Solution+Test+V1+FyBNVC07+Ch+5 7+Circular Rotational+Motion,+Energy,+

8/3/2019 Solution+Test+V1+FyBNVC07+Ch+5 7+Circular Rotational+Motion,+Energy,+

1/16

Suggested solutions Midterm Exam Ch5-7 FyBNVC07 Circular motion, Linear momentum, EnergyNV-College

Midterm Exam Physics B: FyBNVC07Circular-Rotational Motion, Energy and Linear Momentum

Instructions:

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 a

personalized formula sheet which has your name on it. This should be submitted along

with the test.

Time: 8:00-10:20

Grade limits: There are two alternatives to choose. This regards the last problem.

The original test: maximum of 55 points of which 22 are VG points, and 5 points.

The alternative test: maximum of 44 points of which 16 are VG points. points.

Lower limits for examination grade

Pass (G): 18 points

Pass with distinction (VG): 37 points of which at least 7 VG-points

Pass with special distinction (MVG): 40 points of which at least 16 VG-points and you

must show several Pass with Special Distinction qualities in at least three of -marked

qualities.

Problems number 9 and 10are heavily graded and are of greatest importance for both

VG and MVG. You need at least one hour to solve these problems completely. You

may choose to solve these problems before solving the others.

Problem 1a 1b 1c 1d 1e 2 3 4 5a 5b 5c

G 2 1 1 1 3 1 1 1

VG 1 3

G

VG

Problem 6 7a 7b 7c 7d 8 9 10 Sum G VG MVG

G 1 2 1 1 1 4 4 25 19 37 40

VG 3 1 3 6 6 25 8 18

MVG 6

G

VG

MVG

Have Fun!

Behzad

[email protected] to use for educational purposes. Not for sale. 1/16


2/16



1 At which of the indicated positions1 At which of the indicated positions EA the satellite orbiting the Earth in anelliptical orbit1

a. experiences the greatest gravitational

force? Why? Draw the direction of

the force on the satellite at the pointas clear as possible. [2/0]

b. have the greatest velocity? Why?

[1/0]

c. have the greatest acceleration? Why?

[1/0]

d. have the greatest total energy? [1/0]

e. have the greatest potential energy? Why? [0/1]

Suggested solution:

EarththeofMassM ; SatellitetheofMassSm ; distanceSatellite-CM_Earth-CMr ;

SatellitetheofVelocityTangentialousInstanetnev

;

a. Answer: The satelliteexperiences the greatest

gravitational force at theposition D . This is due tothe fact that the satellite isclosest to the Earth at D ,and according to the Newtons Universal Gravitational law thesatellite experiences the force:

2r

mMGF S

= .

Therefore, the smaller r is the larger the force is. [2/0]

b. Answer: At the position D the satellite is moving fastest. This is dueto Keplers second law. We may also drive the same conclusionrealising that according to Newtons second law:

r

vm

r

mMG S

S

2

2=

r

MGv =

Therefore, the smaller r is the larger the force is. [1/0]

1 The figure is not scaled accordingly. Exaggerated.

A

B

C

D

E

D


3/16



c. Answer: At the position D the satellite has its largest possiblecentripetal acceleration:

rSS am

r

mMG =

2

2r

M

Gar = [1/0]

d. Answer: Satellite total Energy is constant. It is the same at allpoints on the orbit.

2

2

1vm

r

mMGE S

Stot +

= [1/0]

e. Answer: The greatest potential energy of the satellite is at theposition A . This is due to the fact that the gravitational potential

energy of the system is given by:

r

mMGE SPE

=

Note that it is negative. The further the satellite is it is lessnegative, and therefore largest. On the other hand we may us theconcept of conservation of energy to statement above. Further thesatellite is smaller its velocity, and its kinetic energy is. Thereforedue to the fact that satellites velocity is smallest at the position A its potential energy is largest at A . [0/1]Alternative solution: KEPE EEE +=

Since E is constant. PEE is its largest when KEE is smallest, i.e. the

point A .

2 Using Newtons Universal Gravitational law derive Keplers third law. [0/3]

Keplers third law: The ratio of the squares of the periods of any two planets

revolving about the Sun is equal to the cubes of their mean distances from the Sun.

