Solution+VMV+Ch3,+5 7+FyBNVCO08+Circular+Motion,+Energy+Momentum

8/3/2019 Solution+VMV+Ch3,+5 7+FyBNVCO08+Circular+Motion,+Energy+Momentum

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Test V1 Ch3, 5-7 FyBNVC08 Projectile, Circular Motion, Linear Momentum, Energy NV-College

Physics BProjectile, Circular Motion, Energy and Linear Momentum

Instructions:

Time: 70 minutes: 11:00-12:10

Note: If your goal is MVG try to solve the problems in the following

order: 3, 4, 5, 2, 1. You should try to solve first those problems that

look familiar. Divide the time accordingly. The time available is

only 11:00-12:10.

The Test Warning! There are more than one version of the test.

At the end of each problem a maximum point which one may get for a correct solution

of the problem is given. (2/3/) means 2 G points, 3 VG points and an MVG quality.

Tools Approved formula sheets, ruler, and graphic calculator. You may use one page of apersonalized formula sheet which has your name on it. This should be submitted along

with the test.

Grade limits: There are 5 problem, 3 of which MVG type. The test gives a maximum of 28 points

of which 22 are VG points.

Lower limits for examination grade

Pass (G): 7 points

Pass with distinction (VG): 15 points of which at least 9 VG-points in addition to G+ in

the G-test.

Pass with special distinction (MVG): 19 points of which at least 9 VG-points in

addition to G+ in the G-test.

Problems number 4 and 5 are heavily graded and are of greatest importance for both

VG and MVG. You may choose to solve these problems before solving the others.

Name Grade

Problem 1a 1b 2a 2b 2c 3 4 5 Total Minum Qualification

G 0 0 0 0 0 2 2 2 6 7

VG 3 2 2 3 2 2 4 4 22 15

MVG 19

G

VG

MVG

[email protected] to use for educational purposes. Not for sale. 1/12


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1 A rocket is shot into the air. At the moment it reaches its highest point , a horizontaldistance d from its starting point, a prearranged explosion separates it into two parts

Im and

IIm .

is stopped in the midair and falls vertically to the Earth.Im

a. If III mm 3= , where then would IIm land? [0/3]b. If III mm 3= , where then would IIm land? [0/2]

Data: III mm 3= , dxCM 2= , dxI =

Problem: ?=IIx ,

Suggested Solution:Assuming the origin of the projectile as the origin of the system: andusing the definition of centre of mass:

III

IIIIIICM

mm

xmxmx

+

+=

3

7

3

838

4

32

3

32

dddxxdd

xdd

mm

xmdmd IIII

II

II

IIII =

=+=+

=/+/

/+/=

Answer: It will land3

7dxII = from the launch site.

Similar to above, but this time with III mm 3=

III

IIIIIICM

mm

xmxmx

+

+=

dddxxdd

xd

dmm

xmdm

d IIIIII

IIII

IIIIII

538384

3

23

3

2==+=

+=

/+/

/+/=

Answer: It will land dxII 5= from the launch site.



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2 A ball of mass m is attached to a horizontalcord of length L whose other end is fixed.

2 A ball of mass m is attached to a horizontalcord of length L whose other end is fixed.

a) If the ball is released, what will be thetension in the cord at the lowest point of

balls path? [0/2]

a) If the ball is released, what will be thetension in the cord at the lowest point of

balls path? [0/2]

b) A peg is located a distance h directlybelow the point of attachment of thecord. If Lh 80.0= , what will be thetension in the cord when the ball reaches

the top of its circular path about the peg?

b) A peg is located a distance h directlybelow the point of attachment of thecord. If Lh 80.0= , what will be thetension in the cord when the ball reaches

the top of its circular path about the peg?

