Review 3

Final Exam

A fi l ti i h d l d f ll ti f Ph i 231• A common final exam time is scheduled for all sections of Physics 231 • Time: Wednesday December 14, from 8-10 pm. • Location for section 002 : BPS 1410 (our regular lecture room). • This information can also be found on our course schedule page

• An alternate exam time will be scheduled for students who have• An alternate exam time will be scheduled for students who have conflicts with the regular time. – Two students have confirmed conflicts with me and will take the

exam thenexam then.

– You must contact me by email and obtain permission from me to take the exam at the alternate time. If you fail to do this, you will b b d f ki h l fi lbe barred from taking the alternate final exam.

– RCPD students should arrange to take their final exam at Bessey.• Alternate time: Tuesday December 13, from 1-3 pm• Location: BPS 1410 (this room)

Exam 1 problem 4• Concepts/equations:Co cep s/equa o s:

fi ""thil itthiframe "stationary" thein velocity theis

t

vvvv

′

+′=

• Solution is posted.frame moving"" theof velocity theis framemoving"" thein velocity theis

tvv

′

• Examples of alternate formulations – Give boat and river speed,

time and ask for distances across and downstream.

– Give boat and water speed, width and ask for distance downstream.– Give boat and water speed, width and ask for total distance.– Give boat speed, river speed and distance downstream and ask for total

distance.– Give boat speed, river speed and distance downstream and ask for

distance across.

Exam 1 problem 5• Concepts/equations:Co cep s/equa o s:

( )PEPEPEΔ

2

21 mvKE =

• Conservation of energy

( )00 yymgPEPEPE ff −=−=Δ

• Solution is posted.

00 PEKEPEKEE ff +=+=

p• Examples of alternate

formulations– Ask for v at BAsk for v at B.– Give v at C and ask for

height at A.Give v at B or C and ask for– Give v at B or C and ask for v0.

Exam 1 problem 6• Concepts/equations:Co cep s/equa o s:• Work:

Kinetic energ :( ) xFsFW xΔ=Δ= θcos

• Kinetic energy:2

21 mvKE =

• Work-energy theorem:

• Solution is posted.0KEKEW ftotal −=

p• Examples of alternate

formulations– Give velocity ask for FGive velocity ask for F.– Give a different shape for

the curve.

Exam 1 problem 7• Concepts/equations:Co cep s/equa o s:• Net force on each object in the

problem vanishes.

N

ii

F 0=

W mg=

• Solution is posted.• Examples of alternate

formulations– Give Tension, mass and ask

for angle. The tension in the two wires supporting the mass of 9 kg is 118 N what is the

– Give T and angle and ask for mass.

the mass of 9 kg is 118 N, what is the angle?

i,yi

F 0 2Tsin mg

9 9 8

θ= = −

( )1 o

mg 9 9.8sin 0.3732T 2 118

sin 0.373 22

θ

θ −

= = =

= =

Exam 1 problem 8• Concepts/equationCo cep s/equa o• Newton’s second law:

tension forcesF ma=

• tension forces• Newton’s 3d Law:

of 2on1 of1on2F F= −

• Solution is posted.• Examples of alternate

formulations– Give Tension, F, M1, M2

and ask for angle.– Give T, F, M2 and angle , , 2 , g

and ask for M1.• Key is that the physical

situation is the same, but the ,question is different.

Exam 1 problem 9• Concepts/equation• Ne ton’s second la :

netF ma=

• Newton’s second law:

• tension forcesW mg=• Gravity:

• Normal force:• Friction forces:

Nf ss μ<• Solution is posted.

.• Examples of alternate formulations

– Give M2, µs and ask for M1.Consider the figure above with M1=143 kg and μs = .17, what is the minimum mass for M2 that would overcome the static friction

– Give M1 and M2 and ask for µs.– Put table and masses on the Moon or some other planet where g is

different, Give M2 µs and ask for M1.

M2 that would overcome the static friction force on M1?

, 2, µs 1

• Key is that the physical situation is the same, but the question is different.

• Concepts/equation( )Fr sin 0τ θ= =

Exam 2 problem 1( )2

i ii

I = m r

I 0

2

I 0L I is conserved L=0

1KE I

τ αω

= == Δ

• Examples of alternate formulations– Make into a conceptual problem

2rotKE I

2ω=

Make into a conceptual problem.– Give the initial velocity and ask the

final velocity.Ask the ratio of initial/final– Ask the ratio of initial/final velocity .

– Ask for the initial velocity and puck masspuck mass

• Key is that the physical situation is the same, but the question is different.

Exam 2 problem 2• Concepts/equation• Conservation of momentum• Conservation of momentum

• Solution is posted.1,i 2,i 1,f 2,fp p p p+ = +

• Examples of alternate formulations– Give final velocities and

ask initial velocity of the boy.

