∣− ∣∣~vi

∣∣)= (0.057 kg)(48 m/s− 50 m/s)= (0.057 kg)(−2 m/s)= −0.11 kg · m/s

Note that in general, ∆ |~p| 6= |∆~p|, as is evident in this example.

1.X.43

15

(a)

∆~p = ~pB− ~p

A

= m(~vB− ~v

A)

= (6.4× 1023

kg)(⟨−2.5× 10

4, 0, 0

⟩m/s−

⟨0, 0,−2.5× 10

4⟩

m/s)

= (6.4× 1023

kg)(⟨−2.5× 10

4, 0, 2.5× 10

4⟩

m/s)

=⟨−1.6× 10

28, 0, 1.6× 10

28⟩

kg · m/s

(b) Sketch a picture showing Sun and the momentum of Mars at locations C and D, like the sketch in Figure ??.

C

DpCpD

Sun x

z

Figure 4: The momentum of Mars at C and D.

(c) Draw the change in momentum vector by sketching the momentum at C and the momentum at D tail to tail. Thechange in momentum is the vector drawn from the head of the momentum at C to the head of the momentum at D, asshown in Figure ??.Note: ∆~p in Figure ?? points up and to the right, towards Sun.

1.X.44 Draw a sketch of the situation, like the one shown in Figure ??.

(a)

∆~p = ~pf− ~p

i

= 〈7, 0, 12〉 kg · m/s− 〈0, 0, 10〉 kg · m/s= 〈7, 0, 2〉 kg · m/s

∆~p‖ is the z-component in this case since ~piwas in the z-direction. So,

∆~p‖ = 〈0, 0, 2〉 kg · m/s

16

pC

pD

p

Figure 5: The change in momentum vector.

x

z

pipf

p

Figure 6: Sketch of puck and momentum vectors

(b) ∆~p⊥ is perpendicular to the +z axis which is the component of ∆~p that is in the y-z plane, in this case.

∆~p⊥ = 〈7, 0, 0〉 kg · m/s

1.X.45

(a) and (b) are true because (a) the speed of light is the same in all reference frames and (b) light will be blueshifted forobservers in the spaceship that is moving toward the light beam. Note that the color of the light will shift toward shorterwavelengths (the blue end of the visible spectrum) for an observer moving toward the light source (or vice versa). That’swhat is meant by the word “blueshifted.”

1.X.46

17

(a) 1. The velocity of a bird changes as it flies through the air, due to interactions with Earth and air.2. The velocity of a gymnast changes as she does a routine on the uneven bars due to interactions with Earth and thebars.

(b) 1. The temperature of coffee in a cup on a desk changes due to interactions with the cup and air.2. The temperature of the air in a car changes due to interactions with the car (such as its windows) and Sun (radiation).

(c) 1. The shape of the space capsule Genesis changed as it collided with Earth, because its parachutes did not deploy.(See http://apod.nasa.gov/apod/ap090705.html for a picture of the capsule.) This change of shape was due to itsinteraction with Earth’s surface (a contact force).2. The shape of an atom (its electron cloud) changes as a result of its electromagnetic interaction with other atomswhen bonded in a molecule.

(d) 1. A free neutron decays into a proton, an electron, and an antineutrino due to the weak nuclear interaction of quarkswithin the neutron.2. A positron and an electron annihilate each other and produce photons, due to the weak nuclear interaction of thepositron and electron.

(e) 1. When a laptop sits on an inclined laptop stand, one might expect the laptop to slide off the stand (with a changingvelocity) due its gravitational interaction with Earth. However, it remains at rest with no net interaction. Thus, thelaptop stand interacts with the laptop, exactly balancing the gravitational force by Earth on the laptop.2. Suppose that water in an aquarium has a constant temperature. One expects the water to cool due to interactionswith air and glass around the aquarium. The fact that there is no net interaction means that there must be a sourceof thermal energy, such as an electric heating coil in the water, that heats the water.

1.X.47

(a), (b), (d), (e) are observations that give conclusive evidence of an interaction.

Note that (c) is not by itself evidence of an interaction. An object can move with constant velocity if it has no interactionswith its surroundings or if there is no net interaction with its surroundings, meaning that all interactions balance one another.

1.X.48

(b), (c), (d), (e), and (f) show evidence of an interaction. In the case of (b), its speed changes (and therefore its velocitychanges). In the case of (c) through (f), its direction of motion changes (and therefore its velocity changes).

(a) has a constant velocity and therefore has no net interaction with its surroundings.

1.X.49

The persons in cases (b) and (d) will observe motion that appears to violate Newton’s first law because their reference framesare accelerating relative to Earth (which can be assumed in these cases to be an inertial reference frame). In cases (a),(c), and (e), the person’s velocity (relative to Earth) is constant, and therefore, the person’s reference frame is an inertialreference frame where Newton’s first law will be valid in all experiments.

1.X.50

(a), (c), and (d) are correct. Using a spaceship far from any large objects (like stars, planets, etc.) is a useful thoughtexperiment because it doesn’t make contact with anything (such as air or a track or a road, for example) and doesn’t interactgravitationally with anything since the gravitational force is quite weak for interactions with objects that are very far away.

http://apod.nasa.gov/apod/ap090705.html

18

1.X.51

Nothing interacts with the spaceship after its thrusters are turned off; therefore, according to Newton’s first law, its velocitywill remain constant, and the spaceship will continue with the same speed of 1× 10

4m/s in the same direction as it had

when its thrusters were turned off.

1.X.52

The ball remains at rest (nearly); therefore, its velocity is constant. Thus, the net force on the ball must be zero. Since weknow that Earth exerts a downward gravitational force on the ball, there must be something exerting an upward force onthe ball so that the net force on the ball is zero.

1.X.53

While you are walking and holding the book, the ball moves with a constant velocity (relative to an observer who is standingat rest). When you stop, the ball continues moving with a constant velocity as it rolls across the book because there is nonet force on the ball to change its velocity, until it rolls off the book and then the net force on the ball is the gravitationalforce by Earth which changes its velocity as it falls.

1.X.54

All of these statements are true. Equal vectors must have equal components and the same direction (i.e. unit vector).

1.X.55

~a is a vector because of the arrow symbol above the variable.

1.X.56

(b) and (c) are vectors. (d) is a vector component and has properties of a vector and can be treated as a vector though it istechnically a vector component when written in this form.

1.X.57

|~v| is a scalar.

1.X.58

(a), (c), (d), and (f) are all vectors. Note that in the case of (f), a scalar times a vector is a vector.

1.X.59

(a), (c), and (d) are all vectors. Multiplying a vector by a scalar results in a vector. Dividing a vector by a scalar also resultsin a vector.

19

1.X.60

~v =⟨8× 106, 0,−2× 107

⟩m/s

|~v| =(√

(8× 106)2 + (0)2 + (−2× 107)2)

m/s

= 2.2× 107m/s

1.X.61

(a) The magnitude of a vector is indicated by the length of the arrow representing the vector. The arrows that have thesame magnitude as ~a have the same length as ~a. Counting gridlines shows that |~a| = 10 units.

Vectors ~c, ~e, and ~f also have a length of 10 units.

To calculate the magnitude of ~b, first determine its components. Then apply the Pythagorean theorem.

∣∣∣~b∣∣∣ =(√

(7.1)2 + (7.1)2 + (0)2)

= 10 units

Thus, ~b has the same magnitude as ~a. In the same way, you can verify that ~d also has the same magnitude as ~a.