[0/3][0/3]

c) Find the position of a peg such that inthe rotation of the ball about the peg the

tension in the cord is zero. [

c) Find the position of a peg such that inthe rotation of the ball about the peg the

tension in the cord is zero. [

0/2]

uggested solution:

0/2]

uggested solution:SSa) Answer: The tension on the cord at the lowest point of the balls

rotation, i.e. point B is mgFBT 3=

Using conservation of en rgy we may calculate the velocity of thee

ball at the lowest point of its path:

gLvgLvmvmgL 221 22 === .BBB2

Considering the fact that the radius of the circularmotion is identical to the length of the cord, Lr= . Ttension in the cord at the lowest point of its path is

he

mgLvv BB +==

22

mFr

mmgF TBTB

( )mgFmg

L

LgmFmg

L

gLmF

gLv

mgL

vmF

BTTBTB

B

BTB

322

2

22

=+/

/=+=

=

+=

b) The speed of the ball when it reaches the top of itst the peg, i.e. pointcircular path abou A , may also be

t:calculated from the conservation of energy statemen

mgLmgLmgLmvLmgmvmgL AA 6.04.02

14.0

2

1 22 ==+=

gLvgLv AA 2.12.12 ==

This is due to the fact that, the potential energy of thelowest point is assumed zero, and its highest point in the

ius n i

sion in the cord at point

circular motion of radius L2.0 is L4.0 above the ground

zero.Considering the fact that the rad of the circular motio

the ten

s Lr 2.0= ,A , , may be calculated as:TA

F

mgL

vmFmgL

vmFr

vmmgF ATA

TAA

TA ==+2.02.0

222A =

Bm

mg

TBF

Ra

ARa

mg

m

TAF


4/12



( )mg

L

LgmFmg

L

gLmF

gLv

mgL

vmF

TATA

ATA

/

/==

=

=

2.0

2.1

2.0

2.1

2.1

2.0

22

mgFmgmgFTATA

56 == Answer: mgFTA 5=

c) The speed of the ball when it reaches the top of its circular pathabout the peg, i.e. at point may als be calcu eC o lated from th,

conservation of energy statement:

( )( ) mghmgLmvmgLhLmgmvmgL CC 222

12

2

1 22 +=+=

gLghvgLghvvghgLgL CCC 2422

1

2

122

222 ===+

gLghvC 242 = gLghvC 24 =

This is due to the fact that, the potential energy of the lowest point is

assumed zer , an its highe

hecircular ion in the

o d st point is( )hL 2 above the ground zero.Considering the fact that the radius of t

motion is ( )hL , the tenscord is :

mghL

vmFmg

hL

vmF

r

vm CTC

CTC

CTC

=

=

222

mgF =+

( )

mghL

gLghmF

gLghv

mghL

vmF

TC

C

CTC

=

=

=2

2

24

24

( )mg

hL

Lhgmmg

hL

gLghmFTC

=

=

2224

( )mg

hL

LhgmFTC

=

22

Note that the condition requires:0TCF

( ) 05

335242422 ++ LhhLhhLLh

TCFLhLhLhL

Therefore at the limit the tension in the cord is zero: .Lh 6.0= 0=TCF

Answer: 06.0 == TCFLh

Ramg

m C

( )mg

hL

LhgmFTC

=

22


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In assessing your work with problems 8, 9 and 10 your teacher will pay extra attentio to:In assessing your work with problems 8, 9 and 10 your teacher will pay extra attentio to:

How well you plan and carry out the task. How well you plan and carry out the task. Which priciples of physics you use and how you justify using them Which priciples of physics you use and how you justify using them How general your solutions are How general your solutions are How well you justify your conclusions How well you justify your conclusions How well you cary out your calculations How well you cary out your calculations How well you present your work How well you present your work How well you use physical and matematical language. How well you use physical and matematical language. How clear your solutions are. How clear your solutions are.

3 Uniform Circular Motion, Linear Momentum, Projectile Motion [2/2/]In a physics lab, a small cube slides down a frictionless

incline as illustrated in the figure below, and eventuallystrikes a cube at the bottom that is only one-half its mass.

The collision is perfectly elastic.

a. If the mass kgM is released from the heightmh and hits the mass kgm and the table is

mH above the floor where does each cube land?

b. If the is cmh 30= height and the table iscm off the floor, where does each cube

land?