– Make man to be at rest initially. Give all velocities of man and b d h f h A k h f h bboy and the mass of the man. Ask the mass of the boy.

– Make the man to be moving initially as well. Give both final velocities and the initial velocity of the boy and ask the initial

l i f hvelocity of the man.– Make the collision in two dimensions.

• Key is that the physical situation is the same, but the question is different.

Exam 2 problem 3• Concepts/equation• Newton’s second law • for rotational motion,

Iτ α=• Rotational kinematics:

Iτ α=

αωωωθ +==Δ 0 tt A toy Ferris wheel with moment of inertia of 476 Nm2 is at rest and

( ) ( )ωωθωωω +=Δ+=

121

21

222

00 tinertia of 476 Nm2 is at rest and then experiences a torque of 18 Nm that lasts for 12 s. What is the final angular velocity?

• Solution is posted.• Examples of alternate formulations

θαωωαωθ Δ+=+=Δ 2 21 2

022

0 tt 2

2

18 0.038rad / sI 476

t 0.038rad / s 12s 0.45rad / s

τα

ω α

= = =

= = =Examples of alternate formulations– Give moment of inertia, torque, time required, ask the final angular speed.– Give moment of inertia, solve for torque.

Gi t f i ti t l f ti i d t t t fi l– Give moment of inertia, torque, solve for time required to get to final speed.

• Key is that the physical situation is the same, but the question is different.

Exam 2 problem 4• Concepts/equation

P d l i l• Pendulum, simple harmonic motion:

1 gf ;ω= =f ; 2 2 Lπ π

= =

L T 1/f 2g

π= =g

max cos(2 ft);θ θ π=• Examples of alternate formulations

k h l h h d l─ Ask for the length of the pendulum. ─ Say that there are two pendula on the two different planets.

Give periods or frequencies and ask for the ratio of the l hlengths.

─ Change gravity AND length by same ratio.─ Change length only.

• Key is that the physical situation is the same, but the question is different.

Exam 2 problem 5• Concepts/equationCo cep s/equa o• Stress strain relationship:

LF ΔΔ

• Solution is posted.0LLY

AF Δ=Δ

• Examples of alternate formulations– For this one, you can

A mass of 40 kg is suspended from a steel wire of length 2.10 m. The

interchangeably solve for either F, m, A, delta L, L, or diameter.

wire stretches by 0.2 mm. What is the diameter of the wire? Young’s modulus for steel is 200x109 N/m2.

• Key is that the physical situation is the same, but the question is different.

modulus for steel is 200x10 N/m .

Exam 2 problem 6• Concepts/equationp q• Newton’s law of universal

gravitation:

GM m

S l ti i t d

planet SarahSarah2

planet

GM mF W

R= =

Sarah weighs 408 N on planet X, which has 4 times the mass of Earth and twice its diameter.

• Solution is posted.

• Examples of alternate formulations─ You are given WX, RX/RE, MX/ME and asked for WE.

What does she weigh on earth?

You are given WX, RX/RE, MX/ME and asked for WE.─ You are given WE, WX and MX/ME and asked for RX/RE.─ You are given WE, WX and RX/RE and asked for MX/ME.

Y i W W R /R M d k d f M─ You are given WX , WE, RX/RE , ME and asked for MX. • Key is that the physical situation is the same, but the question is

different.

Exam 2 problem 7• Concepts/equationp q• centripetal acceleration

22t

cva rr

ω= =

t

rv ; = tr

ω θ ω=• Solution is posted.

• Examples of alternate formulations─ We give r, ac and ask for vt. A supersonic plane follows a─ We give ac, vt and ask for r.

• Key is that the physical situation is the same, but the question is diff

A supersonic plane follows a circular path of radius = 75 km. It experiences an acceleration of 7.5

/ 2 Wh i h d f hdifferent. m/s2. What is the speed of the plane?

gtvv

2 downwards m/s 9.8 g−=

=

( )tvvyyy

gtvv

yoy

yy

0

0

121 +=−≡Δ

=

ygvv

gttvy

y

y

y2

02

20

221

Δ−=−

−=Δ

tvxvv

x

xx

0

0

=Δ=

PHY 231 113

• Concepts/equations:• Concepts/equations:2

21 mvKE =

• Conservation of energy

( )00 yymgPEPEPE ff −=−=Δ

gy

• Work energy theorem

00 PEKEPEKEE ff +=+=

• Work energy theorem

f 0 frictionKE KE W− =

Static equilibrium0 and 0 ==

ii

ii F

τ

Static equilibrium

calculation of torque

( )Fr sinτ θ=

calculation of torque

PHY 231 115

Conservation of momentum

1,i 2,i 1,f 2,fp p p p+ = + Conservation of momentum

PHY 231 116

PHY 231 117