hw#3 vectors and projectile motion - university of...

HW3 Vectors and projectile motion

Problem 1: Answer the following question concerning vectors.

Part (a) From the given list choose all that are examples of vectors.

1) Force. 2) Speed. 3) Velocity. 4) Mass. 5) Volume. 6) Acceleration. 7) Temperature.

Answer: 1) Force 2) Velocity and 6) Acceleration are vectors. All other quantities do not have direction

and they are scalars.

Problem 2: Consider the two vectors A and B. You know the magnitudes of these vectors (1 m and 10 m respectively), but you do not know anything about their directions.

Part (a) If a vector C is defined to be the sum of these two vectors (i.e. C = A + B) which of the following are true about the magnitude of C? Choose all that apply. MultipleSelect : 1) Cmin = 9 2) Cmin = (12 + 102)0.5 3) Cmax = 11 4) Cmin cannot be determined. 5) Cmax = 10 6) Cmax = (12 + 102)0.5 7) Cmin = 0 8) Cmin = -9

1) Correct. C is minimum when

2) Wrong. Cannot be right if 1) is correct.

3) Correct. C is maximum when






AB

AB

Problem 3: Vector A has a given magnitude of |A| and points due East. Vector B of unknown magnitude points directly North. The sum A + B makes an angle of θ degrees North of East. Part (a) Write an expression for the magnitude of vector B in terms of the magnitude of A, |A|, and the angle θ, using any required trigonometric functions. Part (b) What is the numerical value for the magnitude of the difference of vectors A and B, |A - B|? Part (c) What is the numerical value for the magnitude of the sum of vectors A and B, |A + B|? Numeric : A numeric value is expected and not an expression.

Solution:

(a)

tan |A||B| |A|

|B| tan

(b) I redraw the figure for clarity:

|A|

|B|

A

B

|A|

P

O

Q

R

|B|

|A + B| |A ‐ B|

cos

|A||B-A|

|B-A|

|A| cos

so (SAS), QRO tocongruent is PORBut

|B-A|

|A| cos

(c)

cos

|A| |BA|

|BA|

|A| cos

|A|

A+B

Problem 4: Find the following for path B in the Figure shown where length of one block is l = L m. Part (a) Find the total distance traveled in km. Part (b) Find the magnitude of the displacement in meters. Part (c) Find the angle, in degrees, of the direction of the displacement measured North of East.

Solution:

(a) Path B: 4 blocks east, 3 blocks north, and 3 blocks west, total distance traveled = 4+3+3 = 10 blocks.

total distance traveled = 10L meters = 10L/1000 = L/100 km

(b)

3.162LL 1010L L)((3L) |r| 222

(c)

o1- 71.57 3 tan 3 L

3L tan

4L

3L

3L

r 3L

4L‐3L=L

4L

3L

3L

r 3L

4L‐3L=L

|r|

|r|

The direction of the displacement is 71.57o North of East.

m/s 4.409

94 cos 6.72

)22.5(26.5 cos 6.72 Vo

ooTot x

m/s 5.072

94sin 6.72

)22.5(26.5sin 6.72 Vo

ooyTot

Problem 5: The figure depicts the sum of two velocities. Part (a) Find the component of vtot along the x axis in m/s. Part (b) Find the component of vtot along the y axis in m/s. Solution: (a) (b)

m/s 6.354

91 cos 6.72 V oTot x'

m/s 2.188

91sin 6.72 V oy'Tot

Problem 5: The figure depicts the sum of two velocities. Part (a) Find the component of vtot along an x axis rotated 30° counterclockwise relative to those in the Figure. Part (b) Find the component of vtot along a y axis rotated 30° counterclockwise relative to those in the Figure. Solution: Let us do some simple geometry first. My first goal is to figure out the angle between vB and the x‐axis. +30o = 22.5o + 26.6o +30o = 49o = 19o (a) (b)

x'

30o

O D

y'

222y

2x

ooo

yyyy

ooo

xxxx

C)3256.0B9613.0A1305.0(C)9455.0B2756.00.9914A ( DD |D|

C3256.0B9613.0A1305.0

C)3256.0()B9613.0(A)1305.0 (

)19Csin ()Bcos16()sin7.5A (

C BA D

C9455.0B2756.00.9914A

C)9455.0()B2756.0(0.9914A

)19 Ccos()Bsin16(cos7.5A

C BA D

CBA D 0 DCBA that Note

Problem 7: A farmer wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as A, B, and C in the figure, where A = A km, B = B km, and C = C km and then correctly calculates the length and orientation of the fourth side D. Part (a) What is the length of the vector D in kilometers? Part (b) What is the orientation of the vector D, in degrees W of S? (a)

|C3256.0B9613.0A1305.0|

|C9455.0B2756.00.9914A | tan

|D|

|D| tan

|D|

|D|

sideAdjacent

side Opposite tan

1-

y

x1-

y

x

(b)

