Transcript 2053_Lecture_10-08-13

PHY2053 Exam 1
PHY 2053 Fall13 Exam 1
80
Number
60
Number of Students = 638
Average = 11.8
Median = 12
High = 20
Low = 3
40
20
0
Grade
Average = 11.8
High = 20 (6 students)
Low = 3
R. Field 10/08/2013
University of Florida
PHY 2053
Very Good!
Page 1
Estimated Course Grades
26 students have
100 points!
PHY 2053 Fall13 Estimated Course Grades
? 40 D? 45 D
? 50 D+
? 55 C? 60 C
? 65 C+
? 70 B? 75 B
? 82 B+
? 87 A? 92 A
30
Number
25
20
15
10
5
20%
After One Exam
C
Average = 74.2
High = 100
B
A
Percent of Students
35
PHY 2053 Fall13 Estimated Course Grades
15%
After One Exam
Number = 650
A or A- = 25.1%
>=B = 51.8%
>=C = 82.2%
Very Good!
11.5%
3.4%
3.1%
4.3%
15.4%
10.5%
9.5%
9.2%
10%
5%
16.3%
9.7%
5.4%
1.7%
0%
0
0
5
10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100
Points (100 max)
E
D-
D
D+
C-
C
C+
B-
B
B+
A-
A
Grade
• Estimated Grades:
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.
R. Field 10/08/2013
University of Florida
PHY 2053
Page 2
Linear Momentum
Units = kg∙m/s
• Single Particle:
The linear momentum of a particle is its mass times its velocity vector.





dv d (mv ) dp
Fnet  ma  m


dt
dt
dt


p  mv


dp
Fnet 
dt
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:
Center-of-Mass
The overall linear momentum of a system of particles is the vector sum of the

momentums of all the particles.
N







Ptot   pi  m1v1  m2 v2  m3v3    mN v N  M tot vcm
i 1


dPtot
Fnet 
 M tot acm
dt (system of particles)
• Momentum Conservation:
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.
 tot  tot
Pi  Pf
R. Field 10/08/2013
University of Florida
(conservation of linear momentum, isolated system)
PHY 2053
Page 3
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
Initial Mometum
y-axis
Final Mometum
y-axis
Initial Energy
y-axis
Final Energy
v'
V =0
M
m
V
x-axis
h
x-axis
x-axis
v
pi  p f
mv  MV  mv'
m
V
(v  v ' )
M
R. Field 10/08/2013
University of Florida
Ei  E f
2
1
2 MV  Mgh
V 2  m  (v  v ' ) 2
h
 
2g  M 
2g
2
2
 1  (1000m / s  500m / s )

 20.4cm

2
250
2
(
9
.
8
m
/
s
)


2
PHY 2053
Page 4
Exam 2 Spring 2012: Problem 10
q
v
x-axis
• 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 q 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 q 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 ( px )i  ( M car  M Ball )Vi
( p x ) f  M BallVball  M Ball (Vi  Vmuzzle cos q )
% Right: 57%
cos q 
( px )i  ( px ) f
( M car  M Ball )Vi  M Ball (Vi  Vmuzzle cos q )
M carVi
(5000kg)( 2m / s )

 0.6667
M BallVmuzzle (100kg)(150m / s )
R. Field 10/08/2013
University of Florida
PHY 2053
q  48.2
Page 5
Momentum Change: Impulse
The time rate of change of the momentum is equal to the net force acting on an
object.



dp
F (t ) 
dt

dp  F (t )dt

 

p  p f  pi  Ft
• Impulse:
The change in the linear momentum is equal to the impulse.
The impulse J is defined as follows:
 
J  Ft
(constant force)

 

p  p f  pi  J
(definition of impulse for constant force)
• Average Constant Force:
Often one is interested in the average constant force necessary to produce an

impulse J within the time interval t = t2 – t1. 
 
J  Favet
Fave
J

t
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?
R. Field 10/08/2013
University of Florida
PHY 2053
Fave  M (v1  v2 ) / t
Page 6
Impulse: Example Problem
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?
Answer: 14.7 N∙s
1
2
y-axis
v1 = 0
MgH  12 Mv22 
H

1
2M

p22
y-axis
v'1

 
  
J  p  p f  pi  p'1  p2  M

J M
v'2 = 0
h

p2   M 2 gH yˆ
v2
(impulse)
Mv12  MgH  12 Mv22

2 gh  2 gH



1
2
Mv '12  12 Mv '22  Mgh
1
2
Mv '12 

p'12  Mgh


p'1  M 2 gh yˆ

2 gh  2 gH yˆ
 (0.5kg) 2(9.8m / s 2 )( 4.9m)  2(9.8m / s 2 )(19.6m)


1
2M

 (0.5kg) (9.8m / s) 2  2(9.8m / s) 2  3(0.5kg)(9.8m / s)  14.7 N  s
R. Field 10/08/2013
University of Florida
PHY 2053
Page 7