GENERAL PHYSICS PH 221-3A (Dr. S. Mirov) Test 3 (11/7/07)

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keySample Test 3

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 Motion along a straight line with a constant acceleration1. Motion along a straight line with a constant accelerationvaver. 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 plane3. 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; ox ;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

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 ˆy z x yx z

x y zy z x yx z

x y z

i j ka a a aa a

a b b a a a a i j kb b b bb b

b b b

ˆˆ ˆ( ) ( ) ( )

y

y z y z z x z x x y x ya 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 St ti f i ti f N3. 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 ( )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

i n tm e c t hW E E E E 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 hE 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

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

S9. Angular Position: ( r a d i a n m e a s u r e )S

r

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

1 ( )2

12

o

o o

o o

t

t

t t

2 2

2

2 ( )12

o o

o t t

2. Linear and angular variables related:

22 2 2v rT 2; ; ; ; t rs r v r a r a r T

r 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

y y

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

2 2d t

8. Rolling bodies:

2 21 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 p

c o m

c o m c o m

c o m

v R

K I m v

a Rga

2 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

aI M R

9. Torque as a vector: ; s i nr F r F r F r F

1. Angular Momentum of a particle: ( );l r p m r v

g p

sinl rmv rp rmv r p r mv

2. Newton’s Second law in Angular Form: netdldt

n

3. Angular momentum of a system of particles: 1

n

ii

net ext

L l

dLdt

4 A l M t f Ri id B d L I4. Angular Momentum of a Rigid Body: L I

5. Conservation of Angular Momentum: (isolated system)i fL L

6. Static equilibrium: net0; 0if ll th f li i l 0 0 0

netFF F

, , ,if all the forces lie in xy plane 0; 0; 0net x net y net zF F

7. Elastic Moduli: stress=modulus strain

8. Tension and Compression: , E is the Young's modulusF LEA L

A L

9. Shearing: , G is the shear modulusF LGA L

10. Hydraulic Stress: , B is the bulk modulusVp B y ,p

V

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. What is the height above Earth's surface of the center of mass of the two-ball system at +y g ythe end of 2.0 s?

y

m1

0

m2

v1=0v2=15 m/s

y=25m5.4m

10.4m

t=0 t=2.0s( ) Find final y coordinate of the ball 1 after 2.0 s.a

2 2

19.8 225 0 25 5.4

2 2(b) Find final y coordinate of the ball 2 after 2.0 s

fgty t m

2 2

2 29.8 215 2 10.4

2 2(c) Find y coordinate of the c

f ogty v t m

enter of mass of two balls( ) y

1 1 2 2

1 2

(0.5)(5.4) (0.25)(10.4) 7.1(0.5 0.25)com

m y m yy mm m

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)does each skater travel? (Neglect friction)

10m100kg 60kg

0 +x

( ) Since two scaters represent an isolated system and initially theu are at resttheir center of mass will be at rest and they will meet at the center of massa

(100)(0) (60)(10)1 1 2 2100 6

1 2com

m x m xxm m

0 3.75 3.8

(b) The 100 kg skater moves 3.75 m

m m

( ) g(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. (a) find the speed of the block just after impact.(b) find the friction force between block and table.

v1o v20=0

m1 m2

2.7 m

1 10 2 20

(a) Consider inelastic bullet-block collision. At the instant of collison bullet-block system isisolated, hence, we can use law of conservation of linear momentum to describe the collision

m v m v 11 10 2 201 2

(0.005)(150) (2)(0)( ) ; 0.374 4 10 /( ) ( )

m v m vm m v v m s 1 10 2 20 1 2

1 2

( ) ;( ) (0.005 2)

(b) Calculate work done by the net force on a block-bullet system during its motion along the tableThere are 3 forces acting

m m

on our system: Weight Normal reaction and kinetif frictionThere are 3 forces acting on our system: Weight, Normal reaction, and kinetif friction. Only kinetic friction force will perform non zero work. (c) Motion of a block-bullet system after the collision can be describe

net kW f S

2 2 2

d by a work-energy theorem( ) ( ) (0 005 2)(0 374)m m v m m v 21 2 1 2( ) ( ) (0.005 2)(0.374)0 5 10

2 2 2(2.70)net f i k km m v m m vW K K f S f N

S

4. The impact of the head of a golf club on a golf ball can be approximately regarded as anelastic collision. The mass of the head of the golf club is 0.15 kg. The speed of the clubbefore 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?

2(a) For elastic golf club-ball collision cbf ci

mv v

2 22(0.15)(46) 0.15 0.047 4.7 1070

c ci

bf

fc b

m vb cv

m m

m m kg

2(0.047)(70) 0( ) 6600 6.6 100 0005

bf biavg

P PPb F N Nt t

0.0005t t

5.

6. Joe is painting the floor of his basement using a paint roller. The roller has a mass of2.4 kg and a radius of 3.8 cm. In rolling the roller across the floor Joe applies a forceF=16N directed at an angle of 35% as shown. Ignoring the mass of the roller handle, what is the magnitude of the angular acceleration of the roller?g g

35

F N

fx

y

fs Fmg(a) 2nd Newton law x axis: sin

( ) 2 d N t L f t ti b ts comF f ma

b f R I

( ) 2nd Newton Law for rotation about com: (c) recall the relationship between and : ;

( ) b i (b) d ( ) i ( ) i

s

com com

b f R Ia a R

Id F R

( ) substitute eq.(b) and (c) in (a): sin ;d F m RR

sin sin 2 sin 2(16)sin 351 3 3(2 4)(0 038)

F F FI R 1 3 3(2.4)(0.038)

2

67

I mRmR mR mRR

rad

267s

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.identical inclines. Rank the objects according to how high they go, least to greatest.

( )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 a

to thermal ener

2 2

gy because the objects do not slide; 2) normal force is perpendicular to the pass;3) gravitational force is a conservative one.1 1 0 0I Mv Mgh

0 02 2( ) second term is the same for all our objects, is

I Mv Mgh

b

also the same, the larger I the larger maximum height h reached by the object.f l t t t t i t f h i ht tt i d

from least to greatest in terms of height attained: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 inbalance, if

(a) the weight of the seesaw can be ignored (b) th i h M 300N (b) the seesaw weighs Mg=300N

6 m

Mg

x w2 w1

( ) Use 2 nd condition of equilibrium:a

1 2

2

(6 ) 0;6 6(400) 2.18

700 400

W x W xWx m

W W

1 2 700 400

( )Use 2 nd condition of equilibrium:

W W

b

1 2

2

( )Use 2 nd condition of equilibrium:(3 ) (6 ) 0;

6 3 6(400) 3(300) 2 36

bW x Mg x W x

W Mgx m

1 2

2.36700 400 300

x mW W Mg