(b) To be equal, the vectors must have the same magnitude and direction. The only vectors equal to ~a are ~c and ~f.

1.X.62

For each vector, begin at the tail of the vector and count the gridlines to the right or left and the gridlines up or down to getto the head of the vector. These are the x and y components. The results are:

~a = 〈5, 3, 0〉 m~c = 〈6,−9, 0〉 m~g = 〈−10, 3, 0〉 m

1.X.63

No, it is not a meaningful value. A scalar cannot be added to a vector.

1.X.64

(a) See Figure ??

20

Free Plain Graph Paper from http://incompetech.com/graphpaper/plain/

!p

Figure 7: Vector ~p.

(b)

−~p = −〈−7, 3, 0〉= 〈7,−3, 0〉

It points to the right and downward, as shown in Figure ??.


!p

!!p

Figure 8: Vector −~p.

21

1.X.65

2〈2, 6,−3〉 m = 〈4, 12,−6〉 m

Therefore, (a) and (c) are true. (b) is false because multiplying a vector changes its magnitude (i.e. length of the arrow) butNOT its direction. Both vectors have the same direction, thus the same unit vector.

1.X.66

(a) To measure the x-component of the vector, begin at the tail of the vector and count gridlines horizontally from the tailto the head of the vector. In a similar way, count gridlines vertically from the tail to the head of the vector to get they-comopnent. As a result, ~a = 〈−4,−3, 0〉.

(b) ~b = 〈−4,−3, 0〉

(c) This statement is true. ~a and ~b have the same components.

(d) ~c = 〈4, 3, 0〉

(e) Yes, this is true. −〈−4,−3, 0〉 = 〈4, 3, 0〉

(f) ~d = 〈−3, 4, 0〉

(g) This statement is false. −~c = −〈4, 3, 0〉 = 〈−4,−3, 0〉 6= 〈−3, 4, 0〉

1.X.67

(a) ~d = 〈−6, 3, 2〉 m

(b) ~e = −~d = −〈−6, 3, 2〉 m = 〈6,−3,−2〉 m

(c) Take the position of the tail of the vector and add the vector ~d.

〈−5,−2, 4〉 m + 〈−6, 3, 2〉 m = 〈−11, 1, 6〉 m

(d) Take the position of the tail of the vector and add the vector −~d.

〈−1,−1,−1〉 m +−〈−6, 3, 2〉 m = 〈5,−4,−3〉 m

22

1.X.68

f~a = (2.0)(〈0.02,−1.7, 30.0〉)= 〈0.04,−3.4, 60〉

1.X.69

(a) The vector ~f is shown in Figure ??.


Figure 9: Vector ~f

(b)

2~f = < −4, 8, 0 >

The vector 2~f is shown in Figure ??.

(c) The vector 2~f has twice the magnitude (length) as ~f.

(d) The vector 2~f has the same direction as ~f.

(e)

~f/2 = < −1, 2, 0 >

The vector ~f/2 is shown in Figure ??.

23


Figure 10: Vector 2~f


Figure 11: Vector ~f/2

(f) The vector ~f/2 has half the magnitude (length) as ~f.

(g) The vector ~f/2 has the same direction as ~f.

(h) Yes, multiplying a vector by a scalar changes its magnitude.

24

(i)

a~f = −3~f

Thus, a = −3.

1.X.70

This vector points in the −x direction, so no calculation is needed. The unit vector is < −1, 0, 0 >.

1.X.71

For vector < 2, 2, 2 >,

unit vector =< 2, 2, 2 >(√

(2)2 + (2)2 + (2)2)

=< 2, 2, 2 >√

12= < 0.58, 0.58, 0.58 >

For vector < 3, 3, 3 >,

unit vector =< 3, 3, 3 >(√

(3)2 + (3)2 + (3)2)

=< 3, 3, 3 >√

27= < 0.58, 0.58, 0.58 >

Note that its direction is the same as the vector < 2, 2, 2 > because

< 3, 3, 3 > =(

32

)(< 2, 2, 2 >)

and multiplying a vector by a scalar changes its magnitude, but not its direction.

1.X.72

|~a| is

|~a| =(√

(400)2 + (200)2 + (−100)2)

m/s2

= 458 m/s2

25

Its unit vector is

a =~a|~a|

=

(√(400)2 + (200)2 + (−100)2

)m/s2

458 m/s2

= < 0.87, 0.44,−0.22 >

Thus,

~a = (458 m/s2)(< 0.87, 0.44,−0.22 >)

1.X.73A picture of ~g is shown in Figure ??.


!g

Figure 12: Vector ~g

(a)

|~g| =√

(4 m)2 + (7 m)2 + (0 m)2

= 8.06 m

(b)

g =~g|~g|

=〈4, 7, 0〉 m

8.06 m= < 0.50, 0.87, 0 >

26

(c) A picture of g is shown in Figure ??.


!gg

Figure 13: Vector g

It has the same direction as ~g but has a length of 1.

(d) One should expect to get the original vector ~g. Let’s verify:

|~g| g = (8.06 m)(< 0.50, 0.87, 0 >)= 〈4.0, 7.0, 0〉 m

exactly as expected.

1.X.74

(a) The position of the proton is the vector from the origin to the location of the proton. In this case,

~r =⟨3× 10−10,−3× 10−10, 8× 10−10

⟩m

(b)

|~r| =√

(3× 10−10 m)2 + (−3× 10−10 m)2 + (8× 10−10 m)2

= 9.1× 10−10 m

27

(c) Its unit vector is

r =~r|~r|

=

⟨3× 10−10,−3× 10−10, 8× 10−10

⟩m

9.1× 10−10 m= 〈0.33,−0.33, 0.88〉

1.X.75

(a)

magnitude = 9.5direction = < 0, 0, 1 >

< 0, 0, 9.5 > = (9.5)(< 0, 0, 1 >)

(b)

magnitude = 679direction = < 0,−1, 0 >

< 0,−679, 0 > = (679)(< 0,−1, 0 >)

(c)

magnitude =√

(3.5× 10−3 )2 + (0 )2 + (−3.5× 10−3 )2

= 4.9× 10−3

direction =< 3.5× 10−3, 0,−3.5× 10−3 >

4.9× 10−3

= < 0.71, 0,−0.71 >

< 3.5× 10−3, 0,−3.5× 10−3 > = (4.9× 10−3)(< 0.71, 0,−0.71 >)

(d)

magnitude =√

(4× 106 )2 + (−6× 106 )2 + (3× 106 )2

= 7.8× 106

direction =< 4× 106,−6× 106, 3× 106 >

7.8× 106

= < 0.51,−0.77, 0.38 >

< 4× 106,−6× 106, 3× 106 > = (7.8× 106)(< 0.51,−0.77, 0.38 >)

28

1.X.76

unit vector =< 3, 3, 3 >√

(3 )2 + (3 )2 + (3 )2

=< 3, 3, 3 >√

27= < 0.577, 0.577, 0.577 >

The edges of the cube make up the +x, +y and +z axes. As a result, one can use the direction cosine to get the angle thevector makes with any axis. Since the vector is along the diagonal of a cube, then it makes the same angle with respect toeach axis.

Choose an axis, for example the x-axis. The direction cosine is

cos(θx) =

x-component of the vectormagnitude of the vector

=x-component of the unit vector

1= 0.577

Take the inverse cosine of both sides, with your calculator in degree mode.