H 90=

Suggested Solutions:

The problem may be divided to 3 partsPart I:

kgM is released from above the table its velocity just before hitting

the mass may be calculated using conservation of mechanical

energy:

mh

kgm

ghvghvvMghM 222

1 22 ==/=/ ghv 2=

Part II: At the moment of elastic collision. If we assume that the massMcontinues its motion to the right at velocity and the mass

moves also to the right at :

smv M /1 m

smv m /1

Conservation of Linear Momentum: Assuming the direction to the right ispositive:

mM mvMvMv 11 += Using mM 2=

MmmM vvvvvv 1111 2222 =+= ( )Mm vvv 11 2 = conservation of mechanical energy:

2

1

2

1

22

1

2

1

2 222

1

2

1

2

1mMmM vvvmvMvMv +=

/+

/=

/

( ) ( )( MMMm vvvvvvv 1121221 22 +== )


6/12



( )( )

( )( )( )

( ) MmM

MM

m

mDivide

Mm

MMmvvv

vv

vvvv

v

v

vvv

vvvvv11

1

11

1

2

1

11

11

2

1

2

2

2

2+=

/

+/=

=

+= Mm vvv 11 +=

vvvvvvvvv

vvv

vvv

vvv

mmm

added

Mm

Mm

Mm

Mm

3

4

344322

222

22 11111

11

11

11===

=

+=

=

+=

vv m 3

41

=

vvvvvv

vvv

vvv Mm

mM

Mm3

1

3

4

3

4 11

11

11 ==

=

=

+= vv M3

11 =

==

==

=

=

=

3

24

3

4

3

2

3

1

3

4

3

1

2

1

1

1

1

ghvv

ghvv

vv

vv

ghv

m

M

m

M

3

21

ghv M= ,

3

241

ghv m =

Part III: The block M is projected horizontally to the right at3

21

ghv M= ,

we may calculate where it lands after falling mH

hHg

Hghx

g

Ht

g

HtytgHy

tgh

tvx

M

MM

3

22

3

2

220

2

1

3

2

22

1

==

====

==

Similarly

hHg

Hghx

g

Ht

g

HtytgHy

tgh

tvx

m

mm

3

82

3

24

220

2

1

3

24

22

1

==

====

==

Answer: The mass M will land at hHxM3

2= , and the mass will land atm

hHxm3

8= to the right of the edge of the table.

If , and :cmh 30= cmH 90=

( ) ( ) cmmg

hHx

m14039.190.030.0

3

8

3

8===

( ) ( ) cmmg

hHx

M353464.090.030.0

3

2

3

2===

Answer: The mass M will land at cmxM 35 , and the mass will land atm

cmxm 140 to the right of the edge of the table.


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h

xH

gh

xgtgHy

gh

xtt

ghx mmM

MM

=

==

==

4

9

2

3

2

1

2

1

2

3

3

2 22

2

tgh

xM =3

2;

h

xHy mM

=

4

9 2

( ) ( )

( )( )hyH

g

yHght

ghx

g

yHt

g

yHtyHtgtgHy

MM

M

MMMM

=

==

=

===

3

22

3

2

3

2

22

2

1

2

1 222

2

2

1tgHyM = ; ( )hyHx MM =

3

2

Similarly

h

x

Hgh

x

gtgHygh

x

tt

gh

x

mm

m

M

m

=

==

==

64

9

24

3

2

1

2

1

24

3

3

24 22

2

tgh

xm =3

24,

h

xHy mm

=

64

92

( ) ( )

( )( )hyH

g

yHght

ghx

g

yHt

g

yHtyHtgtgHy

mm

m

mmmm

=

==

=

===

3

82

3

24

3

24

22

2

1

2

1 222

2

2

1tgHy

m

= ; ( )hyHxmm

=3

8

0

10

20

30

40

50

60

70

80

90

0 20 40 60 80 100 120 14

x

y

0

h

xHy mm

=

64

9 2

hxHy mM

=4

92


8/12



4 Milky Way Galaxy and the missing mass [2/4/] TheSun is an average middle age star which lies at the outer edge of one of the spirals of

our Milky Way Galaxy. It is about three hundredthousand light-years from the centre of the galaxy,

4 Milky Way Galaxy and the missing mass [2/4/] TheSun is an average middle age star which lies at the outer edge of one of the spirals of

our Milky Way Galaxy. It is about three hundredthousand light-years from the centre of the galaxy,

lyR5100.3 = , and it rotates about the centre of the

galaxy. It takes about two hundred million years tomake one complete revolution about the centre of the

galaxy, yT8

100.2 = .

i Calculate the average tangential velocity at whichthe Sun makes this revolution.

ii Estimate the total mass of the Milky Way galaxy.For simplicity you may assume that all mass of the galaxy is concentrated at the

centre of the super black hole that lies at the centre of the Milky Way galaxy.

iii If all stars had about the same mass of our Sun, i.e. kg , how manystars would there be in the Milky ways.

