Download Test 3

GENERAL PHYSICS PH 221-3A (Dr. S. Mirov)
Test 3 (11/7/07)
key
Test
Test 3
3 (11/03/10)
(11/04/09)
STUDENT NAME: ________________________ STUDENT id #: ___________________________
-------------------------------------------------------------------------------------------------------------------------------------------
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
1. Two 4.0-kg blocks are tied together with a compressed spring between them. They are thrown from the
ground with an initial velocity of 35m/s, 45 above the horizontal. At the highest point of the trajectory they
become untied and spring apart. About how far below the highest point is the center of mass of the twoblock system 2.0 s later, before either fragment has hit the ground?
( a ) The chosen system of interest consists of two blocks and a spring.
spring
The spring forces pushing blocks apart at the top of the trajectory are internal forces
to the system. If we ignore air drag, the net external force Fnet acting on the


system is the gravitational force. Thus from equation Fnet  Ma com , the acceleration

a com  g . This means that the center of mass of two springed apart blocks
follo ws the same pparabolic trajectory
j
y that theyy had being
g tied together.
g
b) Find time of maximum height t h using equation 0  voy  gt h
voy
35 sin 45
 2.53 s
g
98
9.8
c) Find maxiumum height y h
th 

gt h2
9.8  2.53 2
y h  voy t h 
 35 sin 45  2.53 
 31.25
31.25m
m
2
2
d) Find y coordinate of the com 2.0 s after it will reach the highest
point. In another words 4.53 s after the system was thrown.
gt 2
9.8  4.53 2
y  voy t 
 35 sin 45  4.53 
 11.56 m
2
2
e)  y  y h -y  19.69  20 m
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?
640N
20m
3200N
0
+x
( a ) Since hunter-polar bear represent an isolated system and initially they are at rest
their center of mass will be at rest and they will meet at the center of mass
m x  m2 x2 m1 gx1  m2 gx2 (640)(0)(3200)(20)
xcom  1 1


 16.7m
640 3200
m1  m2
m1 g  m2 g
( ) The 640 N hunter moves 16.7 m
(b)
(c) The 3200 N polar bear moves 20-16.7=3.3m
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?
v10=50 m/s
mA
V20=‐25m/s
mB
x
(a) Consider inelastic block-block collision. At the instant of collision block-block system is
isolated, hence, we can use law of conservation of linear momentum to describe the collision
m v  m2 v20 (2.0)(50)  (4.0)(25)
m1v10  m2 v20  (m1  m2 )v;  v  1 10

 0.0m / s
(m1  m2 )
(2 0  4.0)
(2.0
4 0)
(b) Hence, K f  0.
( )C
(c)
Calculate
l l t final
fi l kinetic
ki ti energy.
m1v12 m2 v22 2.0  502 4.0  (25) 2
Kf 



 2500  1250  3750 J
2
2
2
2
(d) Kinetic energy lost during the collision
K  K i  K f  3750  0  3750 J  3.8 103 J
4. In Figure below, block 1 has mass m1=460 g, block 2
has mass m2=500 g, and the pulley, which is mounted
on a horizontal axle with negligible friction, has
radius R=5.00 cm. When released from rest,
block 2 falls 75.0 cm in 5.00 s without the cord
slipping on the pulley.
(a) What is the magnitude of the acceleration
of the blocks?
(b) What is tension T2?
(c) What is tension T1?
((d)) What is the magnitude
g
of the ppulley’s
y
angular acceleration?
(e) What is its rotational inertia?
(a) We use constant acceleration kinematics. If down is taken to be positive and a is the
acceleration of the heavier block m2, then its coordinate is given by y  21 at 2 , so
a
2 y 2(0.750 m)

 6.00  102 m / s2 .
2
2
t
(5.00 s)
Block 1 has an acceleration of 6.00  10–2 m/s2 upward.
(b) Newton’s
Newton s second law for block 2 is m2 g  T2  m2 a , where m2 is its mass and T2 is the
tension force on the block. Thus,
T2  m2 ( g  a)  (0.500 kg)  9.8 m/s2  6.00 102 m/s2   4.87 N.
(c) Newton’s second law for block 1 is m1 g  T1  m1a, where T1 is the tension force on
the block. Thus,
T1  m1 ( g  a)  (0.460 kg)  9.8 m/s2  6.00 102 m/s2   4.54 N.
a
a
m 1g
(d) Since the cord does not slip on the pulley, the tangential acceleration of a point on the
rim of the pulley must be the same as the acceleration of the blocks, so
m 2g
+Y

a 6.00  102 m / s2
. rad / s2 .

