HW 02 - Fa14 - Kinematics in 2D

5/20/2018 HW 02 - Fa14 - Kinematics in 2D

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HW 02 - fa14 - kinematics in 2D:chapter 4; 3, 4, 17, 21, 26, 47, 53, 56, 67, 82.

Chapter 4, problem 3 (9th

ed.) Chapter 4, problem 3 (10th

ed):A positron undergoes a displacement

(2.0 3.0 6.0 )r i j k m

, ending with the position vector (3.0 4.0 )r j k m

. What was the positrons initi

position vector?

Beginning with0 0

r r r r r r

, we write,

0

2.0 2.0

(3.0 4.0 ) (2.0 3.0 6.0 ) 3.0 3.0 6.0 2.0 6.0 10.0

4.0 6.0 10.0

r j k m i j k m m m i j k

(1


ed.) Chapter 4, problem 4 (10th

ed):The minute hand of a wall clock measures

cm from its tip to the axis about which it rotates. The magnitude and angle of the displacement vector of the t

are to be determined for three time intervals. What are the (a) magnitude and (b) angle from a quarter after t

hour to half past, the (c) magnitude and (d) angle for the next half hour, and the (e) magnitude and (f) angle f

the hour after that?

Caution: the question isnt asking for the angle between quarter past and half past (90 ), but rather the

angle between the two representative vectors,

half-past quarte r-past 10( ) 10( ) 10( ) ;f ir r r r r j i cm i j cm

(1

Converting (1.2) to polar form,

1 11010

10( ) 10 2 (tan 180 ) 10 2 (tan 1 180 ) 0.14 (180 45 ) 0.14 (225cmcm

r i j cm m m

WileyPLUS has the solution as 45 degrees, however, and it seems to be mistakenly referring to the following

angle (see how a schematic answers all, and facilitates understanding?),

(1

(c) and (d) repeating (1.2),

1on the hour half past hour 10 10( ) 20 20 90 ; tan 90 ;f ir r r r r j j cm j cm cm

(1

(e) In a full-hour sweep, the hand returns to its starting position, and the displacement is zero. (f) The

corresponding angle for a full-hour sweep is also zero. Proof:

on the hour on the hour 0 0 0 ;f ir r r r r

(1


2/6

Note the use of the zero vector, 0

, rather than the scalar-zero ( 0 0

).

chapter 4, problem 17 (9th

ed.) chapter 4, problem 17 (10th

ed.):A cart is propelled over an xy-plane

with acceleration components 24.0m

x sa , and 22.0

my s

a . Its initial velocity has components 0, 8.0m

x sv an

0,12 m

y sv . In unit-vector notation, what is the velocity of the cart when it reaches its greatest y-coordinate?

Solution:The problem statement fails to either say (1) the particle starts at the origin, or (2) has its greatest ydisplacement. We shall assume either (1) or (2); the maximum y-displacement occurs when the y-velocity is

The timeat which this occurs1is,

2

0,solve for

0,

120 6.0 ;

2.0

myt s

y y y my s

vv v a t t s

a

(1

Hence, the velocity (a vector!) is given by,

4.00,0, 0, 0, 0,0, 2.0

0

0, 0, 0,

8.0 (12) 32 ;

00

x x

y y

a a

y ax x x x y ay

x y

y y yy y

vv a v v vv m mv v i v j v at i

v a va s sv

We see that the y-velocity is 0, but this is precisely consistent with (1.7). Also, note use of column-vector as

well as unit-vector notation.


ed) Chapter 4, problem 21 (10th

ed):A dart is thrown horizontally with an init

speed of 0, 10m

x sv toward point P, the bulls-eye on a dart board. It hits at point Q on the rim, vertically bel

P, 0.19ft s later. (a) What is the distance PQ? (b) How far away from the dart board is the dart released?

Additional question: What is the distance from the throwers hand to the point Q?