Suggested solution:1

2

12

1

1

P

P

P

P

r

vm

r

mMG =

1

2

Pr

MGv = [0/1]

The period of any planet isv

rT PP

11

2 =

. Therefore: ( )

( )2

2

12

1

2

v

rT PP

=

Substitute1

2

Pr

M

Gv = in ( )( )

2

2

12

1

2

v

r

TP

P

=

: ( )( )

1

2

12

1

2

P

PP

r

MG

r

T

=

[0/1]

( ) ( )211

2

1 2 PP

P rr

MGT = ( ) ( )211

2

1 2 PPP rrGMT = ( )3

1

22

1 4 PP rGMT =

Similarly, for the planet number two: ( ) 3222

2 4 PP rGMT =

Dividing these two equations leads to:( )( ) 32

2

3

1

2

2

2

2

1

4

4

P

P

P

P

r

r

GMT

GMT

=

( )( ) 32

3

1

2

2

2

1

P

P

P

P

r

r

T

T

=

3

1

1

2

2

1

=

P

P

P

P

r

r

T

T

Keplers Third Law [0/1] QED


4/16


3 Find the speed at which the Earth revolves about the Sun. You may assume that

Earths orbit is nearly circular, and the speed of the rotation Earth about the Sun is

almost constant.

3 Find the speed at which the Earth revolves about the Sun. You may assume that

Earths orbit is nearly circular, and the speed of the rotation Earth about the Sun is

almost constant. kgM 24Earth 105.98 = ; kgM

30

Sun 1099.1 = ,

kmr 6

Earth-Sun

106.149 = ; 2211 /1067.6 KGmNG = [3/0]

Suggested solution:

SE

E

SE

ES

r

vM

r

MMG

= 2

2

SE

S

r

MGv

=2 SE

S

r

MGv

= [1/0]

smr

MGv

SE

S /1098.21087.8106.149

1099.11067.6 48

9

3011 ==

==

[1/0]

smr

MGv

SE

S /1098.2 4==

[1/0]

Second Method:

The period of any planet isv

rT SEE

=2

. Therefore:E

SE

T

rv

=2

[1/0]

smT

rv

E

SE /1098.236002425.365

106.14922 49

=

=

=

[1/0]

4 " Knocked off " You are standing on a log and a friend is trying to

knock you off. He throws the ball at you. You can catch it, or you can

let it bounce off of you. Which is more likely to topple you, catching

the ball or letting it bounce off? Why? Explain. [1/1]

Suggested solution:

I should try to catch it!

If the ball is moving horizontally at smvb / just before catching it, and its mass is kgm ,

and my mass is smM / , and I am at rest standing on a long

If I catch it, my recoil velocity may be calculated using conservation of the linearmomentum:

( )catchb vMmmv +=

( )smv

Mm

mv b /

+= [1/0]

If I let the ball bounce off me, and if I assume it bounce off at smvb /2 , my recoil

velocity is going to increase. Conservation of linear momentum requires that:

2bb mvvMmv =

( )22 bbbb vvmmvmvvM +=+=

( )2bb vvM

mv +=

Due to the fact that( )Mm

m

M

m

+> , and bbb vvv >+ 2 if I let the ball to bounce off me

then I may be knocked off! [0/1]



5/16



5 A paratrooper whose chute fails to open lands to an open land in snow. She is injured

vaguely. If she had landed on a solid ground, stopping time would be more than 10

folds shorter and she would die. Does the presence of snow increase, decrease, or

leave unchanged the value of

5 A paratrooper whose chute fails to open lands to an open land in snow. She is injured

vaguely. If she had landed on a solid ground, stopping time would be more than 10

folds shorter and she would die. Does the presence of snow increase, decrease, or

leave unchanged the value of

a. the paratroopers change in momentum. Why? [1/0]a. the paratroopers change in momentum. Why? [1/0]

b. the force stopping the paratrooper? [1/0]b. the force stopping the paratrooper? [1/0]c. the impulse on the paratrooper. Why? [0/1]c. the impulse on the paratrooper. Why? [0/1]