WSWS

S

WSSW

VV V

Vfor asking is problem The

VV V

waterof velocity - ship ofVelocity

water torelative ship ofVelocity

sin VV2V V

sin VV2V ) sin cos (V

) sin Vsin VV2V ( cos V

)sin VV () cos (V VV |V|

)V(V sin VV V VV V

)0(V cos V cos V 0 V VV V

WSW2

SWW

WSW2

SW2222

W

222WWSW

2SW

22W

2WSW

2W

2Sy

2SxS

SWSWyWSWSyWySWySy

SWxWWSxWxSWxSx

Problem 8: A ship sets sail from Rotterdam, The Netherlands, heading due north at VSW m/s relative to the water. The local ocean current is VW m/s in a direction north of east. Part (a) What is the magnitude of the velocity of the ship relative to the Earth in m/s? Part (b) What is the angle of the velocity of the ship relative to the Earth in degrees N of E? Solution (a)

VSW

VW

VS

N

E

VSW

VW

VS

E

N

cos V

sin VVtan

cos V

sin VV

|V|

|V| tan

)V(V sin VV V VV V

)0(V cos V cos V 0 V VV V

W

WSW1-

W

WSW

Sx

Sy

SWSWyWSWSyWySWySy

SWxWWSxWxSWxSx

(b)

VSW

VW

VS

E

N

tot

player 1

tot

player

tot

player

puck

puck

player puck

player yplayer ytot playerpuck

yplayer ytot ypuck

tot puck

player xtot tot xtot puck

player xtot xpuck x

playertotpuck

V

Vtan

V

V tan

V

V

cos V

sin V )1/()2(

(2)--- V sin V

)V V and 0(V V sin V

VV V

(1)--- V cos V

0) V and V(V 0 V cos V

VV V

VV V

Problem 9: An ice hockey player is moving at Vplayer m/s when he hits the puck toward the goal. The speed of the puck relative to the player is Vpuck m/s. The line between the center of the goal and the player makes a 90.0° angle relative to his path as shown in the figure. Part (a) What angle must the puck’s velocity make relative to the player (in his frame of reference) to hit the center of the goal in degrees? Solution: The given figure and the way it names variables is actually kind of confusing, so let me clarify here: Vplayer is the velocity of player relative to the Earth, no confusion here. Vpuck is actually the velocity of the puck relative to the player, so be careful here. Actually it is clear if you read the problem text carefully. Vtot is the velocity of the puck relative to the Earth. So Velocity of puck relative to player = velocity of puck – velocity of player will give Vpuck = Vtot ‐ Vplayer.

g

v t gt -v 0 )4(

0point vhighest At

(a)

(4)--- gt - v vat vv

(3)--- tg 2

1 t vy ta

2

1 t vy y

(2)--- vv

(1)--- t v x t v xx

:motion of Equations

0y0y

y

0yyyiy

20y

2yyii

xix

0xxii

Problem 10: A famous golfer strikes a golf ball on the ground, giving it an initial velocity v = v0xi + v0yj. Assume the ball moves without air resistance and its motion is described using a Cartesian coordinate system with its origin located at the ballʼs initial position. Part (a) Determine the maximum height above ground, hmax in meters, attained by the golf ball. Part (b) Express the total horizontal distance, xmax, the ball travels when it returns to ground level in terms of v0x, v0y, and g. Part (c) Evaluate the total horizontal distance, xmax in meters, the ball travels when it returns to ground level. Solution:

x

y

v0x

v0y

g

v2v

g

2v v x (1) into thisSubstitute

g

2v t again when ground thehill willball theso point, launching the toscorrespond 0t

g

2v t 0or 0 t

0 t g 2

1 v0or 0 t

t)g 2

1 v t( 0 tg

2

1 t v 0 )3(

0y ground, toreturns ball golf When the

:(c) and (b)

g

v

2

1

g

v

2

1

g

v

g

v g

2

1

g

v vy (3) into thisSubstitute

0y0x

0y0xrange

0y

0y

0y

0y2

0y

20y

20y

20y

2

0y0y0yhighest

g

sin 2v0

g

sin 2v t ground above football for the time total

g

sin 2v t 0or 0 t

g

2v t 0or 0 t

0 t g 2

1 v0or 0 t t)g

2

1 v t( 0 tg

2

1 t v 0 )3(

0y ground, toreturns football When thec)(

cos v vb)(

sin v v(a)