θx

= cos−1(0.577)= 55◦

1.X.77

Calculate the magnitude of each vector using

∣∣ ~vector∣∣ =

√(vector

x

)2

+(

vectory

)2

+(

vectorz

)2

If the magnitude of the vector is 1 (or close to 1 considering that the numbers are rounded), then it is a unit vector.

(a) magnitude = 1. It’s a unit vector.

(b) magnitude =√

0.5. It’s NOT a unit vector.

(c) magnitude =√

1/3. It’s NOT a unit vector.

(d) magnitude =√

0.82. It’s NOT a unit vector.

(e) magnitude = 3. It’s NOT a unit vector.

(f) magnitude =√

3. It’s NOT a unit vector.

(g) magnitude ≈ 1. It is a unit vector.

29

(h) magnitude ≈ 1. It is a unit vector.

1.X.78

(a), (b), and (c) are true. If two vectors are equal, then they have the same magnitude, direction, and vector components.

1.X.79

(a) A picture of the baseball, tennis ball, and their position vectors is shown in Figure ??.


baseball

tennis ball

!B

!T

Figure 14: The baseball, tennis ball, and their position vectors.

(b) See Figure ??.

(c) ~B = 〈3, 5, 0〉 m

(d) See Figure ??.

(e) ~T = 〈−3,−1, 0〉 m

(f) Think of the phrase “final minus initial.” The relative position vector of the tennis ball relative to the baseball pointsfrom the baseball (initial) to the tennis ball (final). It is shown below in Figure ??.

(g) Starting at the tail of ~r, count the number of gridlines horizontally (x) and vertically (y) to get to the head of ~r. Thus,~r = 〈−6,−6, 0〉 m.

(h)

~T− ~B = 〈−3,−1, 0〉 m− 〈3, 5, 0〉 m= 〈−6,−6, 0〉 m

30


baseball

tennis ball

!B

!T

!r

Figure 15: The position of the tennis ball relative to the baseball, ~r.

(i) Yes, ~r = ~T− ~B. Examine the answers to parts (g) and (h).

(j) Solve for ~T and ~B algebraically.

~T =~r + ~B~B = ~T−~r

(k) ∣∣∣~B∣∣∣ =√

(3 m)2 + (5 m)2 + (0 m)2 = 5.8 m∣∣∣~T∣∣∣ =√

(−3 m)2 + (−1 m)2 + (0 m)2 = 3.2 m

|~r| =√

(−6 m)2 + (−6 m)2 + (0 m)2 = 8.5 m

(l) ∣∣∣~T∣∣∣− ∣∣∣~B∣∣∣ = 3.2 m− 5.8 m = −2.6 m

(m)∣∣∣~T− ~B∣∣∣ is the magnitude of ~r which is 8.5 m. Thus, the answer is NO. −2.8m 6= 8.5m; therefore,∣∣∣~T∣∣∣− ∣∣∣~B∣∣∣ 6= ∣∣∣~T− ~B∣∣∣

31

1.X.80The position of object 2 relative to object 1 is

~r2 rel. to 1

= ~r2−~r

1

= 〈5, 2, 0〉 m− 〈3,−2, 0〉 m= 〈2, 4, 0〉 m

This points to the right and “upward” toward the top of the page. This is consistent with the picture shown in Figure ??.

–4 –2 0 2 4 6 m

–2

0

2

4

6 m

–4

r1

r2

r2 r1

Figure 16: Sketch of ~r2 rel. to 1

The position of object 1 relative to object 2 is

~r1 rel to 2

= ~r1−~r

2

= −(~r2−~r

1)

= −〈2, 4, 0〉 m= 〈−2,−4, 0〉 m

~r1 rel to 2

is opposite ~r2 rel to 1

. It points downward and to the left, as shown in a rough sketch in Figure ??. (Note: this sketchcorresponds to the previous sketch though it is not exact.

1.X.81

(a) A sketch of the given vector ∆~r is shown in Figure ??.

∆~r = 〈4,−13, 0〉 m− 〈9.5, 7, 0〉 m= 〈−5.5,−20, 0〉 m

32

x

y

r2

r1

r1−r2

Figure 17: Sketch of ~r1 rel to 2

x

y

⟨9.5,7,0 ⟩

⟨4,−13,0 ⟩

Figure 18: A sketch of the given vector.

The resulting vector points “down” and to the left, which is consistent with the picture.

(b)

|∆~r| =√

(−5.5 m)2 + (−20 m)2 + (0 m)2

= 20.7 m≈ 21 m

33

1.X.82

(a)

~rtree rel. to head

= ~rtree−~r

head

= 〈−25, 35, 43〉 m− 〈12, 30, 13〉 m= 〈−37, 5, 30〉 m

(b)

|~r| =√

(−37 m)2 + (5 m)2 + (30 m)2

= 47.9 m≈ 48 m

1.X.83

(a)

~rplanet rel. to star

= ~rplanet

−~rstar

=⟨−4× 10

10,−9× 10

10, 6× 10

10⟩

m−⟨

6× 1010, 8× 10

10, 6× 10

10⟩

m

=⟨−10× 10

10,−17× 10

10, 0⟩

m

(b)

~rstar rel. to planet

= −~rplanet rel. to star

= −⟨−10× 10

10,−17× 10

10, 0⟩

m

=⟨

10× 1010, 17× 10

10, 0⟩

m

1.X.84

(a)

~rplanet rel. to star

= ~rplanet

−~rstar

=⟨−1× 10

10, 8× 10

10,−3× 10

10⟩

m−⟨

6× 1010,−5× 10

10, 1× 10

10⟩

m

=⟨−7× 10

10, 13× 10

10,−4× 10

10⟩

m

34

(b) ∣∣∣~rsp

∣∣∣ =(√(

−7× 1010)2 +

(13× 1010

)2 +(−4× 1010

)2 ) m

= 15.3× 1010

m= 1.53× 10

11m

≈ 1.5× 1011

m

(c)

rsp

=~r

sp∣∣∣~rsp

∣∣∣=

⟨−7× 10

10, 13× 10

10,−4× 10

10⟩

m

15.3× 1010m= < −0.46, 0.85,−0.26 >

1.X.85

~rfrom electron to proton

= ~rproton

−~relectron

= < xp, y

p, z

p> − < x

e, y

e, z

e>

= < (xp− x

e), (y

p− y

e), (z

p− z

e) >

~rfrom proton to electron

= ~relectron

−~rproton

= −~rfrom electron to proton

= − < (xp− x

e), (y

p− y

e), (z

p− z

e) >

= < (xe− x

p), (y

e− y

p), (z

e− z

p) >

1.X.86

∆~r is the displacement of a particle. It is also referred to as the change in the position of the particle.

∆t is a time interval. It is also referred to as the change in the clock reading t and is the time elapsed on the clock.