Suggested solutions:Suggested solutions:a. The presence of snow leaves unchanged the paratroopers change in

momentum. The reason is the fact that the initial velocity and finalvelocity of the paratrooper are identical in both cases. She stopsanyway and the change in the momentum is just

a. The presence of snow leaves unchanged the paratroopers change inmomentum. The reason is the fact that the initial velocity and finalvelocity of the paratrooper are identical in both cases. She stopsanyway and the change in the momentum is just mvPP = 012 . [1/0]

b. The presence of snow decreases the force stopping the paratrooper. Ittakes longer time for her to stop and the force on her is much less: i.e.

the force is more than 10 times smaller in landing on snow. [1/0]

c. The presence of snow leaves unchanged the impulse on theparatrooper. The impulse that stops her is exactly equal to the changeof her linear momentum and that is: mvPPI == 012 . [0/1]

6 Tarzan whose mass is kg105 swinging in an arc from a hanging m0.10 vine (rope).

If his arms are capable of exerting a maximum force of

kN90.1 on the vine, calculate the maximum speed he can

tolerate in swinging. [1/3]

Suggested solution: Answer: smv /10.9max

Data: kgm 105= , mr 0.10= , kNFT 90.1max, = ; Problem:

?max =v

At its lowest point the tension is the maximum on thevine. Free-body-diagram of the situation is shown onthe figure.Newtons equation of motion requiresthat:

amF

rr

= CT

mamgF =

r

vmmgFT

2

= [1/0]

gm

F

r

v T =2

rgm

Fv T

=2 [1/0] Free-body-diagram [0/1]

rgm

Fv

T

= max,

max

108.9105

1900

max

=v

smv /10.995.82max=

[1/0]

NFT 1900max, =

?max =vmg

ca


6/16



7 An ice-hockey puck is7 An ice-hockey puck is g160 and is moving at

sm /15 . Magnus stroke the puck with a club. The

force that the club transfers to the puck is illustrated

below as a function of the time.

a. Calculate the impulse transferred to the club.

[2/0]

b. Calculate the velocity of the puck after the

stroke, if the hit is in the opposite direction of the original motion of the puck.

[1/0]

c. Calculate the velocity of the puck

after the stroke, if the hit is in the

direction of the original motion of

the puck. [1/0]

d. Calculate the velocity of the puck

after the stroke, if the hit is normal

to the direction of the original

motion of the puck. [1/1]

Suggested solution: Answer:s

mkgsNI == 55 , b: smv /162 ;

c: smv /362 , d: smv /35 at2 64 with +x-axis

Data: kg6gm 1.0160 == , smv /151 =

a. Ans tr tower: The impulse ansferred the puck is: sNI = 5

t under theThe impulse transferred by the club to the puck is he areacurve of force vs. time graph. It may be estimated as the area of atriangle whose base is s1.0 and its height is N100 . Therefore, the

impulse transferred to the puck is: mNI = 5

( )( )s

m1kgmNsNI === 551.0100

2 [2/0]

b. Answer: If the stroke is in the opposite direction of the original motionof the puck, velocity of the puck immediately after the stroke is

smv /162 :

inal direction of motion of the puck is taken as positIf the orig ive:

12vmvmPPI rrrr == 12

12 vmIvmrr

+=

Imvmv = 12

m

Ivv = 12

smsmv /16/25.1625.31154

12515

16

50015

16.0

5152 =====

Answer: smv /162 i.e. at sm /16 in -x-direction [1/0]

0

20

40

60

80

100

0,00 0,02 0,04 0,06 0,08 0,10

t (s)

F

(N)


7/16



c. Answer: If the stroke is in the direction of the original motion of thepuck, velocity of the puck immediately after the stroke is sm /36 :v2

If the original direction of motion of the puck is taken as positive:

1212 mvmvPPI ==

12mvImv +=

12 vm

Iv +=

smsmv /46/25.461525.31154

12515

16

50015

16.0

52 =+=+=+=+=

Answer: smv /462 in +x-direction [1/0]

d. Answer: If the stroke is normal to the direction of the original motionof the puck, immediately after the stroke, the puck is moving at:

sm /35 . It is deflected byv2 64 from its original direction of

motion:If we name the original direction ofthe motion of the puck x and the

direction perpendicular to it :y

1212 vmvmPPIrrrr

==

12 vmIvmrr

+=

Imv y =2

m

Iv y =2

smv /25.314

125

16

500

16.0

52 ==== [1/0]

