GENERAL PHYSICS PH 221-2A (Dr. S. Mirov) Test 3 (11/06/14)
STUDENT NAME: ________________________ STUDENT id #: ___________________________ -------------------------------------------------------------------------------------------------------------------------------------------
ALL QUESTIONS ARE WORTH 30 POINTS. WORK OUT FIVE PROBLEMS. NOTE: Clearly write out solutions and answers (circle the answers) by section for each part (a., b., c., etc.)
Important Formulas:
1. Motion along a straight line with a constant acceleration vaver. speed = [dist. taken]/[time trav.]=S/t; vaver.vel. = x/t; vins =dx/t; aaver.= vaver. vel./t; a = dv/t; v = vo + at; x= 1/2(vo+v)t; x = vot + 1/2 at2; v2 = vo
2 + 2ax (if xo=0 at to=0)
2. Free fall motion (with positive direction ) g = 9.80 m/s2; y = vaver. t vaver.= (v+vo)/2; v = vo - gt; y = vo t - 1/2 g t2; v2 = vo
2 – 2gy (if yo=0 at to=0)
3. Motion in a plane vx = vo cos; vy = vo sin; x = vox t+ 1/2 ax t2; y = voy t + 1/2 ay t2; vx = vox + at; vy = voy + at;
4. Projectile motion (with positive direction ) vx = vox = vo cos; x = vox t; xmax = (2 vo
2 sin cos)/g = (vo2 sin2)/g for yin = yfin;
vy = voy - gt = vo sin - gt; y = voy t - 1/2 gt2;
5. Uniform circular Motion a=v2/r, T=2r/v
6. Relative motion P A P B B A
P A P B
v v va a
7. Component method of vector addition
key
A = A1 + A2 ; Ax= Ax1 + Ax2 and Ay = Ay1 + Ay2; A A Ax y 2 2 ; = tan-1 Ay /Ax;
The scalar product A = c o sa b a b
ˆ ˆˆ ˆ ˆ ˆ( ) ( )x y z x y za b a i a j a k b i b j b k
= x x y y z za b a b a b a b
The vector product ˆ ˆˆ ˆ ˆ ˆ( ) ( )x y z x y za b a i a j a k b i b j b k
ˆˆ ˆˆˆ ˆ
ˆˆ ˆ( ) ( ) ( )
y z x yx zx y z
y z x yx zx y z
y z y z z x z x x y x y
i j ka a a aa a
a b b a a a a i j kb b b bb b
b b b
a b b a i a b b a j a b b a k
1. Second Newton’s Law ma=Fnet ;
2. Kinetic friction fk =kN;
3. Static friction fs =sN;
4. Universal Law of Gravitation: F=GMm/r2; G=6.67x10-11 Nm2/kg2;
5. Drag coefficient 212
D C A v
6. Terminal speed 2
tm gv
C A
7. Centripetal force: Fc=mv2/r
8. Speed of the satellite in a circular orbit: v2=GME/r
9. The work done by a constant force acting on an object: c o sW F d F d
10. Kinetic energy: 212
K m v
11. Total mechanical energy: E=K+U
12. The work-energy theorem: W=Kf-Ko; Wnc=K+U=Ef-Eo
13. The principle of conservation of mechanical energy: when Wnc=0, Ef=Eo
14. Work done by the gravitational force: c o sgW m g d
1. Work done in Lifting and Lowering the object:
; ; f i a g f i a gK K K W W i f K K W W
2. Spring Force: ( H o o k ' s l a w )xF k x
3. Work done by a spring force: 2 2 21 1 1; i f 0 a n d ; 2 2 2s i o i f sW k x k x x x x W k x
4. Work done by a variable force: ( )f
i
x
x
W F x d x
5. Power: ; ; c o sa v gW d WP P P F v F v
t d t
6. Potential energy: ; ( )f
i
x
xU W U F x d x
7. Gravitational Potential Energy:
( ) ; 0 a n d 0 ; ( )f i i iU m g y y m g y i f y U U y m g y
8. Elastic potential Energy: 21( )2
U x k x
9. Potential energy curves: ( )( ) ; ( ) ( )m e c
d U xF x K x E U xd x
10. Work done on a system by an external force:
F r i c t i o n i s n o t i n v o l v e d W h e n k i n e t i c f r i c t i o n f o r c e a c t s w i t h i n t h e s y s t e m
m e c
m e c t h
t h k
W E K UW E E
E f d
11. Conservation of energy: i n t
i n tf o r i s o l a t e d s y s t e m ( W = 0 ) 0m e c t h
m e c t h
W E E E EE E E
12. Power: ; a v gE d EP Pt d t
