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

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keyTest 3 (11/04/09)Test 3 (11/04/09)

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. = Δx/Δt; vins =dx/Δt; aaver.= Δvaver. vel./Δt; a = dv/Δt; 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 cosθ; vy = vo sinθ; 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 cosθ; x = vox t; ox ;xmax = (2 vo

2 sinθ cosθ)/g = (vo2 sin2θ)/g for yin = yfin;

vy = voy - gt = vo sinθ - gt; y = voy t - 1/2 gt2;

5. Uniform circular Motion a=v2/r, T=2πr/v

6. Relative motion P A P B B A

P A P B

v v va a

= +=

r r r

r r

7. Component method of vector addition

A = A1 + A2 ; Ax= Ax1 + Ax2 and Ay = Ay1 + Ay2; A A Ax y= +2 2 ; θ = tan-1 ⏐Ay /Ax⏐;

The scalar product A = c o sa b a b φ⋅rr

ˆ ˆˆ ˆ ˆ ˆ( ) ( )x y z x y za b a i a j a k b i b j b k⋅ = + + ⋅ + +

rr

r = x x y y z za b a b a b a b⋅ + +

r

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× = + + × + +rr

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× = − × = = − + =

r rr r

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 =μkN;

3 St ti f i ti f N3. Static friction fs =μsN;

4. Universal Law of Gravitation: F=GMm/r2; G=6.67x10-11 Nm2/kg2;

5. Drag coefficient 212

D C A vρ=

6. Terminal speed 2

tm gv

C Aρ=

7. Centripetal force: Fc=mv2/r

8. Speed of the satellite in a circular orbit: v2=GME/r

r9. The work done by a constant force acting on an object: c o sW F d F dφ= = ⋅

rr

10. Kinetic energy: 212

K m v=

11. Total mechanical energy: E=K+U

12. The work-energy theorem: W=Kf-Ko; Wnc=ΔK+ΔU=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 φ=

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 WΔ = − = + = = −

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 tφ= = = = ⋅

Δ

r r

6. Potential energy: ; ( )f

i

x

xU W U F x d xΔ = − Δ = − ∫

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 yΔ = − = Δ = = =

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

= Δ = Δ + Δ

= Δ + ΔΔ =

i n tm e c t hW E E E E= Δ = Δ + Δ + Δ11. 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 EΔ + Δ + Δ =

12. Power: ; a v gE d EP Pt d t

Δ= =

Δ;

13. Center of mass: 1

1 n

c o m i ii

r m rM =

= ∑r r

14. Newtons’ Second Law for a system of particles: n e t c o mF M a=r r

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

= =r

r rr

2. Collision and impulse: ( ) ; ;

f

i

t

a v gtJ F t d t J F t= = Δ∫r r

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 tΔ

= − Δ = − Δ = − ΔΔ Δ Δ

t t tΔ Δ Δ

Impulse-Linear Momentum Theorem: f ip p J− =rr r

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=r r

4. Inelastic collision in one dimension: 1 2 1 2i i f fp p p p+ = +r r r r

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

rθ =

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 )θ θ θΔ = −

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 tθ θω ωΔ

= =Δ

12. Angular acceleration: ; a v gd

t d tω ωα αΔ

= =Δ

1. angular acceleration: 2

1 ( )2

12

o

o o

o o

t

t

t t

ω ω α

θ θ ω ω

θ θ ω ω

= +

− = +

− = +

2 2

2

2 ( )12

o o

o t t

ω ω α θ θ

θ θ ω ω

= + −

− = −

2. Linear and angular variables related:

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

r vθ ω α ω

ω= = = = = = =

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

ω= =

=

∑

∫

y y∫

4. The parallel axes theorem: 2c o mI I M h= +

5. Torque: s i ntr F r F r Fτ φ⊥= = =

6. Newton’s second law in angular form: n e t Iτ α=

7. Work and Rotational Kinetic Energy: ; ( ) f o r ;f

if id W c o n s tW

θ

θτ θ τ θ θ τ= − == ∫

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

ω ω= Δ = − = − = 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

ω

ω

αθ

=

= +

=

=

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 Fτ τ φ ⊥ ⊥= × = = =rr r

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

r r r r r

g psinl rmv rp rmv r p r mvφ ⊥ ⊥ ⊥ ⊥= = = = =

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

τ =r

r

n

3. Angular momentum of a system of particles: 1

n

ii

net ext

L l

dLdt

τ

=

=

=

∑rr

rr

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

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

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

netFF F

τ= =r r

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

7. Elastic Moduli: stress=modulus strain×

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

Δ=

A L

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

Δ=

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

V

( ) The chosen system of interest consists of two blocks and a springa

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 two-block system 2.0 s later, before either fragment has hit the ground?( ) The chosen system of interest consists of two blocks and a 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

a

net external force acting on the netF

system is the gravitational force. Thus from equation , the acceleration . This means that the center of mass of two springed apart blocksfollo

net com

com

F Maa g

==

r r

r

ws the same parabolic trajectory that they had being tied together.p j y y g gb) Find time of maximum height using equation 0

35 sin 45 2.539 8

h oy h

oyh

t v gt

vt s

g

= −

= = =

2

9.8c) Find maxiumum height

35 sin 45 2.

h

hh h

gy

gty v t= − = ⋅29.8 2.5353 31.25m⋅

− =35 sin 45 2.2h oy hy v t 53 31.25

2d) 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.

m

2 29.8 4.5335 sin 45 4.53 11.562 2

e) -

oy

h

gty v t m

y y

⋅= − = ⋅ − =

Δ = 19.69 20 y 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?travel?