Therefore, the velocity of the puck immediately after the stroke is

( ) ( ) ( ) ( ) smsmvvv xy /35/66.341525.311525.3122222

2

2

22 =+=+=+=

Its direction of motion is:

=

= 64

15

25.31tantan

1

2

21

y

y

v

v

Answer: smv /352 at 64 with +x-axis [0/1]

y

x

64

smv /352


8/16


8 A flat puck of mass8 A flat puck of mass M is rotated in a circle on a frictionless air hockey tabletop, and

is held in this orbit by a light string which is connected to a hanging mass m through a

central hole as illustrated in the figure below.

Draw the free body-diagram and prove that the

tangential velocity of the puck is given by

gM

mRv = [0/3]

m

mg

TF

Suggested solution:The tension in the cord is equal to the weight of the mass ,

i.e. .

m

mgFT =

But for the rotating mass M the centripetal force is provided by

the tension in the string, and due to this centripetal force it isrotating on the horizontal frictionless table at velocity .

Newtons second law requires that:

v

R

vMMaF RT

2

== [0/1]

MTF

R

vaR

2

=

Therefore:

M

Rmgvmg

R

vM

mgF

R

vMMaF

T

RT ==

=

== 22

2

[0/1]

MRmgv = QED Free-body-diagrams [0/1]



9/16



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

How well you plan and carry out the task. Which priciples of physics you use and how you justify using them. How well you use physical and matematical language.

How general your solutions are. How well you present your work

How well you cary out your calculations.

How clear your solutions are.

How well you justify your conclusions.

9 A relatively small spherical metal ball of mass 1m fastened to the end of a string of

length l to form a simple pendulum is released from rest in the horizontal position as

illustrated in the figure2 below. The size of the metal ball is negligible compared to

the length of the string. At the bottom of its swing, the metal ball collides with a block

of mass 2m (where 12 mm > ) initially resting on a horizontal long table. [4/6/]

9 A relatively small spherical metal ball of mass 1m fastened to the end of a string of

length l to form a simple pendulum is released from rest in the horizontal position as

illustrated in the figure

a. If the block is also metallic, and

therefore the collision is assumed

perfect elastic, calculate the velocity of

the metallic ball and of the metallic

block immediately after the collision.

a. If the block is also metallic, and

therefore the collision is assumed

perfect elastic, calculate the velocity of

the metallic ball and of the metallic

block immediately after the collision.

b. Find the maximum heightb. Find the maximum height

2 below. The size of the metal ball is negligible compared to

the length of the string. At the bottom of its swing, the metal ball collides with a block

of mass 2m (where 12 mm > ) initially resting on a horizontal long table. [4/6/]

h (with

respect to the surface of the table, and

in terms of 1m , 2m , and l ) to which

the reflected ball rises.

c. If coefficient of friction between the

surface of the table and the block is , find the maximum displacementx (in

terms of 1m , 2m , and l ) of the block with respect to its initial position.

d. If the block is made of a soft material, and therefore the collision is assumed

perfectly inelastic, calculate the height the center of mass of the system rises

after the collision.

Alternative Choice: [Note: This choice gives you only 4 G and 3 Vg points at

maximum]You may choose insteadto solve the problem for the very special case of

, ,gm 0.1501 = gm 0.2502 = cm.120=l , 25.0= . [4/3]

Suggested solutions: Answer: lgv i 21 = , smv i /85.41 = , lgmm

mmv f 2

21

121

+

= ,

smv f /21.11 = ; lgmm

mv 2

2

21

12

+= , smv /64.32 = ; l

+

=

2

21

12

mm

mmh , mh 075.0= ;

mx 7.2= ; Inelastic: smv /82.1= ; l

2

21

1

+=

mm

mh , mh 169.0

2 Not scaled properly

1ml

2m


10/16



Data: , ,kgm 150.01 = kgm 250.02 = m20.1=l , 25.0= .