MSun30102

iv Through the visible light astronomy we estimate there are about 100-200 billionstars in an average galaxy, including Milky Way galaxy. Is there any discrepancy

between your results and that of the observations? If so, discuss the possible reason

for the discrepancy.

Data: ,kgMSun30102 lyR 5100.3 = , yT 8100.2 = , 2211 /1067.6 kgmNG

Suggested Solutions:

( ) mmlyR 21855 108.2360024365103103103 === mR 21108.2 ( ) syT 1588 103.6360024365100.2100.2 == sT 15103.6

i Answer: The average tangential velocity of the Sun about the centre ofthe galaxy is: sm /

6

.v 108.2

( )

sms

m

T

Rv /108.2

103.6

108.222 615

21

==

smv /108.2 6

ii The total mass of the Milky Way galaxy is about kg4410 .MM 4.3 For simplicity we may assume that all mass of the galaxy isconcentrated at the centre of the galaxy, and we denote it by .MM

According to Newtons Universal Gravitation law2

R

mMGF SM= .

kg

G

RvM

R

vm

R

mMG MS

SM 4422

2104.3

== kgMM

44104.3

Answer: If all stars had about the same mass of our Sun there should

be about 14107.1 n stars in the Milky Way.

14

30

44

107.1102

104.3

=

kg

kg

M

Mn

S

M .

iii According to our calculations there should be 14107.1 stars in ourGalaxy, but there are about 100-200 billion observable stars in theMilky way. This is about thousand times larger. The missing mass isthe so called Dark Matter which is the source of recent intensive

research in physics: 850100.2107.1 11

14

nObservatio

Theory

n

n


9/12



5 The Roller Coaster. [2/4/]Illustrated in the figure below is a section of a roller

coaster with a circular vertical loop. Jenny who is

interested of physics and whose mass is kgm takes a

bathroom scale with her to the amusing-park.

Assuming that the radius of the vertical loop is m

and at the highest point of the roller coaster Jenny is

mh above her lowest point in the vertical loop.

Jenny sits on the scale and registers the reading of thescale at some specific points named in the figure as

5 The Roller Coaster. [2/4/]Illustrated in the figure below is a section of a roller

coaster with a circular vertical loop. Jenny who is

interested of physics and whose mass is kgm takes a

bathroom scale with her to the amusing-park.

Assuming that the radius of the vertical loop is m

and at the highest point of the roller coaster Jenny is

mh above her lowest point in the vertical loop.

Jenny sits on the scale and registers the reading of thescale at some specific points named in the figure as

RR

A , B , C, and D . In the calculations below ignore

the friction. Draw in each case the free-body-

diagram.

i If Jenny is to remain on the track, even at the top of the vertical circular loop, findthe minimum height mh in terms of mR the radius of the vertical loop.

ii What is the minimum speed of Jenny at the point A such that she makes the turn?If her speed is above the minimum needed

iii Calculate the reading of the scale at the point A .iv Calculate the reading of the scale at the point B .v Find the difference between the reading of the scale at the top of the loop, A , and at

the bottom of the loop, B . What is your conclusion? Explain.

vi Calculate the reading of the scale at the point C.vii Calculate the reading of the scale at the point D .

Suggested solutionsi In order for the object to pass the top point

the normal force must be larger than zero. Itsminimum value at the top is zero. i.e.: 0NF .

Free-body-diagram for the object at the topmay be represented as:

At the top of the loop, i.e. at point A .Taking the direction of the acceleration as positive, Newtons second law,

amFnetr

r

= for the object passing the top may be written as:

CtopN mamgF =+_

h

R2

A

B

C

D

ARa

m

mgR

hFNA

= 5

2


10/12



Note that the apparent weight of any object, the weight the scaleshowes, is the normal force exerted by the surface on the body, i.e. NF .

R

vmmgF

top

topN

2

_ =+

RmgRFmv topNtop += _2

The total energy of the object at the top is: 2

2

12 toptop mvmgRE +=

( )RmgRFmgRmvmgRE topNtoptop ++=+= _22

12

2

12

At point A RFmgRE topNtop += _2

15.2

The minimum value of the apparent weigh is zero, i.e. 0NF . Therefore,

the minimum height is Rh = 5.2min . Due to the fact that friction, however

small it may be, and additional precautions and safety factors, in the reallife situation, the highest point of the roller coaster is usually muchhigher. Answer: Rh = 5.2min

Note that mgR2 is the potential energy of the object at the top of the loop,

i.e. at point A . The reference level for the potential energy is taken to bethe lowest point of the circle.

ii At point A as demonstrated above CtopN mamgF =+_ . Which resulted in

R

vmmgF topN

2

_ +top

= .