1.X.87

~ri

= 〈0.2,−0.05, 0.1〉 m~r

f= 〈−0.202, 0.054, 0.098〉 m

∆t = 2× 10−6

s

35

(a)

~vavg

=∆~r∆t

=~r

f−~r

i

∆t

=〈−0.202, 0.054, 0.098〉 m− 〈0.2,−0.05, 0.1〉 m

2× 10−6s

=〈−0.402, 0.104,−0.002〉 m

2× 10−6s

=⟨−2.01× 10

5, 5.2× 10

4,−1× 10

3⟩

m/s

(b) Average speed is not always equal to the magnitude of average velocity, except if the object moves in a straight linewithout changing directions. Assume this is the case for the neutron, then only in this special case

average speed =∣∣∣~vavg

∣∣∣=

√(−2.01× 105 m/s

)2 +(5.2× 104 m/s

)2 +(−1× 103 m/s

)2= 2.08× 10

5m/s

1.X.88

~ri

= 〈15, 8,−3〉 m~r

f= 〈20, 6,−1〉 m

∆t = 0.1 s

~vavg

=∆~r∆t

=~r

f−~r

i

∆t

=〈20, 6,−1〉 m− 〈15, 8,−3〉 m

0.1 s

=〈5,−2, 2〉 m

0.1 s= 〈50,−20, 20〉 m/s

1.P.89

(a)

~v = 〈−20,−90, 40〉 m/s~r

i= 〈200, 300,−500〉 m

~rf

= 〈−380,−2310, 660〉 m

~vavg

=∆~r∆t

36

You cannot divide vectors, so

∆t 6= ∆~r~r

avg

You may use

∆t =

∣∣∣∆~ravg

∣∣∣∣∣∣~vavg

∣∣∣Or you may write the velocity in component form and use any one of the components. For instance,

∆t =∆xv

avg x

This method gives

∆t =x

f− x

i

vavg x

=−380 m− 200 m−20 m/s

= 29 s

(b)

∆~r = ~rf−~r

i

= 〈−380,−2810, 660〉 m− 〈200, 300,−500〉 m= 〈−580,−2610, 1160〉 m

|∆~r| =√

(−580 m)2 + (−2610 m)2 + (1160 m)2

= 2914 m

(c)

|~v| =|~r|∆t

=2914 m

29 s= 100 m/s

(d)

v =~v|~v|

=〈−20,−90, 40〉 m/s

100.5 m/s= < 0.2, 0.9, 0.4 >

37

1.X.90

~vavg

=∆~r∆t

=~r

f−~r

i

∆t~r

f−~r

i= ~v

avg∆t

~rf

= ~ri

+ ~vavg

∆t

1.X.91

t1

= 12 s~r

i= 〈84, 78, 24〉 m

~v = 〈4, 0,−3〉 m/st2

= 18 s∆t = t

2− t

1

= 18 s− 12 s= 6 s

~rf

= ~ri

+ ~v∆t= 〈84, 78, 24〉 m + (〈4, 0,−3〉 m/s)(6 s)= 〈84, 78, 24〉 m + 〈24, 0,−18〉 m= 〈108, 78, 6〉 m

1.X.92

~ri

= 〈7, 21,−17〉 m∆t = 3 s~v

avg= 〈−11, 42, 11〉 m/s

yf

= ?

~rf

= ~ri

+ ~vavg

∆t= 〈7, 21,−17〉 m + (〈−11, 42, 11〉 m/s)(3 s)= 〈7, 21,−17〉 m + 〈−33, 126, 33〉 m= 〈−26, 147, 16〉 m

So yf

= 147 m.

38

1.X.93

(a) From t = 6.3 s to 6.8 s:

∆t = 6.8 s− 6.3 s= 0.5 s

~ri

= 〈−3.5, 9.4, 0〉 m~r

f= 〈−1.3, 6.2, 0〉 m

~vavg

=∆~r∆t

=〈−1.3, 6.2, 0〉 m− 〈−3.5, 9.4, 0〉 m

0.5 s

=〈2.2,−3.2, 0〉 m

0.5 s= 〈4.4,−6.4, 0〉 m/s

(b) From t = 6.3 s to 7.3 s:

∆t = 7.3 s− 6.3 s= 1.0 s

~ri

= 〈−3.5, 9.4, 0〉 m~r

f= 〈0.5, 1.7, 0〉 m

~vavg

=∆~r∆t

=〈0.5, 1.7, 0〉 m− 〈−3.5, 9.4, 0〉 m

1.0 s= 〈4,−7.7, 0〉 m/s

(c) The best estimate for ~v at t = 6.3 s is the average velocity during the smallest possible time interval that includest = 6.3 s. Thus, the time interval from t = 6.3 s to 6.8 s gives the best possible estimate in this case for the instantaneousvelocity at t = 6.3 s.

(d) Assume that the bee’s average velocity between t = 6.3 s and 6.33 s is approximately constant. From t = 6.3 s to 6.33 s:

∆t = 6.33 s− 6.3 s= 0.03 s

~ri

= 〈−3.5, 9.4, 0〉 m~v

avg≈ 〈4.4,−6.4, 0〉 m/s

~vavg

=∆~r∆t

∆~r = ~vavg

∆t= (〈4.4,−6.4, 0〉 m/s)(0.03 s)= 〈0.132,−0.192, 0〉 m

39

1.X.94

~ri

= 〈50, 20, 30〉 m~r

f= 〈53, 18, 31〉 m

∆t = 0.1 s

~vavg

=∆~r∆t

=~r

f−~r

i

∆t

=〈53, 18, 31〉 m− 〈50, 20, 30〉 m

0.1 s

=〈3,−2, 1〉 m

0.1 s= 〈30,−20, 10〉 m/s

1.X.95

~ri

=⟨−3× 10

3,−4× 10

3, 8× 10

3⟩

m

~rf

=⟨−1.4× 10

3,−6.2× 10

3, 9.7× 10

3⟩

m

ti

= 18.4 stf

= 21.4 s∆t = 21.4 s− 18.4 s

= 3.0 s

~vavg

=∆~r∆t

=~r

f−~r

i

∆t

=

⟨−1.4× 10

3,−6.2× 10

3, 9.7× 10

3⟩

m−⟨−3× 10

3,−4× 10

3, 8× 10

3⟩

m

3 s

=

⟨1.6× 10

3,−2.2× 10

3, 1.7× 10

3⟩

m

3 s

=⟨

5.33× 102,−7.33× 10

2, 5.67× 10

2⟩

m

1.X.96

40

(a)

~ri

= 〈0.02, 0.04,−0.06〉 m~r

f= 〈0.02, 1.84,−0.86〉 m

∆t = 2 µs

= 2× 10−6

s

~vavg

=∆~r∆t

=~r

f−~r

i

∆t

=〈0.02, 1.84,−0.86〉 m− 〈0.02, 0.04,−0.06〉 m

2× 10−6s

=〈0, 1.8,−0.8〉 m

2× 10−6s

=⟨

0, 9× 105,−4× 10

5⟩

m/s

(b) Now, for this time interval of 5 µs, the initial position of the electron is its position at the end of the previous 2 µsinterval.

~ri

= 〈0.02, 1.84,−0.86〉 m∆t = 5 µs

~v =⟨

0, 9× 105,−4× 10

5⟩

m/s

~rf

= ~ri

+ ~v∆t

= 〈0.02, 1.84,−0.86〉 m + (⟨

0, 9× 105,−4× 10

5⟩

m/s)(5× 10−6

s)

= 〈0.02, 1.84,−0.86〉 m + 〈0, 4.5,−2〉 m= 〈0.02, 6.34,−2.86〉 m

Another way to solve it is to consider the total time interval of 2 µs + 5 µs = 7 µs. In this case, ~riis the electron’s

position at the beginning of the 2 µs interval.