(a) The vertical distance is given by,

2

2 2 2 21 1 1 10 0, 2 2 2 2

( ) 0 0 ( ) (9.81 )(0.19 ) 0.18 ;mf y f y f f f sh y t y v t a t g t gt s m (1

(b) The distance from the dart board is the easier quantity to calculate; we already made reference to it in (1.12 21

0 0, 0, 0,2( ) 0 0 (10 (0.19 )) 1.9 ;mf x f x f x f x f

sd x t x v t a t v t v t s m (1.

(c) The x and y coordinates are functions of time: ( )x x t and ( )y y t for 0 ft t ; hence, we compute the

Pythagorean resultant of these two coordinates,

1The displacement over this time (1.7) is:

2

2 20,solve for2 2 2

0,

(12 )2 0 36 ;

2 2( 2.0 )

myy s

y y y my s

vv v a y y m

a


3/6

2

2 2 2 2 2 21 10 0, 0 0,2 2

exact symbolic answer

2 2 2 2 2 21 10, 0,2 2

2 12

( ) ( ) ( ) ( )

(0 0) (0 0 ( ) ) ( ) ( )

(10 (0.19 )) ( (9.81

PQ f f x f x f y f y f

x f f x f f

m mPQ s s

r x t y t x v t a t y v t a t

v t g t v t gt

r s

2 2)(0.19 ) ) 1.91 ;

s m

(1.

Afterword: Note my rewriting of the problem-statement! Instead of saying with an initial speed of 10m/s, I assigned the symbol 0,xv to the abovementioned initial speed. This facilitates correctly plugging in

right quantities in the right places in the various equations youll encounter.



ed):A stone is catapulted at time 0t , with a

initial velocity of magnitude0,

20.0 mx s

v and at an angle of 40.0 above the horizontal. (Neglect air

resistance.) Find its horizontal and vertical displacements at {1.10 ,1.80 ,5.00 }t s s s after launch.

Letf

t be the time at which the stone strikes the ground. We can parameterize its vector displacement (that is

both components) as a function of time,

2 21 10, 0 0, 02 2

2

2

( ) ( ) ( ) ( ) cos sin

1 20.0 cos 40.0 sin 9.81 ; 0 ;2

x y

f

x t x t x y t j v t x i g t v t y j v t i v t gt j

m mt i v t t j t t

s s

(1.

The time of flight, however, is given by,

2

0, 0, 0,

0, 0,

2 2(20.0 )sin 40.022

sin( )

9.81.62 5.00 ;

my y y s

f y y f f m

s

v v v vv v v g t t

g g gs s

(1.

Therefore, the horizontal and vertical displacements are given by (1.12) not by plugging in 5.00t s , but by

plugging inft . Summarily, in column-vector notation,

27.58 40.1553

8.206 7.248

16.85(1.10 ) ; (1.80 ) ; (5.00 ) ( ) (2.62 ) ;

0.000f

m

m m

m mx s x s x s x t x s

m

(1.

chapter 4, problem 47 (9th

ed) chapter 4, problem 47 (10th

ed):A batter hits a pitched ball when the

center of the ball is 0 1.22y m above the ground. The ball leaves the bat at an angle of 45 with the

ground. With that launch, the ball should have a horizontal range (returning to the launch level) of 107x m (a) Does the ball clear a 7.32fenceh m -high fence that is 97.5fencex m horizontally from the launch point? (

At the fence, what is the distance between the fence top and the ball center?

Let ft be the time at which the ball strikes the ground. Then, the time fencet (for which 0f fencet t ) needed

reach the horizontal coordinate of the fence is given by,

solve for2 21 10 0, 2 2

0 cos (0) ;cos

fencet fence

fence x fence x fence fence fence fence

xx x v t a t v t t t

v

(1.


4/6

The problem also does not give the initial velocity, but does tell us the horizontal range 107x m .2

solve for107 sin 2 ;sin 2

vv g xx m vg

(1.

Then, this time is used to calculate the y-coordinate of the ball at time fencet (again, for which 0f fencet t ).