Minimum speed at the top is associated with minimum normal force,0_ topNF , and therefore Rh = 5.2min .

RgvRgvgmR

vm

R

vmmgF AA

AANA ==/+=/=+ min

2

min

2

min

2

0

Answer: Rgv = Amin Second method:

We may use conservation of energy, instead:

gRghvgRghvmvmgRmgh AAA 4242

2

12 min

2

min

22 ==+=

( ) gRvgRvgRgRvgRRgv AAAA ==== min2

min

2

min

2

min 4545.22

Answer: Rgv = Aminiii If Rh > 5.2 , we may rewrite equations in terms ofh :Conservation of mechanical energy require that

mgR

mgRmghF

mg

R

mvF

R

vmmgF

mgRmghmvmvmgRmgh

NA

ANA

ANA

AA

=

==+

=+=42

422

12

22

22


11/12



mgR

mghmgmg

R

mghmg

R

mgRmghFNA 5

24

242==

=

mgR

hmg

R

RhFNA

=

= 5

252

Answer: At point A , the scale1 will show mgR

hFNA

= 52 .

iv Answer: At point B the scale will show mg+1 R

hFNB

=

2

Free-body diagram for the object at the bottom ofthe loop, i.a. at point B may is illustrated in thefigure to the right:Taking the direction of acceleration, i.e. up,

positive, Newtons second law of motion, amFnetr

r

= ,

for the object at the bottom of the vertical circle

may be expressed as:R

vmmgF BNB

2

=

R

vmmgF BNB

2

+=

We may use conservation of energy to find the tangential velocity ofJenny at the bottom of the loop in terms ofR and .h

2

2

1BBi mvmghEE ==

mgR

Rh

R

ghmmgF

R

vmmgF

ghvmvmgh

NB

BNB

BB

+=+=

+=

==22

22

1

2

22

mgR

hmg

R

RhFNB

+=

+= 1

22

Answer: At point B the scale will show mgR

hFNB

+= 1

2

v Answer: As long as minhh > , the difference between apparent weight ofthe object at the bottom and top of the loop is mgFF NANB 6= and it is

independent from the maximum height of the track, h , and radius of

the track h .

1 Note that a bath-room scale usually shows the weight as

g

mgwhich is the mass, but due to the fact that this

could be misleading, we have chosen scales which shows weight, like that of American-British system scalewhich shows its reading as pound which is the units of weight.

B

mg

Ra m

R

vmmgF BNB

2

+=


12/12



mgmgmgmgR

hmg

R

hFF

mgR

hF

mgR

hF

NANB

NA

NB

6552

12

52

12

=+=

+=

=

+=

Note that as long as Rhh => 5.2min the difference in the apparent weightof the object at the bottom and at the top of the loop, mgFF NANB 6= , is

independent from the original height of the release h , and the radius of

the loop R . It is always mg6 , i.e. just a function of the mass of the

object and the gravitational acceleration of the object: mgFF NANB 6=

vi Answer: At point C the scale shows mg 2 R

hFNC

=

2

At point the scale will show the normal force

which is the centripetal force:

C

R

v

mFC

NC

2

=

Jennys velocity may be calculated usingconservation of energy. Her apparent weight, thevalue the scale shoes, may be determined by rtangential velocity at point C, calculated from the conservation of

mechanical energy in

eplacing value of her

R

vmF CNC = :

2

( )

( ) mgR

hR

gRhmF

R

vmF

gRhvmvmgRmgh

NC

CNC

CC

==

=

=+=

2222

22

2

1

2

22

viiAnswer: The scale will show just Jennys normal weight mgND = .FDue to the fact that the point D is on the flat horizontalsurface of the track, regardless of the fact that the cart isaccelerating or decelerating, due to the fact that there isno acceleration in the vertical direction, the scale shows

the normal weight of Jenny. mg

m

mgFND =

D

mg

Ra

m

C

mgR

hFNC

= 2

2

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Solution+VMV+Ch3,+5 7+FyBNVCO08+Circular+Motion,+Energy+Momentum