 120
2
R
500
.  10 m
(e) The net torque acting on the pulley is   (T2  T1 ) R . Equating this to I we solve for
the rotational inertia:
T  T  R  4.87 N  4.54 N   5.00 10
I 2 1 
2

1.20 rad/s
2
m
 1.38 102 kg  m2 .
10
5. In Figure below a solid ball ( I  25 MR 2 ) rolls smoothly from rest (starting at height
H=6.0 m) until it leaves the horizontal section at the end of the track, at height h=2.0 m.
How far horizontally from point A does the ball hit the floor?
To find where the ball lands, we need to know its speed as it leaves the track (using
conservation of energy). Its initial kinetic energy is Ki = 0 and its initial potential energy
is Ui = M gH. Its final kinetic energy (as it leaves the track) is K f  21 Mv 2  21 I 2 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) the
ball rolls without sliding we can set  = v/R.
v/R Using I  25 MR 2 , conservation of energy
leads to
MgH 
1
1
1
2
7
Mv 2  I  2  Mgh  Mv 2  Mv 2  Mgh  Mv 2  Mgh.
2
2
2
10
10
The mass M cancels from the equation, and we obtain
v
b
g
d
ib
g
10
10
2
g H h 
9.8 m s 6.0 m  2.0 m  7.48 m s .
7
7
Now this becomes a projectile motion. We put the origin at the position of the center of
mass when the ball leaves the track (the “initial” position for this part of the problem) and
take +x rightward and +y downward. Then (since the initial velocity is purely horizontal)
the projectile motion equations become
x  vt and y  
1 2
gt .
2
Solving
g for x at the time when y = h, the second equation g
gives t 
substituting this into the first equation, we find
xv
2  2.0 m 
2h
  7.48 m/s 
 4.8 m.
g
9.8 m/s 2
2 h g . Then,
11
6. A playground merry-go-round has a radius of 3.0m and a rotational inertia of 600 kgm2. 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
l
d hild represent an isolated
i l d system, bbecause the
h net externall torque
about the playground axis of rotaion is zero. Hence, one can use law of conservation of
angular momentum to solve the problem.
 
b) Li  L f ; (I p  mc ri 2 )o  ( I p  mc rf2 ) f
(I p  mc ri 2 )o
(600  20  02 )  0.8
f 

 0.62rad / s
( I p  mc rf2 )
600  20  3.0
3 02
7. In figure below, a climber leans out against a vertical ice wall that has negligible friction.
Distance a is 0.914 m and distance L is 2.10 m. His center of mass is distance d=0.940 m
from the feet ground contact point. If he is on the verge of sliding, what is the coefficient
of static friction between feet and ground?
As shown in the free-body diagram, the forces on the climber consist of the normal forces
wall static frictional
FN 1 on his hands from the ground and FN 2 on his feet from the wall,
force f s and downward gravitational force mg . Since the climber is in static equilibrium,
the net force acting on him is zero. Applying Newton’s second law to the vertical and
horizontal directions, we have
0   Fnet , x  FN 2  f s
0   Fnet , y  FN 1  mg .
In addition,, the net torque
q about O (contact
(
point
p
between his feet and the wall)) must also
vanish:
0    net  mgd cos   FN 2 L sin  .
O
The torque equation gives FN 2  mgd cos  / L sin   mgd cot  / L . On the other hand,
from the force equation we have FN 2  f s and FN 1  mg . These expressions can be
combined to yield
d
f s  FN 2  FN 1 cot  .
L
On the other hand, the frictional force can also be written as f s   s FN 1 , where  s is the
coefficient of static friction between his feet and the ground. From the above equation
and the values given in the problem statement, we find  s to be
 s  cot 
d

L
a
d

2
2 L
L a
0.914 m
0.940 m
 0.216 .
2
2 2.10 m
(2.10 m)  (0.914 m)
13
8. A 400-N uniform vertical boom is attached to the ceiling by a hinge, as shown. An 800-N weight W and a
horizontal guy wire are attached to the lower end of the boom as indicated. The pulley is massless and frictionless.
What is the magnitude of the tension force T of the horizontal guy wire?
Rh
+
Rv
FBD
30
T
W
a) FBD, external forces, axis.
b) Apply second condition of static equilibrium with respect to the z axis
propagating through the hinge perpendiculat to the plane of the screen
WL sin 30  TL  0
T  W sin
i 30  400 N
9.
15