(3)--- tg 2

1 t vy ta

2

1 t vy y

(2)--- vv

(1)--- t v x t v xx


00total

0

0y

0y0y2

0y

00x

00y

0yyyiy

20y

2yyii

xix

0xxii

Problem 11: A quarterback throws a football with an initial velocity v at an angle θ above horizontal. Assume the ball leaves the quarterback’s hand at ground level and moves without air resistance. All portions of this problem will produce algebraic expressions in terms of v, θ, and g. Let the origin of the Cartesian coordinate system be the ballʼs initial position. Part (a) Write an expression for the magnitude of the football’s initial vertical velocity v0y. Part (b) Find an expression for the magnitude of the football’s initial horizontal velocity v0x. Part (c) Write an expression for the total time, ttotal, the football is above the ground. Solution:

x

y

v0x

v0y

gt -sin v sin cos

cos v

gt -sin v vsin gt -sin v v )4( cos

cos v v cos v cos v )2(

sin v vand cos vv

v.is ground respect to with anglean makes arrow theepoint wher at the velocity theIf

(4)--- gt -sin v vat vv

(3)--- tg 2

1 sin t vy ta

2

1 t vy y

(2)--- cosvvv

(1)--- cost v x t v xx


000

0

0000y

0000

yx

00yyiy

200

2yyii

00xix

00xii

Problem 12: An arrow is fired upward with an initial velocity v0 at an angle θ0 above horizontal. Assume the arrow moves without air resistance. Use a Cartesian coordinate system with the origin at the arrowʼs initial position to analyze the arrowʼs motion. Part (a) Some time later the arrow makes an angle of θ degrees with respect to the horizontal. Write an expression for the time, t, that passes between when the arrow was fired and this later point.

Solution:

x

y

v0

0

gcos

) -(sin v t

cos

) -(sin

g

v t

cos

sin cos- cossin

g

v t

sin cos

cossin

g

v t

sin cos

cosv- sin vgt

00

00

000

00

0

0\000

g

gd2sin v sin v- t

2g

g(-2d)4)sin v2( sin v2 t

0 2d-t )sin v2( tg

tg t )sin v2( 2d-

tg 2

1 sin t v d- (3)

d- y cliff, theof bottom thehits stone When the

(4)--- gt -sin v- vat vv

(3)--- tg 2

1 sin t v- y ta

2

1 t vy y

(2)--- cosvvv



2200

200

02

20

20

0yyiy

20

2yyii

0xix

0xii

Problem 13: A student throws a stone from the top of a cliff at a distance d m above a horizontal plane with initial velocity v0 m/s at an angle θ below horizontal. The stone moves without air resistance; use a Cartesian coordinate system with the origin at the stoneʼs initial position. Part (a) With what speed, vf in m/s, does the stone strike the ground? Part (b) If the stone had been thrown from the clifftop in the same manner but upward instead of downward, would its impact velocity be different?

Solution: (a)

y

x O

y= ‐d

v0

1) sin cos ( gd2 v

gd2sin v cos v

)gd2sin v-() cos (v vv |v|

cos v v (2)

gd2sin v- v

)gd2sin v-sin v sin v- v

)gd2sin v sin -v( sin v- v

g

gd2sin v sin v- g sin v- v )4(

g

gd2sin v sin v- t

2220

220

220

2220

20

2fy

2fxf

0fx

220fy

22000fy

22000fy

2200

0fy

2200

, so t has two roots, one positive and one negative. We will take the positive root .

y

x O

y= ‐d

v0

Negative root corresponds to this point Positive root corresponds to this point

sin v gd2sin v 022

0

gd2sin v v

)g

gd2sin v sin vg( -sin v vgt -sin v v )4( New

g

gd2sin v sin v t t,positive Taking

g

gd2sin v sin v t

2g

g(-2d)4)sin v2( sin v2 t

0 2d-t )sin v2( tg

tg 2

1 sin t v d- (3) New


(3)--- tg 2

1 sin t v y ta

2

1 t vy y

220yf

2200

0yf0yf

2200

2200

200

02

20

0yyiy

20

2yyii

(b) If the stone s thrown in similar manner but upward in the of downward, this will only affect the sign of some terms in the equations of motion. Equation (3) and (4) has to be rewritten as: So we have the same vyf as if the stone is thrown downward. We can conclude that the impact velocity should be the same.

