Solution+FyBNVC06+Test+V1+Ch5 8+Circular Rotational+Motion,+Energy

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Suggested solutions Test V1 Ch5-8 FyBNVC06 Rotational motion, Linear momentum, Energy Sjödalsgymnasiet

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

Physics B: FyBNVC06Circular-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 solutionof 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: 120 minutes .

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

give a maximum of 66 points of which 29 are VG points. Only alternative include 6

MVG (¤) points while the other offers only 3 MVG (¤) points.

Lower limits for examination grade

Pass (G): 22 points

Pass with distinction (VG): 44 points of which at least 10 VG-points

Pass with special distinction (MVG): 48 points of which at least 20 VG-points and youmust show several Pass with Special Distinction qualities in at least three of ¤-marked

qualities.

Problems number 8, 9 and 10 are heavily graded and are of greatest importance for

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





1 A g.500 object is moving in a flat horizontal circle of diameter cm0.80 at a

constant speed of sm /00.2 . Calculate the centripetal force acting on the object. [2/0]

Suggested solutions:

Data: kggm 500.0.500 == , mmd

r cmd 400.02

800.0

20.80 ===⇔= , smv /00.2=

( )( )

N N r

vmF C 00.5

400.0

00.2500.0

22

=⋅=⋅= [2/0] Answer: N F C

00.5=

In the multi-choice problems below circle clearly the correct answer. Write clearly your

answer on the box provided for the answer. If the explanation is required write your

reasoning as clear as possible in the space provided below the problem

2 A metallic ball of mass m is attached to the end of a very

light string and is moving at a constant speed of v in a

circular path of radius r on a frictionless horizontal

surface as illustrated in the figure:

a. When the ball is in the position illustrated the direction

of the force is towards [1/0]

i. A

ii. B iii. C

iv. D Answer: ____________

b. If the velocity of the ball is doubled the tension on the string [1/0]

i.

is halvedii. remains the same

iii. is doubled

iv. is quadrupled Answer: ____________

Why? Explain: [0/1]

c. If the length of the string is doubled while the speed of the ball on its horizontal

circular path remains the same. The centripetal acceleration of the ball [1/0]

i. is halved

ii. remains the sameiii. is doubled

iv. is quadrupled Answer: ____________

Why? Explain: [0/1]

d. If the mass of the ball is doubled while the speed of the ball on its horizontal circular

path and the length of the string both remain the same. The centripetal acceleration

of the ball [1/0]

i. is halved

ii. remains the same

iii. is doubled

iv. is quadrupledWhy? Explain: [0/1]

v

m

A

B C

D





Suggested solution:

a. When the ball is in the position illustrated the

direction of the force is towards [1/0]

i. A

ii. B

iii. C iv. D Answer: Alternative iii: C

b. If the velocity of the ball is doubled the tension on

the string [1/0]

i. is halved


iii. is doubled

iv. is quadrupled

Answer: Alternative iv: the force is quadrupled.

r

vmF C

2

⋅=

( )1

22

2 442

C C F r

vm

r

vmF =

⋅⋅=⋅= [0/1]

c. If the length of the string is halved while the speed of the ball on its horizontal

circular path remains the same. The centripetal acceleration of the ball [1/0]

i. is halved


iii. is doubled

iv. is quadrupled

Answer: Alternative iii: The centripetal acceleration of the ball is doubled.

r

vaC

2

=

1

22

2 22

2

C C ar

v

r

va =

⋅== [0/1]

d. If the mass of the ball is doubled while the speed of the ball on its horizontal circular

path and the length of the string both remain the same. The centripetal acceleration

of the ball [1/0]

i. is halved


iii. is doubled

iv. is quadrupled

Answer: Alternative ii: The centripetal acceleration of the ball remains the same. The

centripetal acceleration is independent of the mass of the ball:r

vaC

2

= [1/1]

v

m

A

B C

D





3 A kg00.5 rifle fires a g00.7 bullet at a velocity of sm /.450 . The rifle kicks back

as a result. Calculate the velocity of the kick back of the rifle. [2/0]


