Dr. Mark Stockman GENERAL PHYSICS 101-3 Fall 1995ID #_________________Seat # __________

THIRD HOUR EXAM VERSION A

You are supposed to have a calculator with built-in trigonometric functions and a pen or pencil. A list of necessary formulas is enclosed after the exam-question pages. As scratch paper please use the last (blank) page as well as the remainder of the equation page and backs of the other pages. You should not have anything else with you for this exam. Write numerical answers in blanks after the following problems. You may wish to indicate numbers of Eqs. you have used to obtain the numerical answers (see the equation page), and/or to write down the corresponding formulas with the numerical values of variables.

Important: The number of problems may exceed what can be solved by a student. You will not be graded based on the maximum number of points. You will be graded relatively to your peers, using a grading curve. Therefore, do not attempt to solve each of the problems in the given order, but start with the problems with which you are most comfortable. Having solved these, continue with more complicated (for you) problems.

1. (7 pts) A 12000-kg boxcar traveling at 11 m/s strikes another boxcar at rest. The two stick together and move off with the speed of 3 m/s. What is the mass of the second car?

2. (7 pts) A helium atom with mass of 4 u moving with the speed of 150 m/s collides head-on with a sodium atom (mass 23 u) at rest. The collision is perfectly elastic. What are the final velocities (magnitude and direction) of each atom after the collision?

3. (8 pts) Three cubes, whose sides are in proportion to 1:2:3, are placed next to each other (in contact) with their centers along a straight line and the middle cube in the center (see the figure). The cubes are made of the same uniform material. What is the position of the center of mass of the system with respect to its left edge?

Hint: The mass of a uniform cube is proportional to the third power (i.e., to cube) of its side.

4. (10 pts) The tires of a car make 75 revolutions as the car reduces its speed uniformly from 30 m/s to 16 m/s. The radius of the tires is 0.35 m. (a) What is the angular acceleration?(b) If the car continues to decelerate at this rate, what time will it take to it to stop from the moment it started to slow down?Hint: Rotation angle per revolution is 2 (rad).

2

5. (8 pts) The flywheel of car’s engine is a uniform cylinder of mass 40 kg and radius 20 cm rotating at 1500 rpm. Find how much energy is stored in it.Reminder: Do not forget to find flywheel’s angular velocity in 1/s (or, rad/s), as SI requires.

6. (9 pts) A 60-kg figure skater spins with arms close to her body at 5.0 rev/s. For the sake of this problem, her body can be approximated as uniform cylinder with a height 1.6 m and a radius of 15 cm. (a) What is her angular momentum?(b) How much torque is required to bring her to stop in 4.0 s, if she does not move her arms?Reminder: Do not forget to convert the original data to SI (cm to m, rev/s to rad/s).

3

Dr. Mark Stockman GENERAL PHYSICS 101-3 Fall 1995

LIST OF FORMULAS

1First Exam

Kinematics

average speed distance traveledtime elapsed

_ __

(1.1)

Displacement: x x x 2 1 (1.2)

Velocity: v xt

x xt t

2 1

2 1(1.3)

Acceleration:a v v v

2 1

2 1t t t (1.4)

Uniformly-Accelerated Motionv at v 0 (1.5)

v v v

0

2 (1.6)

x at v t x 12

20 0

(1.7)v v a x x2

02

02 ( ) (1.8)

g ms

9 80 2.(1.9)

Resolution and Operations on Vectors:V V Vx y , cos sin , , = x

2V V V V V V Vx y y 2

(1.10)V V V1 2 , V V V V V V V V Vx x x y y y z z z1 2 1 2 1 2, , (1.11)

V V2 1 a V aV V aV V aVx x y y z z, , , 2 1 2 1 2 1 (1.12)V V V V V V V Vbt ba at ba bt at bt tb ; ; . (1.13)

Projectile MotionFor y axis upward:

v v v gt v

x v t x y gt v t y

v v v v g y y

x x y y

x y

x x y y

0 0

0 02

0 0

20

2 20

20

12

2

;

;

; ( )

(1.14)(For y axis downward, minus sign at g should be changed to plus.) Normally, the coordinate origin is convenient to choose at the starting point, so x y0 00 0 , .

For initial and final points at the same level:

hv

gt

vg

hg

tvg

t

xv v

gvg

vg

ya

yf

ya

mx y

02

0 0

0 0 02

02

22 2

2

22 2

, ,

sin cos sin

= ,

. (1.15)

where h the height of flight, ta is the time of flight to apex, t f is the total time of flight to landing, and xm is the horizontal range.

Second Newton’s Law of Motion:a

FF a m

m, or, (1.16)

kg ms

2 newton = N, pound = lb = 4.45 N(1.17)

Weight: The force of gravity, F gG m (1.18)

Unit conversion:1 mi = 1.61 km, 1 km = 0.621 mi1 mi/h = 1.609 km/h = 0.447 m/s (1.19)1 km/h = 0.278 m/s = 0.621 mi/h

1 m/s = 3.60 km/h

2Second ExamFriction and Inclines

Kinetic friction force F Ffr k N (2.1)Static friction force F Ffr fr N ; F Fs (2.2)

Incline: normal force and acceleration:

.NF mg

a g gx k

cos ;sin cos

(2.3)

Angle of friction: tanf s . (2.4)Circular Motion and Gravitation

Centripetal acceleration and force: avr

m F mvrC C C C

2 2

; ; F a . (2.5)

Newton’s Law of Universal Gravitation: F Gm m

rG

1 22

116 67 10; . N m

kg

2

2 (2.6)

Free-fall acceleration at Earth’s surface: g GMr

ME

EE 2

245 97 10; . kg . (2.7)

Free-fall acceleration at distance r from Earth’s center:

g Fm

GMr

rB

E

EE 2

66 38 10; . m (2.8)

2

Free-fall acceleration at height h above the Earth’s surface:

g g

rr h

gh r

hE

E E

2

21(2.9)

Orbiting velocity of a satellite at distance r from Earth’s center:

vGM

rr

gr

EE . (2.10)

Tr

vr

rrg

rr

GME E

2 2 2 . (2.11)

(For another planet or star, ME should be replaced by the corresponding mass, and RE by the corresponding radius. Note: The revolution period does not depend on satellite’s mass.)