20m640N 3200N

0 +x

( ) 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 massa

1 1 2 2 1 1 2 2

1 2 1com

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

mg

++= =

( )(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 V20=‐25m/s

mAmB

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

(2.0)(50) (4.0)(25)) ; 0.0 /( ) (2 0 4 0)

m v m vm m v v m s− −+ ⇒ = = =1 10 2 20 ( 1 2

1 2

) ;( ) (2.0 4.0)

(b) Hence, 0.( ) C l l t fi l ki ti

f

m m

K

+ +

=

2 2 2 21 1 2 2

(c) Calculate final kinetic energy. 2.0 50 4.0 ( 25) 2500 1250 3750

2 2 2 2(d) Kinetic energy lost dur

fm v m vK J⋅ ⋅ −

= + = + = + =

ing the collision ( ) gy2

g3750 0 3750 3.8 10i fK K K J JΔ = − = − = = ×

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

(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 at= 1

22 , so

a y× −2 2 0750 6 00 10 2 2( . )m m/ sslipping 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 of the pulley’s

a yt

= = = ×500

6 00 102 2( . ). .

sm/ s

Block 1 has an acceleration of 6.00 × 10–2 m/s2 upward. (b) Newton’s second law for block 2 is m g T m a− = where m2 is its mass and T2 is the( ) g p y

angular acceleration? (e) What is its rotational inertia?

(b) Newton s second law for block 2 is 2 2 2m g T m a− = , where m2 is its mass and T2 is the tension force on the block. Thus,

( )2 2 22 2 ( ) (0.500 kg) 9.8 m/s 6.00 10 m/s 4.87 N.T m g a −= − = − × =

(c) Newton’s second law for block 1 is 1 1 1 ,m g T m a− = − where T1 is the tension force on the block. Thus,

( )2 2 21 1( ) (0.460 kg) 9.8 m/s 6.00 10 m/s 4.54 N.T m g a −= + = + × =

aa

(d) Since the cord does not slip on the pulley, the tangential acceleration of a point on therim of the pulley must be the same as the acceleration of the blocks, so

α = =×

=−a 600 10 120

2 22. m/ s rad / s

+Y m2gm1g

a

α× −R 500 10

1202.. .

mrad / s

(e) The net torque acting on the pulley is 2 1( )T T Rτ = − . Equating this to Iα we solve for the rotational inertia:

10

( ) ( )( )22 1 2 2

2

4.87 N 4.54 N 5.00 10 m1.38 10 kg m .

1.20 rad/sT T R

Iα

−−

− ×−= = = × ⋅

5. In Figure below a solid ball ( I MR= 25

2 ) rolls smoothly from rest (starting at heightH=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 (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

22ω and its

final potential energy is M gh. Here we use v to denote the speed of its center of mass andω is its angular speed — at the moment it leaves the track. Since (up to that moment) theball rolls without sliding we can set ω = v/R Using I MR= 2 2 conservation of energy

ball rolls without sliding we can set ω v/R. Using I MR= 5 , conservation of energyleads to

2 2 2 2 21 1 1 2 7 .2 2 2 10 10

MgH Mv I Mgh Mv Mv Mgh Mv Mghω= + + = + + = +

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,

11

g y g gsubstituting 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 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?

) l d hild i l d b h la) 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 problem.

2 2

2 2

2 2

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

L L I m r I m r

I m rrad s

I m r

ω ω

ωω

= + = +

+ + ⋅ ⋅= = =

+ +

r r

( ) 600 20 3.0p c fI m r+ + ⋅

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 F on his hands from the ground and F on his feet from the wall static frictional1NF on his hands from the ground and 2NF on his feet from the wall, static frictionalforce sf 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 andhorizontal directions, we have

net , 2

net , 1

0

0 .x N s

y N

F F f

F F mg

= = −

= = −∑∑

In addition, the net torque about O (contact point between his feet and the wall) must also , q ( p )vanish:

net 20 cos sinNO

mgd F Lτ θ θ= = −∑ .

The torque equation gives 2 cos / sin cot /NF mgd L mgd Lθ θ θ= = . On the other hand,The torque equation gives 2 cos / sin cot /NF mgd L mgd Lθ θ θ . On the other hand, from the force equation we have 2N sF f= and 1 .NF mg= These expressions can be combined to yield

2 1 cots N Ndf F FL

θ= = .

On the other hand, the frictional force can also be written as 1s s Nf Fμ= , where sμ 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 sμ to be

13

2 2 2 2

0.914 m 0.940 mcot 0.2162.10 m(2.10 m) (0.914 m)

sd a dL LL a

μ θ= = = =− −

.

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?

Rv

Rh +

v

FBD30°

WT

30°

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

sin 30 0i 30 400

WL TLT W− + =

Nsin 30 400T W= = N

99.

15