1. No category

GENERAL PHYSICS PH 221-3A (Dr. S. Mirov)
Test
3 (11/7/07)
Sample
Test 3
key
STUDENT NAME: ________________________ STUDENT id #: ___________________________
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ALL QUESTIONS ARE WORTH 20 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
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 = vo2 + 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 = vo2 – 2gy (if yo=0 at to=0)
3
3.
Motion in a plane
4.
Projectile motion (with positive direction )
5.
Uniform circular Motion
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;
vx = vox = vo cos;
x = vox t;;
xmax = (2 vo2 sin cos)/g = (vo2 sin2)/g for yin = yfin;
vy = voy - gt = vo sin - gt;
y = voy t - 1/2 gt2;
a=v2/r,
T=2r/v
6. Relative motion



v PA  v PB  v BA


a PA  a PB
7.
Component method of vector addition
A 
A x2  A y2 ;  = tan-1 Ay /Ax;


The scalar product A a  b = a b c o s 


a  b  ( a x iˆ  a y ˆj  a z kˆ )  ( b x iˆ  b y ˆj  b z kˆ )
 
a  b = a xb x  a yb y  a zbz


The vector product a  b  ( a x iˆ  a y ˆj  a z kˆ )  ( b x iˆ  b y ˆj  b z kˆ )
A = A1 + A2 ; Ax= Ax1 + Ax2 and Ay = Ay1 + Ay2;
iˆ




a  b  b  a  ax
bx
ˆj
a
b
y
y
kˆ
a
a z  iˆ
b
bz
y
a
y
bz
 ( a y b z  b y a z ) iˆ  ( a z b x  b z a x ) ˆj  ( a x b
y
z
ax
bx
 ˆj
 bxa
y
ax
az
 kˆ
bx
bz
a
y
b
y

) kˆ
1.
Second Newton’s Law
ma=Fnet ;
2.
Kinetic friction fk =kN;
3
3.
St ti friction
Static
f i ti
fs =sN;
N
4.
Universal Law of Gravitation: F=GMm/r2; G=6.67x10-11 Nm2/kg2;
5.
Drag coefficient
6.
Terminal speed
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:
10. Kinetic energy:
D 
vt 
K 
1
C  Av
2
2
2m g
C  A
1
m v
2
W
 
 F d cos  F  d
2
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:
W
g
 m gd cos
1.
Work done in Lifting and Lowering the object:
 K

K
f
 K
 W
i
F
a
 W
  k x
; if
g

f
K
Spring Force:
3.
Work done by a spring force:
4.
Work done by a variable force:
W
1
k x
2

s
x


W
x
W
; P
 t
Pavg 
5.
Power:
6.
Potential energy:
7.
d W
d t

 U
i
; W
2
i

 W
a
g
( H o o k 's l a w )
2.
x
K
 W ;  U
2
o
; if x
i

x ; W
s
 
F ( x )d x
i

 
x
x
f


F  v

F ( x )d x
i
 m g ( y
f

y i )  m g  y ; if
y
i
 0 a n d U
1
k x
2
8.
Elastic potential Energy:
U ( x ) 
9.
Potential energy curves:
F ( x )  
10.
Work done on a system by an external force:
F ric tio n is n o t in v o lv e d W
i
 0; U ( y )  m g y
2
d U ( x )
; K ( x ) 
d x
  E
m e c
  K
E
 E
th

 U ( x )
m e c
  U
W h e n k in e tic fric tio n fo rc e a c ts w ith in th e s y s te m
  E
W
m e c
  E
th
  E
in t
fkd
Conservation of energy:
Pavg 
 E
; P
 t
12.
Power:
13.
Center of mass:
14.
f
Gravitational Potential Energy:
 U
11.
 0 a n d x
f
F v c o s 

; P
1
k x
2

rc o m

W
  E
  E
m e c
  E
fo r is o la te d s y s te m

1
M
d E
d t
i 1
  E
in t
m e c
;
n

th
(W = 0 )  E
m
i

ri
Newtons’ Second Law for a system of particles:

F net 

M a
c o m
  E
th
 0
1
k x
2
2
1.
2.




dP
Linear Momentum and Newton’s Second law for a system of particles: P  M v c o m a n d F n e t 
dt

J 
Collision and impulse:

t

F ( t ) d t ; J  F a v g  t ; when a stream of bodies with mass m and
f
ti
F avg  
speed v, collides with a body whose position is fixed

p
Impulse-Linear Momentum Theorem:

 p
f
n
n
 m
 p  
 v
m  v  
 t
 t
 t

 J
i


Pi  P f f o r c l o s e d , i s o l a t e d s y s t e m
3.
Law of Conservation of Linear momentum:
4.
Inelastic collision in one dimension:
5.
Motion of the Center of Mass: The center of mass of a closed, isolated system of two colliding bodies is


p 1i  p

 p1
2 i
f

 p
2 f
not affected by a collision.
6.
Elastic Collision in One Dimension:
7.
Collision in Two Dimensions:
8.
Variable-mass system:
9.
Angular Position:
v
 vi  v
10. Angular Displacement:

m
m

 m
 m
1
1
 p1
2 ix
fx
2
v1i ;
v
2 f
2
 p
2 fx
;

m
p 1 iy  p
2m1
1  m
2 iy
v1i
2
 p1
fy
 p
2 fy
ln
rel
M
M
i
( s e c o n d ro c k e t e q u a tio n )
f
(ra d ia n m e a s u re )
   
