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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
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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
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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
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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]
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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
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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)
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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
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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]
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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
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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
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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=
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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?
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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
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( ) 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
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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
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+=
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?