Data: kg M 00.5= , kggm31000.700.7 −×== , smvb /.450=

The liner momentum of the system is conserved. If we assume the direction of the bullet

positive the rifle kicks in the opposite, i.e. in the negative direction. The total momentum of

the system is zero initially:

smv M

mvmv Mvvmv M br br br /630.0.450

00.5

1000.700

3

=××

==⇔=+−⇔+=−

rr

[2/0]

Answer: smvrifle /630.0=

4 It takes for the Moon days3.27 to rotate about the Earth at a “circular orbit” of radius

m8

1084.3 × . Moon’s mass is kg22

1035.7 × and its radius is m6

1074.1 × . Earth’s

mass is kg24

1098.5 × .

a. Find the speed of the Moon on its orbit about the Earth. [2/0]

b. Find the angular velocity of the Moon. [1/0]

c. What is the source of the centripetal force acting on the Moon? Use two

independent methods to calculate the centripetal force acting on the Moon. [2/1]

d. What would happen to the Moon if its speed was greater than the value

calculated above? Why? [1/1]

e. What would happen to the Moon if its speed was less than the value calculated

above? Why? [1/0]

f. Use two independent methods to calculate the arc (distance) covered by the

Moon in one day. [2/1]

Suggested solution:Data: daysT 3.27= ; m R M E

81084.3 ×=− , mr M

61074.1 ×= , kg M M

221035.7 ×= ,

kg M E

241098.5 ×=

a. Answer: The speed of the Moon on its orbit about the Earth, i.e. its tangential velocity

is: smv /1020≈

T

r v

⋅=

π 2⇔

( )smsmsmv /1020/9.1022/

3600243.27

1084.328

≈=××

×⋅=

π [2/0]

Answer: smv /1020≈

b. Answer: The angular velocity of the Moon is: srad /1048.87−×= π ω [1/0]

T π ω 2= ⇔ srad /

3600243.272

××= π ω

Alternative solution: srad srad m

sm

r

v/1048.8/10664.2

1084.3

/9.1022 76

8

−− ×=×=×

== π ω

c. Answer: The source of the centripetal force on the Moon is the attractive force between

the Earth and the Moon. The centripetal force acting on the Moon is: N F C

20100.2 ×≈

( ) ( ) N N N

r

vmF C

2020

8

2

222

100.210003.21084.3

9.10221035.7 ×≈×=

×⋅×=⋅= [2/0]

Second independent method: Using 211 /1067.6 kgm N G ⋅×≈ −





Initial kinetic energy of the system is:

( ) J vmvm E iB BiR Ri 5.8125152

1250.2

2

1

2

1

2

1 2222 =⋅+=+=

Final kinetic energy of the system is:

( ) J vmvm E fB B fiR R f 5.4677152

1

100.22

1

2

1

2

1 2222

=⋅+=+=

Answer: The mechanical energy is not conserved! f i E J J E =>= 5.4675.812 .

Therefore, the collision is inelastic. [0/1]

6 A rescue plane is going to drop emergency supplies to isolated group of mountain

climbers m240 below. If the plane is flying horizontally with a speed of

hkm /.288 , how far (horizontal distance) in advance of the recipients must the

supplies be dropped? [1/3]Suggested solutions:

Data: m y 275= , smsmhkmvv x x /0.80/3600

1000.288/.2880 ====

The supplies have exactly the same initial velocity as the plane, i.e.: smvv x x /0.800 ==

During the time that the supplies fall freely the move in the horizontal direction according to

( ) mt xst t t t

mt vt v x x x

.56070.800.8000.7

90.4

24090.48.9

2

10275

0.80

222

0

=⋅=⋅=⇔⎪⎩

⎪⎨

⎧

≈⇔=⇔⋅=⋅⋅+=

⋅==⋅=

Answer: The plane must drop the supplies m x .560= in advance of the recipient.