11. Angular velocity and speed:
12. Angular acceleration:
f
 M a (firs t ro c k e t e q u a tio n )
rel
S
r
 
p 1 ix  p
R v
f
v1
avg


2
 
1
(p o s itiv e fo r c o u n te r c lo c k w is e ro ta tio n )

 
;
 t
 
;
 t
 
avg
 
d 
dt
d
dt
(p o s itiv e fo r c o u n te rc lo c k w is e ro ta tio n )
  
1.
angular acceleration:
o

  
o
  ot 
2
 
  
o
  )t
o
1
 t2
2
 2  (   o )
2
o
  t 
1
 t
2
v   r;
a
  r;
t
a
r
1
I
2

I 

2
r 2d m
; I 

2
r;
T

2 r
2

v

m i ri 2 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 ;
f o r a b o d y w ith c o n tin u o u s ly d is trib u te d m a s s .
 I
 M h
4.
The parallel axes theorem: I
5.
Torque:
6.
Newton’s second law in angular form:

7.
Work and Rotational Kinetic Energy:
W
P 
dW
dt
com
2
  r F t  r F  r F s i n 
;  K
 K
v
Rolling bodies:
f
 K
i
2
f

net
 I

1
I
2


2
i
f
 d ; W
  (
f
  i ) fo r   c o n st;
i
 W
w o rk e n e rg y th e o re m fo r ro ta tin g b o d ie s
  R
com
a
com
a
com


Torque as a vector:
1
I
2

1
I com 
2
  R
K
9.
v 2
 
r

Rotational Kinetic Energy and Rotational Inertia:
K
8.
2
Linear and angular variables related:
s   r;
3.
1
(
2
  

2.
  t
o


2

1
m v
2
g s in 
1  I com / M R


 r  F ;
2
2
com
fo r ro llin g s m o o th ly d o w n th e ra m p
  r F s in   r F

 r F
1.
2.
Angular
g
Momentum of a p
particle:
  
 
l  r  p  m( r  v );
l  rmv sin   rp  rmv  r p  r mv

Newton’s Second law in Angular Form:  net

dl

dt
 n 
L   li
3.
Angular momentum of a system of particles:
i 1

 net ext

dL

dt
L  I
 
5. Conservation of Angular Momentum: Li  L f (isolated system)
4
4.
A
Angular
l M
Momentum
t
off a Rigid
Ri id Body:
B d


Fnet  0;  net  0
6.
Static equilibrium:
7.
Elastic Moduli: stress=modulus  strain
8.
Tension and Compression:
9.
Shearing:
if all
ll the
th forces
f
lie
li in
i xy plane
l
Fnet , x  0;
0 Fnet , y  0;
0  net , z  0
L
F
E
, E is the Young's modulus
A
L
F
L
G
, G is the shear modulus
A
L
y
Stress:
10. Hydraulic
pB
V
, B is the bulk modulus
V
+yy
1. At the same instant that a 0.50-kg ball is dropped from 25m above Earth, a second ball,
with a mass of 0.25 kg, is thrown straight upward from Earth's surface with an initial
speed of 15m/s. They move along nearby lines and pass each other without colliding.
g above Earth's surface of the center of mass of the two-ball system
y
at
What is the height
the end of 2.0 s?
m1
v1=0
0
y=25m
10.4m
v2=15 m/s
m2
5.4m
t=0
t=2.0s
( a ) Find final y coordinate of the ball 1 after 2.0 s.
gt 2
9.8  2 2
y1 f  25  0  t 
 25 
 5.4 m
2
2
(b) Find final y coordinate of the ball 2 after 2.0 s
gt 2
9.8  2 2
y 2 f  v2ot 
 15  2 
 10.4 m
2
2
((c)) Find y coordinate of the c enter of mass of two balls
m y  m 2 y 2 (0.5)(5.4)  (0.25)(10.4)
y com  1 1

 7.1m
m1  m 2
(0.5  0.25)
2.
Two skaters with masses of 100 kg and 60 kg, respectively, stand 10.0 m apart; each
holds one end of a piece of rope. If they pull themselves along the rope until they meet, how far
does each skater travel? (Neglect friction)
100kg
10m
0
60kg
+x
( a ) Since two scaters represent an isolated system and initially theu are at rest
their center of mass will be at rest and they will meet at the center of mass
m x  m2 x2 (100)(0)(60)(10)
xcom  1 1