Remember, 0 1.22 0y m , since the pitch is some distance off the ground2,

2

2

21 10 0, 02 2

sin2

exact symbolic answer

2

0 2

( ) ( ) sin ( )cos cos

(9.81 )(97.5sin 2tan 1.22 97.5 tan 45

2 cos

fence fence

fence y fence fenceg x

mfence s

fence

x xy t y v t g t y v g

v

mxy x m m

x

2

2 2

)

2(107 ) cos9.

4877

5m

m

(1.

Yes, the ball clears. Notice that the exact symbolic answer is independent of gravitational field g!

chapter 4, problem 53 (9thed) chapter 4, problem 53 (10thed):In the figure, a baseball is hit at aheight 1.00h m and then caught at the same height. It travels alongside a wall, moving up past the top of th

wall 1.00t s after it is hit and then down past the top of the wall 4.00t t s later, at distance 50.0D

farther along the wall. (a) What horizontal distance is traveled by the ball from hit to catch? What are the (b)

magnitude and (c) angle (relative to the horizontal) of the balls velocity just after being hit? (d) How high is

the wall?

(1.

(a) Letf

t be the time when the ball is caught. Then, by symmetry,

wall above wall2 2 2(1.00 ) 4.00 6.00 ;f belowt t t t t t s s s (1.

The ball has a constant horizontal velocity of,

50.00 12.5 ;

4.00x x

x D m ma v

t t t s s

(1.

Hence, the horizontal distance travelled by the ball from hit to catch is the time (1.19) times the velocity (1.20

0,

210 0, 0,2

[horizontal range] 0 0 12.5 6.00 75.0 ;

x xv v

x f x f x f x f

mR x v t a t v t v t s m

s

(1.

(b) and (c) The magnitude of the initial velocity follows after calculating the y-component of the initial veloc

using the timeft in the equation

0, 0,( )

f y y y fv t v v a t , and subsequently using the Pythagorean formula,

2Again, we calculate an exact symbolic answer; you should do this regularly, as it is a good exercise of your algebra skills.

(Moreover, you should do this because I design quiz questions asking for exact symbolic answers).


5/6

22 2 2 2 2 2 2 250.01 1

0, 0, 2 4.00 2(0) ( ) ( ) ( ) ( (9.81 )(6.00 )) 31.97(12.5) (29.4)m m mDx y f

mt t s ss s

v v v gt s

0,1 1

0,

29.4tan tan

12.566.97

my s

mx s

v

v (1.

(d) this is given by the y-coordinate of the ball at either 1.00t s or 5.00t s ,

2

2 21 1 1 1 1

0 0, 2 2 2 2 2

( ) ( ) ( ) 1.00 (9.81 )(1 )(6 1) 25.53my y f f s

y t y v t a t h gt t gt h gt t t m s s m



ed):An Earth satellite moves in a circular orb

640 km above Earths surface with a period of 98.0 min. What are the (a) speed and (b) magnitude of the

centripetal acceleration of the satellite?The radius of the earth is 66.378 10ER m ,

5 6

2

2 232 ( )2 2 (6.40 10 6.378 10 ) 7.499 10 ;

(98.0 608.01 ;

)

ER

E

h RC r m m v vv a

T T T s s R

m

h R s

chapter 4, problem 67 (9

th

ed) chapter 4, problem 67 (10

th

ed):A boy whirls a stone in a horizontalcircle of radius 1.5 m and at height 2.0 m above ground level. The string breaks, and the stone flies horizonta

and strikes the ground after traveling a horizontal distance of 10 m. What is the magnitude of the centripetal

acceleration of the stone while in circular motion?

Solution:This is a projectile problem grafted onto a centripetal acceleration problem. Do the projectile par

first. Compute the amount of time the stone falls the height of 1.5 m, for 0 initial y-velocity,

2

solve for2 21 10 0, 2 2

2 2( 2.0 m) 41.5 m 0 0 ( ) 0.553 s;

9.81 3

t

y y m

s

yy y v t a t g t t

g

(1.