~ri

= 〈0.02, 0.04,−0.06〉 m∆t = 5 µs

~rf

= 〈0.02, 0.04,−0.06〉 m + (⟨

0, 9× 105,−4× 10

5⟩

m/s)(7× 10−6

s)

= 〈0.02, 0.04,−0.06〉 m + 〈0, 6.3,−2.8〉 m= 〈0.02, 6.34,−2.86〉 m

which agrees with the same answer obtained using the 5 µs time interval.

1.P.97

41

(a)

~vavg AB

=∆~r∆t

=~r

B−~r

A

∆t

=〈22.3, 26.1, 0〉 m− < 0, 0, 0 >

1.0 s− 0.0 s= 〈22.3, 26.1, 0〉 m/s

(b) From t = 1.0 s to t = 2.0 s, assuming it travels with a constant velocity of 〈22.3, 26.1, 0〉 m/s,

~rf

= ~ri

+ ~vavg

∆t= 〈22.3, 26.1, 0〉 m + (〈22.3, 26.1, 0〉 m/s)(2.0 s− 1.0 s)= 〈22.3, 26.1, 0〉 m + 〈22.3, 26.1, 0〉 m= 〈44.6, 52.2, 0〉 m

(c) ~r at point C is 〈40.1, 38.1, 0〉 m which is not the same as what we predicted. We assumed constant velocity when makingour prediction; however, in reality the velocity was not constant, but was decreasing in both the x and y directions.An approximation of constant velocity is only valid for small time intervals. For this projectile, ∆t = 1.0 s was not asmall enough time interval to reasonably assume constant velocity.

1.P.98

ti

= 6 stf

= 10 s~r

i= 〈6,−3, 10〉 m

~rf

= 〈6.8,−4.2, 11.2〉 m~r at t = 8.5 s = ?

Assume that the butterfly travels with a constant velocity. Calculate its velocity.

~vavg

=∆~r∆t

=~r

f−~r

i

∆t

=〈6.8,−4.2, 11.2〉 m− 〈6,−3, 10〉 m

10 s− 6 s

=〈0.8,−1.2, 1.2〉 m

4 s= 〈0.2,−0.3, 0.3〉 m/s

42

Now calculate its position at t = 8.5 s, if it starts at t = 6.0 s.

~rf

= ~ri

+ ~vavg

∆t= 〈6,−3, 10〉 m + (〈0.2,−0.3, 0.3〉 m/s)(8.5 s− 6 s)= 〈6,−3, 10〉 m + (〈0.2,−0.3, 0.3〉 m/s)(2.5 s)= 〈6,−3, 10〉 m + 〈0.5,−0.75, 0.75〉 m= 〈6.5,−3.75, 10.75〉 m

1.X.99

(a) and (e) are correct.

(a) is correct because momentum and velocity are proportional and therefore have the same unit vector and the same direction.

(e) is correct because in the limit as ∆t approaches zero, the displacement vector becomes tangent to the path. Since momen-tum is proportional to velocity which is proportional to displacement, then instantaneous velocity (and thus instantaneousmomentum) is also tangent to the object’s path.

1.X.100

The approximate formula for momentum may be used for (a), (b), (c) and (e) because in all of these cases, the object orparticle is moving with a speed much less than 3× 10

8 ms . In case (e), the electron’s speed is one-hundredth the speed of

light. If a highly precise calculation is not needed, then even in this case, the approximate formula for momentum may beused.

As a rule of thumb, if an object’s speed is less than about 10% of the speed of light, then the approximate formula may beused, except in cases where high precision (i.e. many significant figures) is needed.

1.X.101

(a) γ is a scalar.

(b)

γ =1√

1− |~v|2

c2

The minimum value of γ occurs when the speed of the object is zero. Then γ = 1.

(c) The minimum value of γ is when the object’s speed is a low value (i.e. zero).

(d) There is no maximum value of γ because as the object’s speed approaches the speed of light c, γ becomes infinite.

(e) γ becomes large for large speeds, i.e. speeds close to the speed of light. Note that speed cannot become arbitrarilylarge. The maximum possible speed is the speed of light. However, γ approaches infinity as the speed approaches thespeed of light.

(f) γ ≈ 1 for low speeds.

43

1.X.102

|~v| << c therefore

|~p| = m |~v|= (0.155 kg)(40 m/s)= 6.2 kg · m/s

1.X.103

mproton

= 1.67× 10−27

kg

|~p| = γm |~v|

=m |~v|√1− |~v|

2

c2

=(1.67× 10

−27kg)(0.88)(3× 10

8m/s)√

1− (0.8c)2

c2

=(1.67× 10

−27kg)(0.88)(3× 10

8m/s)√

1− (0.8)2

= 7.35× 10−19

kg · m/s

1.X.104

Note: |~v| << c

m = 0.4 kg~v = 〈38, 0,−2〉 m/s~p = m~v

= (0.4 kg)(〈38, 0,−2〉 m/s)= 〈15.2, 0,−10.8〉 kg · m/s

|~p| =√

(15.2 kg · m/s)2 + (0 kg · m/s)2 + (−10.8 kg · m/s)2

= 18.6 kg · m/s

44

1.X.105

m = 155 g= 0.155 kg

|~v| = (100 mph)(1 m/s

2.2369 mph)

= 44.7 m/s

Note: ~v << c

|~p| = m |~v|= (0.155 kg)(44.7 m/s)= 6.93 kg · m/s

1.X.106

m = 1000 kg

|~v| = (500 mph)(1 m/s

2.2369 mph)

= 224 m/s

Note: ~v << c

|~p| = m |~v|= (1000 kg)(224 m/s)

= 2.24× 105kg · m/s

1.X.107

~p = 〈4,−5, 2〉 kg · m/s

|~p| =√

(4 kg · m/s)2 + (−5 kg · m/s)2 + (2 kg · m/s)2

= 6.7 kg · m/s

1.X.108

melectron

= 9.11× 10−31

kg|~v| = 0.95c

45

|~p| = γm |~v|

=m |~v|√1− |~v|

2

c2

=m(0.95c)√1− (0.95c)2

c2

=m(0.95c)√1− (0.95)2

=(9.11× 10

−31kg)(0.95)(3× 10

8m/s)√

1− (0.95)2

= 8.31× 10−22

kg · m/s

1.X.109

melectron

= 9.11× 10−31

kg|~p| = γm |~v|

=m |~v|√1− |~v|

2

c2

=m(0.9999c)√1− (0.9999c)2

c2

=m(0.9999c)√1− (0.9999)2

=(9.11× 10

−31kg)(0.9999)(3× 10

8m/s)√

1− (0.9999)2

= 1.93× 10−20

kg · m/s

1.X.110

(a)

mproton

= 1.67× 10−27

kg

|~v| = 0.99c

If |~p| ≈ m |~v|, then

|~p| = (1.67× 10−27

kg)(0.99)(3× 108m/s)

= 4.96× 10−19

kg · m/s

46

(b) Since |~v| is not small compared to c,

|~p| = γm |~v|

=m |~v|√1− |~v|

2

c2

=m(0.99c)√1− (0.99c)2

c2

=m(0.99c)√1− (0.99)2

=(1.67× 10

−27kg)(0.99)(3× 10

8m/s)√

1− (0.99)2

= 3.52× 10−18

kg · m/s

(c) ∣∣~pcorrect

∣∣∣∣∣~papprox

∣∣∣ =3.52× 10

−18kg · m/s

4.96× 10−19kg · m/s

= 7.1

Thus, the approximate calculation for the momentum of the particle is about 7 times too small.

