Transcript Unit 06 Momentum and Collisions

Unit 06
Momentum and
Collisions
Class Plans and Demos
Equations List – Literal Equations “What if’s?”
(later slide)
Egg throw into sheet
Draw momentum and impulse vector
Finding Vf of egg “What if?” example
Air track collisions
Bouncing balls of diff sizes toy - astroblaster
Basketball and tennis ball launch – same thing
Read “Attractive Force of Glass” article from
intuitor.com
Math examples: Holt # 2 p. 213
– and # 1 p. 214
HW: Momentum FR Quiz #1 and HW and Review
Handout
Class Plans Day 4: Monday 01/11/10
BW: Astroblaster Question
1st pd : Red pen grade take home quiz: FR with
spring, 6th grade the projectile one
Check HW and Review Handout for grade
Give answers and explanations sheet for study
1st pd : discuss change in lab report calc’s
Report due tomorrow
Give out Midterm practicum (Pendulum lab)
Momentum clicker quiz tomorrow
Get names of students who plan on taking the
AP exam this spring
Simple Harmonic motion (SHM) starts tomorrow
Momentum exam Thursday, w/ a bit of SHM
Ranking Tasks for impulse if time
Videos
Julius Sumner Miller – 2 part episode on
energy and momentum
Podcasts:
Collision (funny, anticipatory event) – Best
of YouTube #298 Mic Mishap
Bouncing (use with Conceptual Physics
section on bouncing and impulse) – Best
of Youtube #392 – “Beach ball ownage”
Unit Plans
Day 1: Egg throw, mic mishap vid, Notes, ‘quiz
1’ for HW
Day 2: Lab – discuss HW, “bumper cars” lab,
Quiz 2 for HW
Day 3: bouncing vid, Astroblaster problem,
finish/discuss lab, discuss quiz 2 (HW), partner
quiz: ‘quiz 3’, Jocko the clown (BW)
Introductory BW: Energy Review
A ballistic pendulum apparatus has a 40-g ball that is caught by a
500-g suspended mass. After impact, the two masses rise a
vertical distance of 45 mm. Find the velocity of the combined
masses just after impact.
m1 = 0.04 kg
m2 = 0.500 kg
h = 0.045 m
KE0 = PEf
PEf = (m1+m2) ghf
= (0.04+0.500)(9.8)(0.045)
= 0.24 J
KE0 = PEf = 0.24 J
1
KE0  mT v02
2
v0 
2 KE

mT
2(0.24)
= 0.94 m/s
0.54
Bellwork: Day 3
Bellwork
BW’s Day 3- Astroblaster
 The Astroblaster is a toy version of what’s called a
“multiple collision accelerator”
 It quickly transfers momentum from larger objects
to smaller and smaller ones, giving them faster
and faster speeds.
 We will calculate the velocity of the last and
smallest ball as it is launched from the
astroblaster
 First we must know how much momentum is put
into the system of balls
 If you drop the astroblaster from ½ m above the
ground, how fast will the smallest (red) ball be
going when it is launched from the astroblaster?
BW’s Day 3- Astroblaster Data
 Measurements:
 Drop height: 0.5 m
 g = 10 m/s2
 Ball masses:




Red ball: mr = 4 g = 0.004 kg
Yellow ball: my = 9 g = 0.009 kg
Green ball: mG = 30 g = 0.030 kg
Blue ball: mB = 47 g = 0.047 kg
 Assumptions:
 linear momentum is conserved in all collisions (pi = pf)
 Mass of earth is so large compared to mass of astroblaster that
earth’s mass (mE )may be approximated as infinite and therefore
the velocity of earth before and after colliding with the astroblaster
is negligible (0 m/s)
IMPULSE AND MOMENTUM
The impulse (F t) is a vector quantity equal in
magnitude to the product of the force and the time
interval in which it acts. Its direction is the same as that
of the force. It is abbreviated with a capital “J”…
J = F t
Units: newton.second (N.s)
The momentum (p) of a particle is a vector quantity
equal in magnitude to the product of its mass m and its
velocity v. Abbreviated with a lower case “p”.
p = m v Units: (kg.m/s)
Impulse Example:
A 3 N baseball moving toward the batter with a velocity of 15 m/s is
struck with a bat which causes it to move in a reversed direction
with a velocity of 30 m/s. Find the impulse and the average force
exerted on the ball if the bat is in contact with the ball for 0.01 s.
Fg = 3 N
Fg
3
m