We find from this 0, /x xv v x t and2

tangential /ra v r that we can compute the centripetal acceleration from

this, given 10 mx , and2 2

tangential 0,xv v

r r ra . Also, take a moment to introduce the exact formula for arin

symbols/problem-givens,

2

222 2 2 2 20, 2tan

2 22 2

(9.81 )(10 m)( / )1.6 10 ;

2 2( 2.0 m)(1.5 m)( )

mx s

ry

g

vv x t x x g x ma

r r r t r yr sr

(1.



ed):A 200W m wide river has a uniform flo

speed of 1.1m

water sv

through a jungle and toward the east, 1.1m

water sv i

. An explorer wishes to leave a

small clearing on the south bank and cross the river in a powerboat that moves at a constant speed of 4.0 ms

w

respect to the water.Note that speed with respect to the water means, in contrast to the boats velocity with

respect to the un-moving3ground,

2 2

, ,4.0 ( ) ; 1.1 ;m mboat water rel x boat water y boat water s sv v v v v v v

(1.

3Well, actually, the ground ismoving: around the Sun and through the Milky Way galaxy, but thats not relevant to this problem.


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There is a clearing on the north bank 82d m upstream from a point directly opposite the clearing on the

south bank.Drawing a schematic,

(1.

(a) In what direction must the boat be pointed in order to travel in a straight line and land in the clearing on north bank?

Letf

t t be the arrival time of the boat, and let 0t be the departure time. Then, the x and y coordinates

the boat with respect to the un-moving ground4, and in terms of these, are,

, ,used statement ( )

, ,1 1 1

, ,

,

( )( ) ( ) ; ( ) ; 0 ; tan tan tan

( )

x water x boat fv v t d

f x boat f x water f

water x boat y boat f

f y boat f

x t v t d v tx x t v v t y y t v t t t

y t v t W

Hence, we are motivated to calculate ft , which follows from using the equation 210 0, 2( ) r rr t r v t a t for r

being the direction parallel to5the boats velocity,

2 2,2 2 2

, 2 2 2 2

, ,

2( ) ( ) 0;

x boat

f boat water f rel f x boat f f f

rel x boat rel x boat

dv W dd t v v t v t d v t W t t

v v v v

(1.

One solves (1.31) for ft using22 4

20 B B AC

AAx Bx C x (a.k.a., the quadratic formula), first by

calculating the (cumbersome) coefficients , ,A B Cin (1.31),

2 2 2 2 2,2 2

2 2 2 2 2 2 2 2, ,

2 2(82 )(1.1 ) ( ) (200 82 )

1 1; 12.2 ; 31(4.0 ) (1.1 59) (4.0 ) ( . .1 1 ) 1

mx boat s

m m m mrel x boat rel x boat s s s s

dv m W d m

A s B s Cv v v v s

Note that I indicated unitson the quadratic-formula coefficients, , ,A B C. We quickly look at the discriminan

and compare it to the value of B , 2 113.07 1 .24 2s BAC sB , indicating we take the upperof t

quadratic formula. Subsequently, we have 12.2 113.072(1)

62.63f

t s . Placing this time into (1.30), we have,

,1 1 8 37.032 (1.1 )(62.63 )

ta 320

;n tan0

mx water f s

d v t m s

W m

(1.

(b) How long will the boat take to cross the river and land in the clearing? 12.2 113.072(1)

62.63ft s

4Alternatively: these are the x and y coordinates the boat wouldhave if the stream was still.5I actually tried to avoid using the simplification this assertion avoids, but it produces a giant mess of nested square roots:

,

,

2 2,

,

1

, ,

2,1 1 2 2

, ,

,

tan tan ; tan ;

tan tan ; ( ) ;

y boat water

x boat

boat y boat

y boat

v sv dx boat y boat waterv W

v vx boat

boat water x boat water y boat v

y boat

v v sv

vv v v v v

v

HW 02 - Fa14 - Kinematics in 2D

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Transcript of HW 02 - Fa14 - Kinematics in 2D