1.X.111

m = 1.6 kg~p = 〈0, 0, 4〉 kg · m/s

(a)

|~p| = 4 kg · m/s

(b)

p =~p|~p|

= < 0, 0, 1 >

(c)

|~p| = m |~v|

|~v| =|~p|m

=4 kg · m/s

1.6 kg= 2.5 m/s

47

1.X.112

|~v| = 0.9996c

γ =1√

1− |~v|2

c2

=1√

1− (0.9996c)2

c2

=1√

1− (0.9996)2

= 1250

1.X.113

melectron

= 9× 10−31

kg|~v| = 0.996cv = < −.655,−0.492,−0.573 >

(a)

γ =1√

1− |~v|2

c2

=1

1− (0.996)2

= 11.2

(b)

|~v| = 0.996c

= (0.996)(3× 108m/s)

= 2.988× 108m/s

(c)

|~p| = γm |~v|= (11.2)(9× 10

−31kg)(2.988× 10

8m/s)

= 3.0× 10−21

kg · m/s

(d)

~p = |~p| p

48

Since ~p is proportional to ~v, then their unit vectors p and v are the same. Thus,

~p = (3× 10−21

m/s) < 0.655,−0.492,−0.573 >

=⟨

1.97× 10−21

,−1.48× 10−21

,−1.72× 10−21⟩

kg · m/s

1.X.114

|~p| = γm |~v||~p|m

= 0.85c

Thus,

γ |~v| = 0.85c|~v|√

1− |~v|2

c2

= 0.85c

|~v|2

1− |~v|2

c2

= (0.85c)2

|~v|2 = (0.85c)2(

1− |~v|2

c2

)|~v|2 = (0.85c)2 − (0.85)2 |~v|2

|~v|2 + (0.85)2 |~v|2 = (0.85c)2

|~v|2 (1 + 0.852) = (0.85c)2

|~v|2 =(0.85c)2

1 + (0.85)2

|~v| =0.85c√

1 + (0.85)2

= 0.65c

1.X.115

(a) Draw a sketch of the situation, like the one shown in Figure ??.

Sketch the change in momentum vector by drawing the initial and final momentum vectors tail to tail and drawing thechange in momentum from the head of the initial momentum to the head of the final momentum, as shown in Figure??.

49

pi

pf

Figure 19: Momentum vectors before and after the collision.

pipf

p

Figure 20: The change in momentum of the ball.

~vi

= < vx, 0, 0 >

~vf

= < −vx, 0, 0 >

∆~p = ~pf− ~p

i

= m(~vf− ~v

i)

= m(< −vx, 0, 0 > − < v

x, 0, 0 >)

= m(< −2vx, 0, 0 >)

= < −2mvx, 0, 0 >

The change in momentum is in the −x direction which is consistent with the picture.

(b)

∆ |~p| =∣∣∣~p

f

∣∣∣− ∣∣~pi

∣∣= m

∣∣∣~vf

∣∣∣−m ∣∣~vi

∣∣= mv

x−mv

x

= 0

50

Note |∆~p| 6= ∆ |~p|.

1.X.116

In going all the way around, in one revolution, ~pfis the same as ~p

i. Thus, |∆~p| = 0.

In going half a revolution (180◦), sketch ~piand ~p

fat opposite sides of the circle, as shown in the example in Figure ??. (You

may choose any two points on the circle, as long as they are on opposite sides of the circle. Also, you may assume that themerry-go-round is rotating clockwise. Your choice of points or direction of rotation does not affect the final answer for themagnitude of the change in momentum.)

rotation

Figure 21: The initial and final momenta of the child for half a revolution.

To find the change in momentum, sketch ~pfand ~p

itail to tail. ∆~p is the vector from the head of ~p

ito the head of ~p

f. See

Figure ??.

Figure 22: The change in momentum of the child for half a revolution.

As you can see, ∆~p = 2~pf. Thus,

51

|∆~p| = 2∣∣∣~p

f

∣∣∣= 2m

∣∣∣~vf

∣∣∣= 2(50 kg)(5 m/s)= 500 kg · m/s

1.P.117

(a)

∆~pBC

= ~pC− ~p

B

= 〈2.55, 0.97, 0〉 kg · m/s− 〈3.03, 2.83, 0〉 kg · m/s= 〈−0.48,−1.86, 0〉 kg · m/s

∆~pCD

= ~pD− ~p

C

= 〈2.24,−0.57, 0〉 kg · m/s− 〈2.55, 0.97, 0〉 kg · m/s= 〈−0.31,−1.54, 0〉 kg · m/s

∆~pDE

= ~pE− ~p

D

= 〈1.97,−1.93, 0〉 kg · m/s− 〈2.24,−0.57, 0〉 kg · m/s= 〈−0.27,−1.36, 0〉 kg · m/s

∆~pEF

= ~pF− ~p

E

= 〈1.68,−3.04, 0〉 kg · m/s− 〈1.97,−1.93, 0〉 kg · m/s= 〈−0.29,−1.11, 0〉 kg · m/s

(b) To sketch ∆~pBC

, sketch ~pCtail-to-tail at the location of ~p

Band sketch ∆~p

BCfrom the head of ~p

Cto the head of ~p

B.

Do this for each of the other vectors as well. The results are shown in Figure ??.

(c)∣∣∣∆~p

BC

∣∣∣ is greatest because both ∆pxand ∆p

yare greatest (in magnitude) for the interval from B to C.

1.P.118

(a) Below is a sample program. Note that the initial position of the ball is at the left side of the origin. The linescene.autoscale=0 turns off autoscaling so that the camera will not zoom while viewing the simulation. The statementrate(100) is used to slow down the animation.

52

B

C D

EB

p

Cp

Dp

Ep

!pC

!!pBC

!!pCD

!!pDE!p

D!p

E

Figure 23: The change in momentum between successive points along the path of the projectile is drawn at each location.

1 from __future__ import d i v i s i o n2 from v i s u a l import ∗3

4 ba l l=sphere ( pos=(−1 ,0 ,0) , r ad iu s =0.1 , c o l o r=co l o r . ye l low )5

6 scene . au to s ca l e=07

8 vx=0.49 v=vecto r ( vx , 0 , 0 )

10

11 dt=0.0112

13 while 1 :14 r a t e (100)15 ba l l . pos = ba l l . pos + v∗dt

(b) In the sample program below, the wall is placed at x = 1.1 m and the width of the wall is 0.2 m. This places thesurface of the wall at x = 1 m. Use an if statement to detect the moment that the ball crosses x = 1 m. Note thatthe position of the ball refers to the location of the center of the ball; therefore, one must check whether the surface ofthe ball crosses x = 1 m.

At first you can put numbers into the if statement. However, it’s best to use attributes of the ball and wall, such asthe radius of the ball and size of the wall, in your if statement. This way, if you change the location of the wall or theradius of the ball, your simulation will still work.


