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.