 3.75m  3.8m
100 60
m1  m2
( ) The 100 kg
(b)
g skater moves 3.75 m
(c) The 60 kg skater moves 10-3.75=6.25m=6.2m
3. A 2 kg block of wood rests on a long tabletop. A 5 g bullet moving horizontally with a
speed of 150 m/s is shot into the block and sticks in it. The block then slides 270 cm
along the table and stops.
(a) find the speed of the block just after impact.
(b) find the friction force between block and table.
v1o
m1
v20=0
m2
2.7 m
(a) Consider inelastic bullet-block collision. At the instant of collison bullet-block system is
isolated, hence, we can use law of conservation of linear momentum to describe the collision
m v  m2v20 (0.005)(150)  (2)(0)
m1v10  m2v20  ( m1  m2 )v;  v  1 10

 0.374  4  101 m / s
( m1  m2 )
(
(0.005
 2))
(b) Calculate work done by the net force on a block-bullet system during its motion along the table
There are 3 forces acting on our system: Weight,
Weight Normal reaction
reaction, and kinetif friction.
friction
Only kinetic friction force will perform non zero work.  Wnet   f k S
(c) Motion of a block-bullet system after the collision can be described by a work-energy theorem
Wnet
(m1  m2 )v 2
( m1  m2 )v 2 (0.005
(0 005  2)(0
2)(0.374)
374) 2
 K f  Ki   f k S  0 
 fk 

 5  102 N
2
2S
2(2.70)
4. The impact of the head of a golf club on a golf ball can be approximately regarded as an
elastic collision. The mass of the head of the golf club is 0.15 kg. The speed of the club
before the collision is 46 m/s. The ball acquires a speed of 70 m/s after the collision. The
golf club and a ball are in contact for about 0.5 ms.
a) What must be the mass of the ball?
b) What is the average force exerted by the club on the ball?
(a) For elastic golf club-ball collision vbff 
mb 
2 mc vci
vbf
(b) Favg
 mc 
2mc
vci 
mc  mb
2(0.15)(46)
 0.15  0.047  4.7  102 kg
70
P Pbf  Pbi (0.047)(70)  0



 6600 N  6.6  102 N
t
t
0 0005
0.0005
5.
6. Joe is painting the floor of his basement using a paint roller. The roller has a mass of
2.4 kg and a radius of 3.8 cm. In rolling the roller across the floor Joe applies a force
F=16N directed at an angle of 35% as shown. Ignoring the mass of the roller handle,
g
of the angular
g
acceleration of the roller?
what is the magnitude
y
35
N
F
fs
(a) 2nd Newton law x axis: F sin   f s  macom
F
mg
(b) 22nd
dN
Newton
t L
Law for
f rotation
t ti about
b t com:   f s R  I
(c) recall the relationship between acom and  : acom   R;
I
 m R;
R
F sin 
F sin 
2 F sin 
2(16)sin 35
 




I
1
3mR
R
3(2 4)(0 038)
3(2.4)(0.038)
mR 
mR  mR
R
2
rad
 67 2
s
( d ) substitute
b i
eq.(b)
(b) andd (c)
( ) in
i (a):
( ) F sin
i 
x
7.
A hoop (Ih=MR2), a uniform disk (Id=1/2MR2), and a uniform sphere (Is=2/5MR2), all
with the same mass and outer radius, start with the same speed and roll without sliding up
identical inclines. Rank the objects according to how high they go, least to greatest.
(a )During rolling up the incline motion the mechanical energy of the hoop-earth, disk-earth,
and sphere-earth systems is conserved since 1)frictional force does not transfer any energy
to thermal energy because the objects do not slide; 2) normal force is perpendicular to the pass;
3) gravitational force is a conservative one. 
1 2 1
I   Mv 2  0  0  Mgh
2
2
(b) second term is the same for all our objects,  is also the same,
 the larger I the larger maximum height h reached by the object.
 from
f
lleastt to
t greatest
t t in
i terms
t
off height
h i ht attained:
tt i d
sphere, disk, hoop.
9. A uniform seesaw of length 6 m has two youngsters of weights w1=700N and w2=400 N
sitting on the ends. Find the proper location of the pivot for the seesaw to be just in
balance, if
(a) the weight of the seesaw can be ignored
(b) th
the seesaw weighs
i h Mg=300N
M 300N
6m
x
w1
w2
Mg
( a ) Use 2 nd condition of equilibrium:
W1 x  W2 (6  x )  0;
6W2
6(400)
x

 2.18m
W1  W2 700  400
(b)Use 2 nd condition of equilibrium:
W1 x  Mg (3  x )  W2 (6  x )  0;
6W2  3Mg
6(400)  3(300)
x

 2.36
2 36m
W1  W2  Mg 700  400  300