Plan: [0/1/]We may divide the problem into three parts:

Part I deals with the events taking place just before the collision. Wemay use the principle of conservation of energy to find the velocity ofthe ball just before the collision.

Part II studies the events taking place at the moments just before andjust after the collision. We may use principle of conservation ofmomentum, and conservation of energy for perfectly elastic collision,but only conservation of linear momentum for the perfectly inelasticcollision to find the velocity of the ball and that of the block

immediately after the collision.

Part III studies the events taking place after the collision. We may useconcept of conservation of mechanical energy, and the work doneagainst friction to solve to find the maximum height h the reflected

ball rises, as well as to find x the maximum distance the block slides

on the table.

Part I:Using conservation of mechanical energy to find velocity of the ball

just before the collision:

iv1 1m

2

1112

1ivmgm =l

lgv i 22

1 =

lgv i 21 = ( )( ) smv i /85.420.18.921 == [1/0] Answer: smi /85.41 =v

Part II: The Elastic Collision:

Use conservation of linear momentum and conservation of mechanical

energy at the elastic collision. Assuming the original direction of themotion of the ball, i.e. to the left is positive:

+=

+=

221111

2

22

2

11

2

112

1

2

1

2

1

vmvmvm

vmvmvm

fi

fi [0/1]

=+

=

221111

2

22

2

11

2

11

vmvmvm

vmvmvm

fi

fi

( )

( )

=+

=

22111

2

22

2

1

2

11

vmvvm

vmvvm

fi

fi


11/16



Divide the first equation by the second one:

( ) 22

2

22

111

2

1

2

11

vm

vm

vvm

vvm

fi

fi

/

/=

+/

/ [0/1]

fi vvv 112 = [1/0]

We may substitute fi vvv 112 = in 221111 vmvmvm fi += to find the velocity of

the ball after the collision:

fifi vvmvmvm 1121111 +=

iiff vmvmvmvm 11121211 =+

( ) ( 121211 mmvmmv if )=+

if vmm

mmv 1

21

121

+

= Answer: lg

mm

mmv f 2

21

121

+

= [1/0]

( ) smv f /21.18497.4150.0250.0

150.0250.01 =

+

= Answer: smf /21.11 =v

iiifiv

mm

mmvvvv

11

21

121112

+

==

ii vmm

mmv

mm

mmv 1

21

121

21

212

+

+

+=

ivmm

mmmmv 1

21

12212

+

++=

ivmm

mv 1

21

12

2

+= Answer: lg

mm

mv 2

2

21

12

+= [0/1]

( ) ( ) smv /64.38497.4150.0250.0 150.022 =+= Answer: smv /64.32 =

Part III: The Elastic Collision:

Use the concept of conservation of the mechanical energy to find themaximum height the reflected ball rises to:

2

1112

1fvmghm /=/

2

12

1fv

gh = substitute smv f /21.11 = in

2

12

1fv

gh =

( ) mvg

h f 075.021.1

82.92

1

2

1 221 =

== mh 075.0=

Substitute lgmm

mmv f 2

21

121

+

= in 21

2

1f

vg

h = :

2

21

12 22

1

+

= lg

mm

mm

gh

2

21

12

2

2

+

////=

mm

mm

g

gh

lAnswer: l

+

=

2

21

12

mm

mmh [1/0]

( ) mh 075.020.1150.0250.0

150.0250.02

=

+

= (as expected) Answer:

mh 075.0=


12/16



The maximum displacement of the block on the table may be calculatedusing the concept of conservation of energy, and the work done againstthe friction:

xgmvm = 22

222

1 Answer:

g

vx

=

2

2

2 [0/1]

( )( ) ( )

mx 7.28.925.02

6373.32

=

=

For the case of perfectly inelastic collision, Part II and part III of theproblem are different. At the moment of collision only the linearmomentum of the system is conserved. Naming the velocity of thecombined block-ball system conservation of the linear momentum

requires:

v

( ) vmmvm i += 2111 ivmm

mv 1

21

1 +

= Answer: lgmm

mv 2

21

1

+= [0/1]

( ) smv /82.18497.425.015.0

15.0=

+= Answer: smv /82.1=

Part III-inelastic collision: Conservation of energy:

( ) ( ) hgmmvmm +=+ 212

212

1

g

vh

2

2

=

2

21

1

2

21

1

2

22

2

1

+=

+=

mm

m

g

gg

mm

m

gh

ll Answer: l

2

21

1

+=

mm

mh [0/1]

( ) mmm

mh 169.020.1

150.0250.0

150.022

21

1

+=

+= l mh 169.0

MVG-Quality Specifications Comments

Has the student well planed, and carried

out the task?