;
13. Center of mass: 1
1 n
c o m i ii
r m rM
14. Newtons’ Second Law for a system of particles: n e t c o mF M a
1. Linear Momentum and Newton’s Second law for a system of particles: a n d c o m n e td PP M v Fd t
2. Collision and impulse: ( ) ; ;
f
i
t
a v gtJ F t d t J F t
when a stream of bodies with mass m and
speed v, collides with a body whose position is fixed a v gn n mF p m v v
t t t
Impulse-Linear Momentum Theorem: f ip p J
3. Law of Conservation of Linear momentum: f o r c l o s e d , i s o l a t e d s y s t e mi fP P
4. Inelastic collision in one dimension: 1 2 1 2i i f fp p p p
5. Motion of the Center of Mass: The center of mass of a closed, isolated system of two colliding bodies is
not affected by a collision.
6. Elastic Collision in One Dimension: 1 2 11 1 2 1
1 2 1 2
2; f i f im m mv v v vm m m m
7. Collision in Two Dimensions: 1 2 1 2 1 2 1 2; i x i x f x f x i y i y f y f yp p p p p p p p
8. Variable-mass system:
( f i r s t r o c k e t e q u a t i o n )
l n ( s e c o n d r o c k e t e q u a t i o n )
r e l
if i r e l
f
R v M aMv v vM
9. Angular Position: ( r a d i a n m e a s u r e )Sr
10. Angular Displacement: 2 1 ( p o s i t i v e f o r c o u n t e r c l o c k w i s e r o t a t i o n )
11. Angular velocity and speed: ; ( p o s i t i v e f o r c o u n t e r c l o c k w i s e r o t a t i o n )a v gd
t d t
12. Angular acceleration: ; a v gd
t d t
1. angular acceleration: 2
2 2
2
1 ( )2
12
2 ( )12
o
o o
o o
o o
o
t
t
t t
t t
2. Linear and angular variables related:
2
2 2 2; ; ; ; t rv rs r v r a r a r Tr v
3. Rotational Kinetic Energy and Rotational Inertia:
2 2
2
1 ; f o r b o d y a s a s y s t e m o f d i s c r e t e p a r t i c l e s ;2
f o r a b o d y w i t h c o n t i n u o u s l y d i s t r i b u t e d m a s s .
i iK I I m r
I r d m
4. The parallel axes theorem: 2c o mI I M h
5. Torque: s i ntr F r F r F
6. Newton’s second law in angular form: n e t I
7. Work and Rotational Kinetic Energy:
; ( ) f o r ;f
if id W c o n s tW
2 21 1; w o r k e n e r g y t h e o r e m f o r r o t a t i n g b o d i e s2 2f i f i
d WP K K K I I Wd t
8. Rolling bodies:
2 2
2
1 12 2
s i n f o r r o l l i n g s m o o t h l y d o w n t h e r a m p1 /
c o m
c o m c o m
c o m
c o mc o m
v R
K I m v
a RgaI M R
1. Angular Momentum of a particle: ( );
sinl r p m r vl rmv rp rmv r p r mv
2. Newton’s Second law in Angular Form: netdldt
3. Angular momentum of a system of particles: 1
n
ii
net ext
L l
dLdt
4. Angular Momentum of a Rigid Body: L I
5. Conservation of Angular Momentum: (isolated system)i fL L
6. Static equilibrium: net
, , ,
0; 0if all the forces lie in xy plane 0; 0; 0
net
net x net y net z
FF F
7. Elastic Moduli: stress=modulus strain
8. Tension and Compression: , E is the Young's modulusF LEA L
9. Shearing: , G is the shear modulusF LGA L
10. Hydraulic Stress: , B is the bulk modulusVp BV
7
1. Three identical rods are rigidly connected to a disk, as shown in Fig. Find thecenter of mass of the system.
20 cm 20 cm
R=8.0 cm m1=40 g
m2=40 g
m4=20 g
20 cm
m3=40 g
x
y
Considering the center of the disk as (0,0) :( 18)(40) (18)(40) 0 0 0;
1400 0 0 (18)(40) 5.1
140
com
com
x
y cm
2. A 640-N hunter gets a rope around a 3200-N polar bear. They are stationary, 20m apart, on frictionless level ice. When the hunter pulls the polar bear to him, how far does the polar bear travel?
