R. Field 10/08/2013 University of Florida

PHY 2053 Page 1

PHY2053 Exam 1

Average = 11.8

High = 20 (6 students)

Low = 3

Very Good!

PHY 2053 Fall13 Exam 1

0

20

40

60

80

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Grade

Nu

mb

er

Number of Students = 638Average = 11.8Median = 12High = 20Low = 3

R. Field 10/08/2013 University of Florida

PHY 2053 Page 2

Estimated Course Grades

• Estimated Grades:

26 students have 100 points!

1. Assumes that you get the same grade on Exam 2 and the Final Exam that you received on Exam 1.

2. Include the first 4 quizzes and assumes that you get the same average on all your remaining quizzes that you have for the first 4 quizzes.

3. Includes the first 5 WebAssign HW assignments and assumes that you get the same average on all your remaining homework assignments that you have for the first 5 assignments.

4. Includes your HITT scores through 9/26/13 and assumes you maintain the same average.5. Includes your first 4 Sakai HW assignments and assumes you maintain the same

average.

PHY 2053 Fall13 Estimated Course Grades

0

5

10

15

20

25

30

35

0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100

Points (100 max)

Num

ber

Average = 74.2High = 100

? 40 D-? 45 D? 50 D+? 55 C-? 60 C? 65 C+? 70 B-? 75 B? 82 B+? 87 A-? 92 A

After One Exam

ABC

PHY 2053 Fall13 Estimated Course Grades

3.1%1.7%

3.4%4.3%

5.4%

9.2%

11.5%

9.5%

16.3%

10.5%9.7%

15.4%

0%

5%

10%

15%

20%

E D- D D+ C- C C+ B- B B+ A- A

Grade

Pe

rce

nt

of

Stu

de

nts

After One ExamNumber = 650

A or A- = 25.1%>=B = 51.8%>=C = 82.2%

Very Good!

R. Field 10/08/2013 University of Florida

PHY 2053 Page 3

Center-of-Mass

Linear Momentum

vmp

dt

pdFnet

• Momentum Conservation:

• Single Particle:

dt

pd

dt

vmd

dt

vdmamFnet

)(

cmtotNN

N

iitot vMvmvmvmvmpP

332211

1cmtot

totnet aM

dt

PdF

The overall linear momentum of a system of particles is the vector sum of the momentums of all the particles.

• System of Particles:

The time rate of change of the momentum is equal to the net force acting on the particle. If Fnet = 0 then p is constant in time (i.e. does not change).

(system of particles)

If the net external force acting on a system of particles is zero (i.e. isolated system) then Ptot is constant in time (i.e. does not change) and hence the total momentum at some initial time ti is equal to the total momentum at some final time tf.

totf

toti PP

(conservation of linear momentum, isolated system)

Units = kg∙m/s

The linear momentum of a particle is its mass times its velocity vector.

R. Field 10/08/2013 University of Florida

PHY 2053 Page 4

Momentum Conservation: Example• Example:Near the surface of the Earth, a bullet with mass m moving directly upward at speed v = 1,000 m/s strikes and passes through the center of mass of a block of mass M initially at rest as shown in the figure. The bullet then emerges from the block moving directly upward at speed v‘ = 500 m/s. If the mass of the block is 250 times the mass of the bullet, to what maximum height does the block then rise above its initial position? Answer: 20.4 cm

)'(

'

vvM

mV

mvMVmv

pp fi

cmsm

smsm

g

vv

M

m

g

Vh

4.20)/8.9(2

)/500/1000(

250

1

2

)'(

2

2

22

222

V =0

v

x-axis

Initial Mometum y-axis

m

M

V x-axis

Final Mometum y-axis Initial Energy

v'

MghMV

EE fi

2

21

h x-axis

y-axis Final Energy

R. Field 10/08/2013 University of Florida

PHY 2053 Page 5

Exam 2 Spring 2012: Problem 10

• A cannon on a railroad car is facing in a direction parallel to the tracks as shown in the figure. The cannon can fires a 100-kg cannon ball at a muzzle speed of 150 m/s at an angle of above the horizontal as shown in the figure. The cannon plus railway car have a mass of 5,000 kg. If the cannon and one cannon ball are travelling to the right on the railway car a speed of v = 2 m/s, at what angle must the cannon be fired in order to bring the railway car to rest? Assume that the track is horizontal and there is no friction.

Answer: 48.2o % Right: 57%

iBallcarix VMMp )()(

v

x-axis

)cos()( muzzleiBallballBallfx VVMVMp

2.48

fxix pp )()( )cos()( muzzleiBalliBallcar VVMVMM

6667.0)/150)(100(

)/2)(5000(cos

smkg

smkg

VM

VM

muzzleBall

icar

R. Field 10/08/2013 University of Florida

PHY 2053 Page 6

Momentum Change: Impulse

dttFpd )(

tFppp if

tFJ

• Average Constant Force:

The change in the linear momentum is equal to the impulse.

dt

pdtF

)(

tFJ ave

Jppp if

The impulse J is defined as follows:

• Impulse:

The time rate of change of the momentum is equal to the net force acting on an object.

(definition of impulse for constant force)

t

JFave

Often one is interested in the average constant force necessary to produce an impulse J within the time interval t = t2 – t1.

Example: A ball of mass M drops vertically onto the floor, hitting with speed v1. If the ball is in contact with the floor for a time t and if it rebounds with speed v2, what is the magnitude of the average force on the floor from the ball?

tvvMFave /)( 21

(constant force)

R. Field 10/08/2013 University of Florida

PHY 2053 Page 7

Impulse: Example Problem

Answer: 14.7 N∙s

A 0.5-kg rubber ball is dropped from rest a height of 19.6 m above the surface of the Earth. It strikes the sidewalk below and rebounds up to a maximum height of 4.9 m. What is the magnitude of the impulse due to the collision with the sidewalk?

222

1212

1 MvMgHMv

(impulse)

y-axis

H

v1 = 0

v2

y-axis

h v'2 = 0 v'1

222

1222

1 pMvMgH M

ygHMp ˆ22

MghMvMv 222

1212

1 ''

MghpMv M 212

1212

1 ''

yghMp ˆ2'1

ygHghMpppppJ if ˆ22' 21

sNsmkgsmsmkg

msmmsmkg

gHghMJ

7.14)/8.9)(5.0(3)/8.9(2)/8.9()5.0(

)6.19)(/8.9(2)9.4)(/8.9(2)5.0(

22

22

22