7 Pilots may be tested for the stresses of flying high-speed jets in a whirling “human

centrifuge” that takes s0.60 to turn through 20 complete revolutions before reaching

its final speed

a. What is the angular acceleration of the “human centrifuge” during the first

s0.60 of the acceleration? [1/1]

b. What is the final angular velocity of the system? [1/1]

Suggested solutions: Data: ss

T st 00.320

0.600.6020 ==⇔=

a. Answer: Assuming a uniform acceleration, the angular acceleration of the “human

centrifuge” during the first s0.60 of the acceleration is:2

/45 srad

π

α = . [1/1]

2

22

2/

4560

2022

2

1srad

t t

π π θ α α θ =

×==⇔⋅= ; ( ) srad t /

3

460

450

π π ω α ω ω =⋅=⇔⋅+=

b. Answer: The final angular velocity of the system is srad /3

4π ω = [1/1]

Second method: srad T

f /3

42

3

222 0

π ω ω ω

π π π ω =+=⇔==⋅=

200 /

4560

3

4

srad

t

t π

π

ω ω α α ω ω ==

−=⇔⋅+=





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.

∗ Which priciples of physics you use and how you justify using them

∗ How general your solutions are

∗ How well you justify your conclusions∗ How well you cary out your calculations

∗ How well you present your work

∗ How well you use physical and matematical language.

∗ How clear your solutions are.

8 A g0.20 rifle bullet traveling at sm /0.240 buries itself in a kg00.5 wooden block

sitting on a flat horizontal surface. The coefficient of friction between the surfaces in

contact is 4.0 . How far will the block slide? [2/4/¤]Suggested Solutions:

Data: kggmb 020.00.20 == , smvb /0.240= , kg M Wb 00.5= , 0=Wbv , 4.0=

We may divide the problem to two parts.

Part I may deal with the part that takes care of the moment of penetration of the bullet into

the wood. At this point the linear momentum of the system is conserved while its mechanical

energy is not. This is due to the fact that part of the mechanical energy of the system is used

in breaking chemical bound between the atoms of the wood and as a result heating of it:

( ) f WbbWbiWbbib vmmvmvmrrr

+=++ ..

( ) f v00.5020.0240020.0 +=× [1/1]

f v02.580.4 =

smsmv f /956.0/02.5

80.4≈= [0/1]

Part II: In this part, the kinetic energy of the wooden block that initially is moving at

smv f /956.0≈ is wasted against the friction that results in the full stop. Lets assume the

block comes to full stop after moving m x . Conservation of energy requires that:

( ) ( ) xgmmvmm Wbb f Wbb⋅+=+ μ

2

2

1⇔

g

v x

f

μ 2

2

= [0/1]

Note that the friction force acting on the wooden block is

( )gmmF Wbb f +=

μ and the work

done against the friction is: ( ) xgmm xF W Wbb f f ⋅+=⋅= .

( )cmm

g

v x

f 6.11116.0

82.94.02

956.0

2

22

=≈××

==μ

[0/1]

Answer: The block moves cm6.11 on the flat horizontal surface where the coefficient of

friction between the two surfaces in contact is 4.0 . [1/0]

Grading:

MVG point, ¤, goes to the strategy and plan of how to solve problem, how it is presented and

solved, as well as its clarity and the mathematical and physical terms and principles used in

the process.





9 A kg0.3 object moving towards the east at sm /0.5 collides head-on with a heavier

object of mass kg0.4 moving at sm /5.3 toward the west. Find the final velocity of

each mass if:

a. the bodies stick together. [2/0]

b. the collision is perfectly elastic. [1/3/¤]

c. the kg0.4 object stops as a result of the collision. [1/1]

d. The kg0.3 object has a velocity of sm /5.3 toward the west after the

collision. [1/2]

e. Are the results obtained above reasonable? Why? [0/3]


kgm 0.31 = , smv /0.51 = , kgm 0.42 = , smv /5.32 −= . East is positive!

a. Answer: The bodies that have stuck together move toward the west at smv f /14.0≈ .

If the bodies stick together, the collision is perfectly inelastic. The linear momentum of the

system is therefore conserved, while part of the kinetic energy of the system is lost as heat

and other forms of energy:( ) f ii vmmvmvmrrr

.... 212211 ++=++

( ) f v0.40.35.30.40.50.3 +=×−×

15.30.40.50.37 =×−×=⋅f v [1/0]

smsmv f /14.0/7

1≈= [1/0]

b. If the collision is perfectly elastic, both the linear momentum and mechanical energy of

the system must be conserved. Lets assume that the kgm 0.31 = object bounces back

and move at f v1 toward west while the kgm 0.42= also bounce and move to the east

at f v2 :