sense?) (make v- v

2v - v v

g

2vg - v v (3) into

g

2v t Substitute (a)

:(a)part Now

g

2v0

g

2v of period afor ground abovestay willball The

g

2v t again when ground thehill willball theso point, launching the toscorrespond 0t

g

2v t 0or 0 t

0 t g 2

1 v0or 0 t t)g

2

1 v t( 0 tg

2

1 t v 0 )3(

0y ground, toreturns football When the(b)

:first (b)part Answer


(3)--- tg 2

1 t vy ta

2

1 t vy y

(2)--- v v vv

(1)--- t v x t v xx


0yy

0y0yy

0y

0yy

0y

0y0y

0y

0y

0y0y2

0y

0yyyiy

20y

2yyii

x0x xix

0xxii

Problem 14: A ball is kicked with an initial velocity of V0x m/s in the horizontal direction and V0y m/s in the vertical direction. Part (a) At what speed does the ball hit the ground in m/s? Part (b) For how long does the ball remain in the air in seconds? Part (c) What maximum height is attained by the ball in meters? ` Solution

v0y

x

y

v0x

vv

)-v(v vv |v| speed a with ground hit the willball the

v v (2)

20y

2x0

20y

2x0

2y

2x

x0x

(c)

g

v

2

1

g

v

2

1

g

v

g

v g

2

1

g

v vy (3) into thisSubstitute

g

v t gt -v 0 )4(

0point vhighest At

20y

20y

20y

2

0y0y0yhighest

0y0y

y

00

o0

0

20

0000

20

002

0

000

0

00

00

00002

00

00yyiy

200

2yyii

00xix

00xii

sin2

gR v

)90 ,0( sin2

gR v

) cossin 2 (sin2 g

sin2 v R

g

cossin 2v R

cos g

sin 2v v R )1(

g

sin 2v t again when ground hit the willball theso point, launching the toscorrespond 0 t

g

sin 2v or t 0 t

0 t g 2

1 sin or v 0 t t)g

2

1 sin v t( 0 tg

2

1 sin t v 0 )3(

0y again, ground thehits ball When thea)(


(3)--- tg 2

1 sin t vy ta

2

1 t vy y

(2)--- cosvvv



o

Problem 15: A football player punts the ball at a 0 angle. Without an effect from the wind, the ball would travel R m horizontally. Part (a) What is the initial speed of the ball in m/s? Part (b) When the ball is near its maximum height it experiences a brief gust of wind that reduces its horizontal velocity by v m/s. What distance does the ball travel horizontally? Solution

v0

x

y

0

tg 2

1

4cos

R 0 )3(

0y again, ground thehits ball When thea)(

(8)--- gt - vat vv

(7)--- tg 2

1

4cos

R y tg

2

1 t 0

4cos

Ry ta

2

1 t vy y

(6)--- v- cosvvv

(5)--- v)t - cos(v x t v xx


2

0

yyiy

2

0

2

0

2yyii

00xix

00xii

4cos

R

cos 2sin

gR

g

sin

2

1

(a))part from sin2

gRv,90 ,0(

sin2

gR

g

sin

2

1

g

sin v

2

1

g

sin v

2

1

g

sin v

g

sin v g

2

1

g

sin v vy (3) into thisSubstitute

g

sin v t gt -v 0 )4(

0point vhighest At

0

00

0

0

20

o0

0

0

02

0

02

002

0

2

00000yhighest

000y

y

o

0 4cos

R

(b) Redefine coordinates and reset time after reaching the highest point:

x

y

v0x‐x

v0

0 R/2 R’

00

00

00

0

00

2

00

00

0000

00

000

00

0

0

2

0

2

2gcos

Rv)-

tan 2

gR(

2

R

R'2

R football by the travelleddistance Horizontal

2gcos

Rv)-

tan 2

gR(

2gcos

Rv)-

2sin

gRcos(

2gcos

Rv)- cos

cos2sin

gR(

2gcos

Rv)- cos

sin2

gR(

2gcos

Rv)- cos(v R'

sin2

gR v(a)part fromBut

2gcos

Rv)- cos(v x

v)t - cos(v x )5(

2gcos

R t

2gcos

R t

4cos

R tg

2

1

hw#3 vectors and projectile motion - university of...

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