4 ba l l=sphere ( pos=(−1 ,0 ,0) , r ad iu s =0.1 , c o l o r=co l o r . ye l low )5 wal l=box ( pos =(1 .1 ,0 ,0 ) , s i z e =(0 . 2 , 1 . 0 , 1 . 0 ) , c o l o r=co l o r . white )6


9 vx=0.410 v=vecto r ( vx , 0 , 0 )11

12 dt=0.0113

14 while 1 :15 r a t e (100)

53

16 ba l l . pos = ba l l . pos + v∗dt17 i f ba l l . pos . x + ba l l . r ad iu s > wal l . x−wal l . s i z e . x/2 :18 v = −v

(c) To make the velocity change at a constant rate, for instance, you must make the velocity change by a constant dvduring each time step. Thus, update the velocity inside the while loop.

In the following example, the x-velocity changes with each time step. The constant, 2 m/s2 in this example, specifiesthe change in velocity per second and is called the x-acceleration of the object. Basically, it tells you that the x-velocitywill increase an amount 2 m/s in each second. Since dt is less than a second, the change in the x-velocity during onetime step is 2dt.




8 vx=09 v=vecto r ( vx , 0 , 0 )

10

11 dt=0.0112

13 while 1 :14 r a t e (100)15 v . x = v . x + 2∗dt16 ba l l . pos = ba l l . pos + v∗dt

If you start the ball from rest, it’s easy to see that the object’s velocity is changing. However, if the initial velocity isnot zero and if the x-acceleration is small, say 0.1 m/s2 in this example, then the change in velocity is not perceptible.In the following example, the initial x-velocity is 0.2 m/s and the x-acceleration is only 0.1 m/s2.




8 vx=0.29 v=vecto r ( vx , 0 , 0 )

10

11 dt=0.0112

13 while 1 :14 r a t e (100)15 v . x = v . x + 0.1∗ dt16 ba l l . pos = ba l l . pos + v∗dt

One can hardly notice that the ball is accelerating.

(d) In VPython, use the graph package to create a graph and plot a function on the graph. You will need to create a newvariable t that is the clock reading (i.e. time). Add a statement to update the clock reading after each time step, usingt = t + dt. The following program plots x vs. t for the object.

54

1 from __future__ import d i v i s i o n2 from v i s u a l import ∗3 from v i s u a l . graph import ∗4



9 vx=010 v=vecto r ( vx , 0 , 0 )11

12 dt=0.0113 t=014

15 xGraph=gd i sp l ay ( x t i t l e= ' t ␣ ( s ) ' , y t i t l e= 'x␣ (m) ' , x=400 , y=0, width=400 , he ight=200)16 xPlot=gcurve ( c o l o r=co l o r . ye l low )17

18 while 1 :19 r a t e (100)20 v . x = v . x + 2∗dt21 ba l l . pos = ba l l . pos + v∗dt22

23 xPlot . p l o t ( pos=(t , b a l l . x ) )24

25 t=t+dt

You will notice that the graph of x vs. t is quadratic (i.e. a parabola).

(e) The previous program can easily be modified to plot x-velocity by simply editing the plot statement. It’s also a goodidea to change the graph labels, names of the objects, and plot color to reflect that it’s a graph of x-velocity vs. time.

1 from __future__ import d i v i s i o n2 from v i s u a l import ∗3 from v i s u a l . graph import ∗4



9 vx=010 v=vecto r ( vx , 0 , 0 )11

12 dt=0.0113 t=014

15 vGraph=gd i sp l ay ( x t i t l e= ' t ␣ ( s ) ' , y t i t l e= 'v_x␣ (m/ s ) ' , x=400 , y=0, width=400 , he ight=200)16 vPlot=gcurve ( c o l o r=co l o r . cyan )17

18 while 1 :19 r a t e (100)20 v . x = v . x + 2∗dt21 ba l l . pos = ba l l . pos + v∗dt22

23 vPlot . p l o t ( pos=(t , v . x ) )24

55

25 t=t+dt

You will notice that in this case, the x-velocity vs. time graph is linear.

1.P.119

(a) In the following program, the initial position of the object is in the lower, left quadrant and moves with a positivex-velocity and positive y-velocity.


4 ba l l=sphere ( pos=(−1,−1,0) , r ad iu s =0.1 , c o l o r=co l o r . ye l low )5


8 v=vecto r ( 0 . 4 , 0 . 4 , 0 )9

10 dt=0.0111

12 while 1 :13 r a t e (100)14 ba l l . pos = ba l l . pos + v∗dt

(b) Increasing the x-velocity and keeping the y-velocity the same makes the object move at a smaller angle with respect tothe +x axis. It also increases the speed of the object.

Increasing the y-velocity and keeping the x-velocity the same makes the object move at a larger angle with respect tothe +x axis. It also increases the speed of the object.

(c) In the following example, upon colliding with the wall, the x-velocity of the ball reverses direction, but its y-velocityremains constant. Note that the if statement checks to see if the position of the surface of the ball is greater than thelocation of the surface of the wall, and when this occurs, reverses the x-velocity of the ball.


4 ba l l=sphere ( pos=(−1,−1,0) , r ad iu s =0.1 , c o l o r=co l o r . ye l low )5 wal l=box ( pos =(1 .1 ,0 ,0 ) , s i z e =(0 . 2 , 1 . 0 , 1 . 0 ) , c o l o r=co l o r . white )6


9 v=vecto r ( 0 . 8 , 0 . 4 , 0 )10

11 dt=0.0112

13 while 1 :14 r a t e (100)15 ba l l . pos = ba l l . pos + v∗dt16 i f ba l l . pos . x + ba l l . r ad iu s > wal l . x−wal l . s i z e . x/2 :17 v . x = −v . x

56

1.P.120

The program will need a counter that counts up to N and then creates a sphere at the position of the object. When creatingthe sphere, it’s necessary to reset the counter to 0 so that it can count up to N again. An example is shown below. Thecounter variable is n.




8 v=vecto r ( 0 . 2 , 0 , 0 )9

10 dt=0.0111 N=10012 n=013

14 while 1 :15 r a t e (100)16 ba l l . pos = ba l l . pos + v∗dt17

18 n=n+119 i f n==N:20 sphere ( pos=ba l l . pos , r ad iu s =0.1 , c o l o r=co l o r . white )21 n=0

1.P.121

Here is a sample program that meets the requirements. Note that part (f) should be done by manually editing the parametersin part (c).


4 # The scene . mouse . g e t c l i c k ( ) s ta tement causes the program to pause5 # un t i l the user c l i c k s the mouse . This i s handy f o r s ee ing each6 # part o f the program in succe s s i on .7

8 # Problem 1.P.1219

10 # part ( a )11 bba l l = sphere ( pos=vecto r (−5 ,2 ,−3) , r ad iu s =0.5 , c o l o r=co l o r . red )12 scene . mouse . g e t c l i c k ( ) # pause u n t i l mouse i s c l i c k e d13

14 # part ( b )15 ba = arrow ( pos=vector ( 0 , 0 , 0 ) , ax i s=bba l l . pos , c o l o r=co l o r . green )16 scene . mouse . g e t c l i c k ( )17

18 # part ( c )19 vba l l = sphere ( pos=vecto r (−3 ,−1 ,3.5) , r ad iu s =0.5 , c o l o r=co l o r . b lue )

57

20 scene . mouse . g e t c l i c k ( )21

22 # part ( d )23 va = arrow ( pos=vecto r ( 0 , 0 , 0 ) , ax i s=vba l l . pos , c o l o r=co l o r . cyan )24 scene . mouse . g e t c l i c k ( )25

26 # part ( e )27 vb = arrow ( pos=bba l l . pos , ax i s=vba l l . pos − bba l l . pos , c o l o r=co l o r . white )28 scene . mouse . g e t c l i c k ( )29

30 # part ( f )31 # The problem a c t u a l l y wants you to e d i t the b a s k e t b a l l ' s p o s i t i o n and32 # v o l l y b a l l ' s p o s i t i o n above .33

34 # part ( g )35 print " ba sk e tba l l ' s ␣ po s i t i o n ␣ i s ␣" , bba l l . pos36 print " v o l l e y b a l l ' s ␣ po s i t i o n ␣ i s ␣" , vba l l . pos37 print "white ␣arrow ' s ␣ t i p ␣ i s ␣ at ␣" , vba l l . pos38 print "white ␣arrow ' s ␣ t a i l ␣ i s ␣ at ␣" , bba l l . pos

1.P.122

(a) Begin your program in VPython with import statements. Then, create a box to represent the track. By default inVPython, x is horizontal, y is vertical (i.e. toward the top and bottom of the screen), and z is out of (and into) theplane of the computer screen. This is important for getting the dimensions and orientation of the track correct.