..... 22112211 ++=++f f ii vmvmvmvm

rrrr

2

22

2

11

2

22

2

112

1

2

1

2

1

2

1 f f ii vmvmvmvm

/+

/=

/+

/

⎪⎩

⎪⎨⎧

+=+

+−=−

2

22

2

11

2

22

2

11

22112211

f f ii

f f ii

vmvmvmvm

vmvmvmvm [0/1]

⎪⎩

⎪⎨⎧

−=−

+=+

222222211211

22221111

i f f i

f i f i

vmvmvmvm

vmvmvmvm

Reorganizing the equations and dividing them to each other:

( ) ( )( ) ( )⎪⎩

⎪⎨⎧

+=+

−=−

f i f i

i f f i

vvmvvm

vvmvvm

222111

2

2

2

22

2

1

2

11

( )( )

( )( )

f i

i f

f i

f i

vvm

vvm

vvm

vvm

222

2

2

2

22

111

2

1

2

11

+/

−/=

+/

−/ [0/1]

i f f i vvvv 2211 −=−

i f i f vvvv 2112 +−=

f f ii vmvmvmvm 22112211 +−=−





)i f i f ii vvvmvmvmvm 2112112211 +−+−=−

i f i f ii vmvmvmvmvmvm 221212112211 +−+−=−

( ) ( )211121222 mmvmmvvm f ii +=−+

( )

( )21

12122

1

2

mm

mmvvm

v

ii

f +

−+

= [0/1]

( )( )

smsmsmv f /14

66/

7

33/

43

3455.3421 ==

+

−+××= To the left smv f /

14

661 =

i f i f vvvv 2112 +−=

smv f /14

535.3

7

3352 =+−= Toward the east smv f /

14

532 =

Check:

⎪⎩

⎪⎨⎧

+=+

+−=−

2

22

2

11

2

22

2

11

22112211

f f ii

f f ii

vmvmvmvm

vmvmvmvm

( )⎪⎪

⎩

⎪⎪

⎨

⎧

=⇔⎟ ⎠

⎞⎜⎝

⎛ ×+⎟

⎠

⎞⎜⎝

⎛ ×=×+×

=⇔×+×−=×−×

OK

OK

12412414

534

14

6635.3453

1114

534

14

6635.3453

22

22

Answer : After the collision the kgm 0.31 = object will move at smv f /14

661 = toward the

west and kgm 0.42 = object will move at smv f /

14

532 = toward the east. [1/0/¤]

Alternative solutions (simpler solution, looses generalization ¤ point):

kgm 0.31 = , smv /0.51 = , kgm 0.42 = , smv /5.32 −= . East is positive!

⎪⎩

⎪⎨⎧

+=+

+−=−

2

22

2

11

2

22

2

11

22112211

f f ii

f f ii

vmvmvmvm

vmvmvmvm

( )⎪⎩

⎪⎨⎧

+=×+×

+−=×−×

2

2

2

1

22

21

435.3453

435.3453

f f

f f

vv

vv

⎪⎪

⎩

⎪⎪

⎨

⎧

=+++⇔=⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ++⇔=+

+=⇔+=⇔=+−

496961121244

314312443

431314143

2

11

2

1

2

12

1

2

2

2

1

121221

f f f

f

f f f

f f f f f f

vvvv

vvv

vvvvvv

0495621 1

2

1 =−+ f f vv

smsmvv f f /7.4/

14

66

21

2472133

21

49521332

1

2

1 ≈=×+±−

=⇔×+±−

=

smsmsmsmvv f f /

1453/

56212/

5619814/

414

6631

431 1

2 ==+=

⎟

⎠

⎞⎜

⎝

⎛ +

=+

=





c. If the kg0.4 object stops as a result of the collision, the collision is not necessarily

perfectly elastic. Therefore, we may have conservation of linear momentum but not

necessarily that of conservation of mechanical energy. We should, of course, be

conscious of not ending up in a situation of creating energy in the process.