Division of the

problem to 3 parts

Which priciples of physics are usedconsiousely in solving the problem.

Have their use justified by thestudent? How well does the student

use physical and mathematical

languages?

Proper use of:

Conservation of

energy, and of linearmomentum;

work of friction

centripetal forces and

acceleration

How general are the solutions? Uses algebraic method

Is the solution clear, and are thecalculations carried out well

structured? Are the calculations are

carried out

Easy to follow

Clear language

Clear steps

Are the results analyzed, evaluated

and verified?

Are the results

acceptable? Logical?Correct Units?


13/16



10 A bullet of mass10 A bullet of mass bm is fired at initial velocity biv into a wooden block of mass Bm

suspended from a light string of length l . The bullet stops in the block which was

initially at rest. The size of the wooden block is negligible compared to the length of

the string l . [4/6/]

a. Calculate the impulse exerted by the bullet on the block.

b. Calculate the mechanical energy (in terms of bm , Bm ,

and biv ) lost during the penetration of the bullet into

the block.

c. Find the height h (in terms of bm , Bm , and biv ) to

which the bullet-block system rises. Explain in detail

and as clear as possible your solutions. Discuss the

physical concepts used in your solution.

d. Find the minimum speed of the bullet biv if the block will swing through a

complete circle.

Alternative Choice: [Note: This choice gives you only 3 G and 2 Vg points at

maximum]You may choose insteadto solve the problem for the very special case of

gmb 0.30= , smvbi /0.250= , kgmB 00.2= , cm0.50=l ; [4/3]

Suggested solution:Data: gmb 0.30= , smvbi /0.250= , kgmB 00.2= , cm0.50=l ;

a.

Calculate the impulse exerted by the bullet on the block.

Suggested solution:Conservation of the linearmomentum requires that:

fi pprr

=

( )vmmvm Bbbib +=

( ) biBbb vmm

mv

+=

( )( ) smv /25000.203.003.0

+=

smv /69.3=

if ppprrr

==Impulse

( ) biBbb

BB vmm

mmvm

+==Impulse

Answer: The impulse exerted bythe bullet on the block is:

( ) biBbBb vmm

mm

+=Impulse to the right.

[1/0]Numerical value:

( )smkg /250

00.2030.0

00.2030.0Impulse

+=

smkg /39.7Impulse =

b. Calculate the mechanical energy (in terms of bm , Bm , and biv ) lost during the

penetration of the bullet into the block.

Suggested solution:Mechanical energy lost during thepenetration of the bullet into the

block may be calculated as thedifference between the energy of

the bullet befor the collision andthe energy of the system

immediately after the collision:

bm

biv

Bm

l


14/16



( ) 222

1

2

1vmmvmE BbbibLost +=

( )( )

2

2

2

1

2

1

++= bi

Bb

bBbbibLost v

mm

mmmvmE

( )2

22

2

1

2

1bi

Bb

bbibLost v

mm

mvmE

+= [1/0]

( )2

22

2bi

Bb

bBbbLost v

mm

mmmmE

+

/+/=

Answer: The system loses

( )2

2bi

Bb

BbLost v

mm

mmE

+=

mechanical energy during thepenetration of the bullet into thewooden block. [0/1]Numerical value.

( )( )2250

00.203.02

00.203.0

+

=

LostE

JELost 924

c. Find the height h (in terms of bm , Bm , and biv ) to which the bullet-block system

rises. Explain in detail and as clear as possible your solutions. Discuss the physicalconcepts used in your solution.