4 t rack=box ( pos =(0 ,0 ,0) , s i z e =(2 , 0 . 05 , 0 . 1 ) , c o l o r=co l o r . green )

(b) The track is at y = 0 and has a height of 0.05 m. Thus the top surface of the track is at y = 0.025 m. You must placethe cart above y = 0, otherwise, it will be hidden by the track.


4 t rack=box ( pos =(0 ,0 ,0) , s i z e =(2 , 0 . 05 , 0 . 1 ) , c o l o r=co l o r . green )5 ca r t=box ( pos =(0 ,0 .05 ,0) , s i z e =(0 . 1 , 0 . 04 , 0 . 0 6 ) , c o l o r=co l o r . white )

(c) The y position of the bottom of the cart must be 0.01 m greater than the top surface of the track which is at y = 0.025 m.Since the cart’s height is 0.04 m, then the center of the cart must be at y = 0.01 m + 0.02 m + 0.025 m = 0.055 m.

The x position of the left end of the track is x = −1 m. The cart’s length is 0.1 m. Thus, the x-position of the cartmust be at x = −1 m + 0.05 m = −0.95 m. Change the initial position of the cart to these values.


4 t rack=box ( pos =(0 ,0 ,0) , s i z e =(2 , 0 . 05 , 0 . 1 ) , c o l o r=co l o r . green )5 ca r t=box ( pos =(−0.95 ,0 .055 ,0) , s i z e =(0 . 1 , 0 . 0 4 , 0 . 06 ) , c o l o r=co l o r . white )

(d) Define a variable v for the initial velocity of the cart.

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4 t rack=box ( pos =(0 ,0 ,0) , s i z e =(2 , 0 . 05 , 0 . 1 ) , c o l o r=co l o r . green )5 ca r t=box ( pos =(−0.95 ,0 .055 ,0) , s i z e =(0 . 1 , 0 . 0 4 , 0 . 06 ) , c o l o r=co l o r . white )6

7 v=vecto r ( 0 . 2 , 0 , 0 )

(e) In the following example, the rate() statement is used inside the while loop to slow down the animation.



7 v=vecto r ( 0 . 2 , 0 , 0 )8

9 dt=0.0110

11 while 1 :12 r a t e (1000)13 ca r t . pos = car t . pos + v∗dt

(f) Change the position of the cart to the right side of the track, and make its x-velocity negative.


4 t rack=box ( pos =(0 ,0 ,0) , s i z e =(2 , 0 . 05 , 0 . 1 ) , c o l o r=co l o r . green )5 ca r t=box ( pos =(0 .95 ,0 . 055 ,0 ) , s i z e =(0 . 1 , 0 . 04 , 0 . 0 6 ) , c o l o r=co l o r . white )6

7 v=vecto r (−0.2 ,0 ,0)8

9 dt=0.0110


(g) The cart will travel at an upward angle (+y direction) if vy is positive or at a downward angle (−y direction) if vy isnegative. Try using ~v = 〈0.2, 0.1, 0〉 m/s with the cart starting at the left end of the track, for example.



7 v=vecto r ( 0 . 2 , 0 . 1 , 0 )8

9 dt=0.0110


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(h) To show a trail, define the trail as a curve, preferably with the same color as the cart. Then, inside the while loop,append the position of the cart to the trail. See the example below for the proper syntax in VPython.



7 v=vecto r ( 0 . 2 , 0 , 0 )8

9 dt=0.0110

11 t r a i l=curve ( c o l o r=car t . c o l o r )12

13 while 1 :14 r a t e (1000)15 ca r t . pos = car t . pos + v∗dt16 t r a i l . append ( pos=car t . pos )

You may want the cart to stop when it reaches the end of the track. In this case, add an if statement inside the whileloop that checks to see if the right edge of the cart has exceeded the right edge of the track or if the left edge of the cart hasexceeded the left edge of the track, and if so, break out of the loop. Here is an example.


4 t rack=box ( pos =(0 ,0 ,0) , s i z e =(2 , 0 . 05 , 0 . 1 ) , c o l o r=co l o r . green )5 ca r t=box ( pos =(0 .95 ,0 . 055 ,0 ) , s i z e =(0 . 1 , 0 . 04 , 0 . 0 6 ) , c o l o r=co l o r . white )6

7 v=vecto r (−0.2 ,0 ,0)8

9 dt=0.0110

11 while 1 :12 r a t e (1000)13 ca r t . pos = car t . pos + v∗dt14 i f ( ca r t . pos . x + car t . s i z e . x/2 > track . pos . x + track . s i z e . x/2) \15 or ( ca r t . pos . x − ca r t . s i z e . x/2 < track . pos . x − t rack . s i z e . x/2) :16

17 break

The / in the condition of the if statement in the above example is used to signify that this is a multiline statement. It wasadded in this case to make the code more readable, but typically the condition of the if statement is on one line.

1.P.123

(a) It helps to place the ice just below y = 0 so that you can put the puck at y = 0 and not have it hidden by the ice. Hereis an example program that creates a 5 m long and 5 m wide piece of ice. The puck is placed in one corner.


4 i c e = box ( pos=(0 ,−0.05 ,0) , s i z e =(5 ,0 .1 ,5 ) , c o l o r=co l o r . white )5 puck = cy l i nd e r ( pos =(−2.4 ,0 ,2 .4) , ax i s =(0 ,0 .04 ,0) , r ad iu s =0.1 , c o l o r=co l o r . red )

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(b) Set values for the position of the puck to put it in one corner of the ice. Any corner is fine.

(c) Define a vector for the velocity and give it an x-component and z-component. For example, use v=vector(1,0,-1).

(d) Use a rate() statement to slow down the animation. An example program showing the puck move in the +x and -zdirection is given below.


4 i c e = box ( pos=(0 ,−0.05 ,0) , s i z e =(5 ,0 .1 ,5 ) , c o l o r=co l o r . white )5 puck = cy l i nd e r ( pos =(−2.4 ,0 ,2 .4) , ax i s =(0 ,0 .04 ,0) , r ad iu s =0.1 , c o l o r=co l o r . red )6

7 v=vecto r (1 ,0 ,−1)8

9 dt = 0.00110

11 while 1 :12 r a t e (100)13 puck . pos = puck . pos + v∗dt

chabay ch 1 solutions

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