..... 22112211 ++=++ f f ii vmvmvmvmrrrr

f v10.35.30.40.50.3 ×=×−×

f v10.30.1 ×=

smv f /0.3

0.11 = [1/0]

( ) J E i 62

2

5.34

2

5322

=×

+×

=

J E f 3

1

3

13

2

12

=⎟

⎠

⎞⎜

⎝

⎛ ×⋅=

Even though in this problem energy is not created! Still the problem is weird. How

could an object collide head on with another and continue moving in the same direction

while the other object that was originally moving to the left stops as a result of the

collision. This problem can not happen in reality and solid objects! [0/1]

d. The kg0.3 object has a velocity of sm /5.3 toward the west after the collision.

Similar to the problem above. We may have conservation of linear momentum but not

necessarily conservation of mechanical energy! Still, we must check if the results are

physically acceptable.

..... 22112211++=++

f f ii vmvmvmvm

rrrr

f v20.45.30.35.30.40.50.3 ⋅+×−=×−× [0/1]

f v20.45.100.1 ×+−=

5.115.100.10.4 2 =+=×f v

smsmv f /875.2/0.4

5.112 == [1/0] Answer: smv f /875.22 =

( ) J E i 62

2

5.34

2

5322

=×

+×

=

( ) ( ) J E f 9.342

875.24

2

5.3322

=

×

+

×

=

In contrast with the part “c”, this situation may take place. No violation of principles of

physics or reality takes place. The 3.o kg ball traveling to the right collides with another ball

traveling left and as a result of collision it changes its direction of motion. The other ball

must also change its direction of motion too. [0/1]

e. Situations “a”, “b”, and “d” are reasonable, but not “c”. See above. [0/3]





10 Show that on a roller coaster with a circular vertical loop, the difference in your

apparent weight at the top of the loop and at the bottom of the loop is g⋅6 . Ignore

friction. Show also that as long as your speed is above the minimum needed, this

answer is independent of the size of the loop or how fast you go through it. [2/4/¤¤¤¤]


In order for the object to pass the top point the normal force must be larger than zero. Its

minimum value at the top is zero. i.e.: 0≥ N F . Free-body-diagram for the object at the top

may be represented as:

Taking the direction of the acceleration as positive, Newton’s second law,

amF net

r

r

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

C top N mamgF =+ _

Note that the apparent weight of any object is the normal force exerted by the

surface on the body, i.e. N F .

R

vmmgF

top

top N

2

_ =+

Rmg RF mv top N top ⋅+⋅= _

2

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

212 toptop mvmgR E +=

( ) Rmg RF mgRmvmgR E top N toptop ⋅+⋅+=+= _

2

2

12

2

12

RF mgR E top N top ⋅⋅+⋅= _ 2

15.2

Note that mgR2 is the potential energy of the object at the top. The reference point for the

potential energy is taken to be the lowest point of the circle.

Free-body diagram for the object at the bottom may be illustrated as:

Taking the direction of acceleration, i.e. up, positive, Newton’s second law of motion,

amF net

r

r

= , for the object at the bottom of the vertical circle may be expressed as:

h

R2

N F

mg

C a

m

+





R

vmmgF bottom

bottom N

2

_ =−

R

vmmgF bottom

bottom N

2

_ +=

mg R

vmF bottombottom N +=

2

_

Using the conservation of energy:

topbottom E E =

Using RF mgR E top N top ⋅⋅+⋅= _ 2

15.2 calculated above we may find the

velocity of the object at the bottom to its energy at the top:

RF mgR E mv E top N topbottombottom ⋅⋅+⋅=== _

2

2

15.2

2

1

RF mgRmv top N bottom ⋅+⋅= _

2 5

Using mg R

mvF bottom

bottom N +=2

_ and replacing RF mgRmv top N bottom ⋅+⋅= _

25 in it results in:

mgF mgF mgmg R

RF mgRmg

R

mvF top N top N

top N bottombottom N ⋅+=++⋅=+

⋅+⋅=+= 65

5 _ _

_ 2

_

Answer: mgF F top N bottom N ⋅=− 6 _ _ QED.

Note that the difference in the apparent weight of the object is mgF F top N bottom N ⋅=− 6 _ _ and

it is independent from the speed of the object and the radius of the ring. It is always mg⋅6

just a function of the mass of the object and the gravitational acceleration of the object.

N F

mg

C a

m

+

Solution+FyBNVC06+Test+V1+Ch5 8+Circular Rotational+Motion,+Energy

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