Keppler’s Third Law:TT

rr

12

22

13

23 . (2.12)

Work and EnergyW F d W F d ll , cos or (2.13)

J = joule = N m =kg m

s

2

2 (2.14)

Work done be the external force when moving mass m distance d up an incline :W dmge K ( cos sin ) (2.15)Work done by the gravity force: W mg d mg hgr sin , (2.16)“Hook’s law”: F kxrest . (2.17)

Kinetic energy of a particle: E mv12

2 (2.18)

Energy-work theorem: W mv mv Wnet net 12

12

202 , or KE . (2.19)

Here v is the final speed of the object, v0 is the initial speed, and Wnet is the work of the net force acting on the object.Definition of potential energy: PE PEB A W e (2.20)Potential energy in the gravitation field: PEgr mgy (the y axis is vertically up). (2.21)

Elastic potential energy (e.g., of a spring) PE el 12

2kx (where x is compression or extension) (2.22)

Generalized energy-work theorem: W WC NCKE - (2.23)

Mechanical energy of a particle: E mv12

2 + PE (2.24)

Conservation of mechanical energy: E E mv mv2 1 22

121

212

, or + PE PE2 1 (2.25)

Power: PW

tm

s

energy transformedtime elapsed

W = watt =Js

kg hp = horsepower = 746 W; ;

2

3 (2.26)

Power needed to go up a grade at an angle to horizontal, with velocity v:P mg v sin (2.27)

3

3Third ExamLinear momentum

P vm (3.1)For a many-particle system: P v v1 m m1 2 2 ... (3.2)

Second Newton’s Law:

P F P Ft

t or (3.3)

Change of momentum: P F P F(net) (ext) t t or (3.4)Conservation of momentum (no external force): P P 0, or const (3.5)

Two-particle collisions:Conservation of momentum: m v m v m v m v1 1 2 2 1 1 2 2 (3.6)For fusion (completely inelastis collision) of two particles:

vm v m v

m m1 1 2 2

1 2(3.7)

Energy loss (dissipation) due to fusion: E E Em m

m mv v

12

1 2

1 21 2

2(3.8)

Elastic collisions (scattering):

Energy-momentum conservation:m v m v m v m v

m v m v m v m v

1 1 2 2 1 1 2 2

1 12

2 22

1 12

2 221

212

12

12

,(3.9)

Relative velocity is conserved: v v v v1 2 2 1 (3.10)Elastic collisions of two particles with equal masses:

Nontrivial solution is exchange of velocities: v v v v1 2 2 1, , (3.11)

For the elastic collision of a particle with velocity v1 with another particle at rest, the final velocities are

v m mm m

v v mm m

v11 2

1 21 2

1

1 21

2, (3.12)

Center of Mass (CM)

CM coordinates: x m x m xm m

y m y m ym mCM CM

1 1 2 2

1 2

1 1 2 2

1 2

......

, ......

(3.13)

Momentum in terms of CM: P v M CM (3.14)Second Newton’s Law in terms of CM: (net)M CMa F (3.15)

Rotational motionRadian measure: (rad) =

ArcRadius

lr

(3.16)

Angular velocity and acceleration:

t t

, (3.17)

Linear velocity for a rotating particle: v r (3.18)

4

Before the collision After the collision

Linear acceleration for non-uniform rotation in terms of centripetal and tangent accelerations:

a a a C T C Ta vr

a r, , 2

(3.19)

Kinematics of uniformly-accelerated angular motion: t 0 (3.20)

12

20 0 t t (3.21)

202

02 ( ) (3.22)

0

2(3.23)

Rotational dynamics: Torque and rotational inertia:Torque: r F rF rFsin , [torque] = N m, (3.24)Second Newton’s law for rotation: = where I I mr, 2 (3.25)

Moment of inertia: I m r m r 1 12

2 22 ... (3.26)

Object Axis IR

IMG

Thin hoop of radius R Through center MR2 RUniform cylinder of

radius RThrough center 1

22MR R

2Uniform sphere of

radius RThrough center 2

52MR 2

5R

Uniform rod of length L Through center 112

2ML L12

Uniform rod of length L Through end 13

2ML L3

rotational KE I12

2 (3.27)

For simultaneous rotation and translation: KE Mv ICM CM

12

12

2 2(3.28)

Here, vCM is the (linear) velocity of the CM, ICM is the moment of inertia about an axis through CM, and M is the total mass of the body.Rotational angular momentum: L I (3.29)

“Second Law for rotation” in terms of L: Lt

I , or, = (3.30)

In the absence of external torque, angular momentum is conserved:For or, 0, ,L const I const (3.31)

Work W done on a rotating object by an external torque:W , (3.32)

where is the angle through which the body is rotated. The power P of this external torque isP . (3.33)

5

This page is intentionally left blank

6