 0.31 kg
g 9.8
v0 = - 15 m/s
vf = 30 m/s
t = 0.01 s
This is why follow through is important in batting, golf swings,
lacrosse and hockey shots: more time in contact with the ball gives
greater change in momentum!
FΔt = mΔv
=m(vf - v0)
=0.31(30-(-15))
= 13.95 Ns
m = 0.31 kg
v0 = - 15 m/s
vf = 30 m/s
t = 0.01 s
F
m(v f  v0 )
t
0.31(30  (15))

= 1395 N
0.01
Area =
Impulse
CONSERVATION OF LINEAR MOMENTUM
According to the law of conservation of linear
momentum, when the vector sum of the external forces
that act on a system of bodies equals zero (a.k.a.: the
system is isolated), the total linear momentum of the
system remains constant no matter what momentum
changes occur within the system.
Although interactions within the system may change the
distribution of the total momentum among the various
bodies in the system, the total momentum does not
change. Such interactions can give rise to two general
classes of events:
a. explosions, in which an
original single body flies
apart into separate bodies
b. collisions, in which two or more bodies collide and
either stick together or move apart, in each case with a
redistribution of the original linear momentum.
For two objects interacting with one another, the
conservation of momentum can be expressed as:
m1v1  m2v2  m v  m v
'
1 1
v1 and v2 are initial velocities,
velocities
v
' and
1
'
2
v
'
2 2
are final
ELASTIC AND INELASTIC COLLISIONS
If the Kinetic energy remains constant in a collision, the
collision is said to be completely elastic.
If the colliding bodies stick together and move off as a
unit afterward, the collision is said to be completely
inelastic. In inelastic collisions only the momentum is
conserved.
•In elastic collisions no permanent
deformation occurs; objects elastically
rebound from each other.
In head-on elastic collisions between
equal masses,velocities are exchanged.
•Kinetic energy is conserved as well as
momentum
•Inelastic collisions are characterized by
objects sticking together and permanent
deformation.
•Kinetic energy is lost but momentum still
conserved.
Inelastic Collision Example:
A 12-g bullet is fired into a 2-kg block of wood suspended from a
cord. The impact of the bullet causes the block to swing 10-cm
above its original level. Find the velocity of the bullet as it strikes
the block.
m1 = 0.012 kg
m2= 2 kg
h = 0.1 m
KE0 = PEf
m1v1= ( m1+m2 )v'
m1 = 0.012 g
m2= 2 kg
h = 0.1 m
KE0 = PEf
1
 m1  m2  v 2   m1  m2  gh
2
v  2gh
 2(9.8)(0.1) = 1.4 m/s
m1v1= ( m1+m2 )v'
(m1  m2 )v '
v1 
m1
(0.012  2)1.4

= 234.7 m/s
0.012
Egg “What if” Question
What if someone smacked the egg
in mid-flight with a force straight
up of 50 N for 0.025 s?
1. Draw impulse vector
2. Get final momentum (add impulse
to initial momentum)
3. Find Vf and angle of egg
“What if’s”
1. What happens to momentum if
mass is doubled?
2. What about kinetic energy?
3. What happens to momentum if
velocity is halved?
4. What about kinetic energy?
Collisions Lab
“Bumper Cars”

Using carts and tracks, create 3 different situations described where momentum is conserved

Measure each cart’s initial and final velocity using any equipment necessary

Describe each event as either elastic or inelastic and show whether K and p is conserved
using measurements and calculations.

Using conservation of momentum, find the ratio of one cart’s mass to the other (mp = X mc ,
where ‘X’ is the ratio)

Ideally, this ratio should be exactly 1.00, however friction will cause this to be slightly higher or
less, giving your experiment an unavoidable source of error.

Find out how much error friction introduced in each case

One report per group.

Should describe your materials, and how they were used (procedure), you may want to use
labeled diagrams to avoid writing alot

Organize your measurements into a data table(s)

Show all calculations clearly and neatly in a separate section (points awarded for neatness)

Extra credit: Discover an accurate method for finding the time and/or force involved in each
collision/event!!!
Collisions Lab
“Bumper Cars”
 Photogate setup tips
 Don’t let friction become a large factor: setup your gates so
they measure velocities just before and after the event
 LabQuest tips: Plug in both gates and make sure they read
blocked or unblocked on the meter screen
 Check the timing mode: it should be “Photogate timing” and
“Gate” enter the diameter of the metal column on the carts as
the object length
 Don’t use the graph screen, use the data table screen
 Scroll the screen over and down to find the velocities recorded
by each gate.