0 +x
20m640N 3200N
1 1 2 2 1 1 2 2
1 2 1
( ) Since hunter-polar bear represent an isolated system and initially they are at resttheir center of mass will be at rest and they will meet at the center of mass
com
a
m x m x m gx m gxxm m m g m
(640)(0) (3200)(20)
640 32002
16.7
(b) The 640 N hunter moves 16.7 m(c) The 3200 N polar bear moves 20-16.7=3.3m
mg
3. Blocks A and B are moving toward each other. A has a mass of 2.0kg and a velocity of 50m/s, while B has a mass of 4.0kg and a velocity of -25m/s. They suffer a completely inelastic collision. What is the kinetic energy lost during the collision?
mAmB
v10=50 m/s V20=‐25m/s
x
1 10 2 20
(a) Consider inelastic block-block collision. At the instant of collision block-block system isisolated, hence, we can use law of conservation of linear momentum to describe the collision
(m v m v 1 10 2 201 2
1 2
2 2 2 21 1 2 2
(2.0)(50) (4.0)(25)) ; 0.0 /( ) (2.0 4.0)
(b) Hence, 0.
(c) Calculate final kinetic energy. 2.0 50 4.0 ( 25) 2500 1250 3750
2 2 2 2(d) Kinetic energy lost dur
f
f
m v m vm m v v m sm m
K
m v m vK J
3
ing the collision 3750 0 3750 3.8 10i fK K K J J
10
4.
11
To find where the ball lands, we need to know its speed as it leaves the track (usingconservation of energy). Its initial kinetic energy is Ki = 0 and its initial potential energyis Ui = M gH. Its final kinetic energy (as it leaves the track) is K Mv If 1
22 1
22 and its
final potential energy is M gh. Here we use v to denote the speed of its center of mass and is its angular speed — at the moment it leaves the track. Since (up to that moment) theball rolls without sliding we can set = v/R. Using I MR 2
52 , conservation of energy
leads to
2 2 2 2 21 1 1 2 7 .2 2 2 10 10
MgH Mv I Mgh Mv Mv Mgh Mv Mgh
The mass M cancels from the equation, and we obtain
v g H h 107
107
9 8 6 0 2 0 7 482b g d ib g. . . . .m s m m m s
Now this becomes a projectile motion. We put the origin at the position of the center ofmass when the ball leaves the track (the “initial” position for this part of the problem) andtake +x rightward and +y downward. Then (since the initial velocity is purely horizontal)the projectile motion equations become
x vt y gt and 12
2.
Solving for x at the time when y = h, the second equation gives t h g 2 . Then,substituting this into the first equation, we find
2
2 2.0 m2 7.48 m/s 4.8 m.9.8 m/s
hx vg
6. A playground merry-go-round has a radius of 3.0m and a rotational inertia of 600 kgm2. It is initially spinning at 0.80 rad/s when a 20-kg child crawls from the center to the rim. When the child reaches the rim what is the angular velocity of the merry-go-round?
a) Playground-child represent an isolated system, because the net external torqueabout the playground axis of rotaion is zero. Hence, one can use law of conservation ofangular momentum to solve the pr
2 2
2 2
2 2
oblem.
b) ; ( ) ( )
( ) (600 20 0 ) 0.8 0.62 /( ) 600 20 3.0
i f p c i o p c f f
p c i of
p c f
L L I m r I m r
I m rrad s
I m r
13
7. A uniform horizontal beam of length 8.00 m and weight 200 N is attached to a wall by a pin connection.Its far end is supported by a cable that makes an angle of 53.0 with the horizontal. If a 600 N man stands2.00 m from the wall, find the tension in the cable and the force exerted by the wall on the beam.