Suggested solution:Even though the collision isperfectly inelastic and thereforethe mechanical energy is notconserved during the penetrationof the bullet into the block, themechanical energy of the system

must be conserved immediatelyafter the penetration of the bulletinto the block:The total energy of the block-bullet system is just that of theirkinetic energy, which will totallyconverted to the gravitationalpotential energy of the system atthe end of the process when thependulum is raised to its highest

and therefore totally stoppedbefore swinging back. Taking theoriginal level of the block (i.e. thelowest point of the swing) as thezero potential energy level, wemay write the conservation ofenergy as:

( ) ( ghmmvmm BbBb +=+2

2

1)

[0/1/]

g

vh

2

2

=

Using

( )

bi

Bb

b vmm

mv

+= , we may

express h only in terms of bm ,

Bm , and biv :

( )

22

2

1

2

+== bi

Bb

b vmm

m

gg

vh

2

2

2

1bi

Bb

b vmm

m

gh

+= [0/1]

( ) mmh 89.12501025.0

025.0

82.92

1 22

=

+

=

mh 89.1


15/16



d. Find the minimum speed of the bullet biv if the block will swing through a complete

circle.

Suggested solution:The block will swing through acomplete circle if the tension inthe string at the top of the swingis larger than zero:At the highest point of the swing,taking the direction of thecentripetal acceleration Ra as the

positive direction, Newtons

second law of motion, amFnetrr

= ,

may be expressed as:

RT mamgF =+ [0/1/]mgmaF

RT=

The requirement that 0TF

implies that:0= mgmaF

RT

0 gaR

gaR

using the expression for the radial

acceleration:

r

vaR

2

= , we may

rewrite this expression as:

gr

vtop

2

or

rgvtop 2

But the radius of the circle isidentical to the length of thestring l , therefore, we may

conclude that the block will swing

through a complete circle if itsvelocity at the top of the swing is:

l gvtop [1/0]

Considering that fact that at thetop of the swing, l= 2h and

l gvtop , we may express the

concept of conservation ofmechanical energy as:

( ) 22

2

12

2

1top

vmgmvm /+/=/ l

where ( )Bb mmm +

22 4topvgv += l

Using the expression for the initialvelocity of the block-bullet systemin terms of the initial velocity of

the bullet:( ) biBb

b vmm

mv

+= , we

may rewrite the equation aboveas:

22

2

4topbi

Bb

b vgvmm

m+=

+l

[ 22

2 4top

b

Bbbi vg

mmmv +

+= l ] [1/0]

Using l gvtop , we may express

the minimum velocity of thebullet required to the blockswinging through a completecircle:

[ ]ll +

+= gg

m

mmv

b

Bbbi

4

2

min

2

2

min

2 5

+=

b

Bbbi

m

mmgv l

+=

b

Bbbi

m

mmgv l5

min

Answer: The complete circleswing is possible only if

+

b

Bbbi

m

mmgv l5 [0/1]

Using l gvtop , we may express

the minimum velocity of thebullet required to the blockswinging through a completecircle:

[ ]ll +

+= gg

m

mmv

b

Bbbi 4

2

min

2

2

min2 5

+=

b

Bbbim

mmgv l


16/16



+=

b

Bbbi

m

mmgv l5

min

Answer: The complete circleswing is possible only if

+

b

Bbbi

mmmgv l5 [0/1]

Numerical values:

( ) ( )

+

03.0

00.203.05.082.95biv

smvbi

/474

Answer: The block will swing a

complete swing only if thevelocity of the bullet is at least

sm /474 .i.e.: smvbi /474

MVG-Quality Specifications Comments

Has the student well planed, andcarried out the task?

Division of theproblem todifferent parts

Which priciples of physics are usedconsiousely in solving the problem.Have their use justified by thestudent?How well the student does usesphysical and mathematicallanguages?

Use of:Conservationof energyConservationof linearmomentum

How general are the solutions? Uses algebraicmethod

Is the solution clear, and are the

calculations carried out wellstructured? Are the calculations arecarried out

Easy to follow

Clear languageClear steps

Are the results analyzed, evaluatedand verified?

Are the resultsacceptable?Logical?Correct Units?

Solution+Test+V1+FyBNVC07+Ch+5 7+Circular Rotational+Motion,+Energy,+

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