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PHY131H1F - Class 15
Today, we are finishing Chapter 9 on Momentum:
Impulse and Momentum
Energy in Collisions
Totally Inelastic Collisions
Elastic Collisions
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Quick Quiz 1 of 3:
⢠Two objects collide. All external forces
on the objects are negligible.
⢠If the collision is âelasticâ, that means it
conserves
A. Momentum p=mv
B. Kinetic energy E = ½ mv2
C. Both
D. Neither
2
Quick Quiz 2 of 3:
⢠Two objects collide. All external forces on the
objects are negligible.
⢠If the collision is âinelasticâ, that means it
conserves
A. Momentum p=mv
B. Kinetic energy E = ½ mv2
C. Both
D. Neither
Quick Quiz 3 of 3:
⢠Two objects collide. All external forces on the
objects are negligible.
⢠If the collision is âtotally inelasticâ, that means
A. momentum is not conserved.
B. the final kinetic energy is zero.
C. the objects stick together.
D. one of the objects ends with zero velocity.
3
Class 15 Preclass Quiz on MasteringPhysics
This was due this morning at 8:00am
90% of students got: In an inelastic collision momentum is
conserved.
50% of students got: The total momentum of a system is
conserved if the net outside forces sum to zero.
71% of students got: A perfectly elastic collision is defined as
one for which mechanical energy is conserved.
Where is the centre of mass of
a solid semicircle?
4
Class 15 Preclass Quiz Student Comments
âDoes the explosive collision always need external forces?â
Harlow answer: No. Explosive collisions involve only
internal forces, but some chemical energy is released so that
the total kinetic energy is greater before the collision than
after.
âwhen one moving ball collide with a ball at rest, it is said
that they stick together at end, this is an example of totally
inelastic collision, but for the ball at rest, its velocity (i.e,
kinetic energy) is increased after the collision. as inelastic
collision states that the final kinetic energy should be less
than the initial kinetic energy. I am confused...â
Harlow answer: Good point. For the ball initially at rest, its
kinetic energy is increased. But if you sum up the K of both
balls and compare before and after, the total decreases.
5
Monday
Nov. 16
Nov. 23
Nov. 30
Dec. 7
Tuesday
Nov. 17
Nov. 24
Dec. 1
Dec. 8
Wednesday
Nov. 11
Nov. 18
Nov. 25
Dec. 2
Thursday
Nov. 12
Nov. 19
Nov. 26
Dec. 3
Friday
Nov. 13
Nov. 20
Nov. 27
Dec. 4
7
8
9
10
Harlow answer: âWhat day of the
week are your Practicals on?â
Class 15 Preclass Quiz Student Comments
âhow do practicals schedule work again?
.................................................................haha just kiddingâ
Class 15 Preclass Quiz Student Comments
âwhat kind of collision is for car accident?
Harlow answer: Always inelastic. If the cars âstick togetherâ
during the collision, then this is totally inelastic.
âI am slightly confused about whether mechanical energy is
conserved in inelastic collisions or not because even if the
kinetic energy is being transferred to other types of energy
such as heat or sound energy, is overall mechanical energy
still not conserved?
Harlow answer: âMechanical Energyâ only includes K + U.
It does not include Eth, or damage or sound energy etc. So
mechanical energy is not conserved during an inelastic
collision. But total energy is conserved.
âcould you please make the second term test easy?â
Harlow answer: No. Sorry.
6
⢠The second term test in PHY131H1F will be written on
Tuesday, Nov 17, from 6:10 to 7:30 pm in Room TBA
⢠It will be based on:
â Uncertainties Reading
â Wolfson Chapters. 5-9 and Sections 10.1-10.3
â Classes 7-16
â Practicals 3-7
â MasteringPhysics problem sets 3-7
⢠The alternative sitting is on the same day from 4:30 - 6:00pm. If you
need to write at the alternative sitting due to conflicts please go to the
portal left hand side menu and click the link âAlternative Sitting
Registration Test 2â. The deadline to register for the alternative sitting
is 4:00pm on Thursday Nov. 12, 2015.
Class 15 Preclass Quiz Student Comments
âWhat topics do the test 2 cover?â
Class 15 Preclass Quiz Student Comments
âI don't know why and I CANNOT USE THE PORTAL !!!
hence i am not able to watch the preclass video which is
driving me crazy.â
Harlow comment: For future reference, there is a âback
doorâ to much of my physics content. Just google âJason
Harlow Teachingâ and youâll get to my teaching page. Click
on PHY131 Fall 2015. Links to all the preclass videos are
just sitting there.
7
Class 15 Preclass Quiz Student Comments
âwe never learned integrals in high school, but were
expected to just know them now for physics?â
Harlow answer: Yes, you should learn them. Itâs not too
hard.
Alyssa: âGive me hope by posting this in lecture!â
Paul: âI kinda made a deal with my friend that I'd get my
comment up on the board since her comment was posted. If
this does goes up I guess I GOT BRAGGING RIGHTS!!!
WOOT WOOT!â
Caroline: âI want to make it to the board too...â
âIn Prof Thomson's book, he mentions "the spherical cow" as
a "classic" joke about physicists approaching real world
problems. elaborate.â
Consider the two integrals below. Whatâs the difference?
đ˝ = đšđđĄ
đ = đš â đ đ
Last day I asked at the end of class:
Impulse (Ch.9)
This is the force integrated
over time, which gives the
change in momentum.
[Units: kg m / s]
Work (Ch. 6)
This is the force integrated
over distance, which gives
the change in energy.
[Units: Joules]
8
Recall from Ch.4: Momentum
Momentum is the product of a particleâs mass and
velocity, has units of kg m/s, and is given by
An object can have a larger momentum if it is:
⢠moving faster or,
⢠has more mass
Note: Momentum is a vector quantity. It has both x and
y components.
đ = đ đŁ
A basketball with mass 0.1 kg is traveling down and
to the right with vxi = +5 m/s, and vyi = â5 m/s.
It hits the horizontal ground, and then is traveling up
and to the right with vxf = +5 m/s, and vyf = +4 m/s.
What is the change in the x-component of its
momentum?
A. +0.5 kg m/s
B. â0.5 kg m/s
C. +0.1 kg m/s
D. +0.9 kg m/s
E. zero
x
iv
fv
y
9
A basketball with mass 0.1 kg is traveling down and
to the right with vxi = +5 m/s, and vyi = â5 m/s.
It hits the horizontal ground, and then is traveling up
and to the right with vxf = +5 m/s, and vyf = +4 m/s.
What is the change in the y-component of its
momentum?
A. +0.5 kg m/s
B. â0.5 kg m/s
C. +0.1 kg m/s
D. +0.9 kg m/s
E. zero
iv
fvx
y
ImpulseThe impulse upon a
particle is:
⢠Impulse has units of N s, but you should be able to show
that N s are equivalent to kg m/s.
⢠The impulse-momentum theorem states that the change
in a particleâs momentum is equal to the impulse on it:
đ˝đĽ = đĄ1
đĄ2
đšđĽđđĄ
âđđĽ= đ˝đĽ âđđŚ= đ˝đŚ
đ˝đŚ = đĄ1
đĄ2
đšđŚđđĄ
10
A 0.50 kg cart rolls to the right
at +1.2 m/s.
It collides with a force sensor.
A plot of force versus time (with
positive force defined as
towards the right) gives an area
of â1.0 N s.
What is the velocity of the cart
immediately after the collision?
Was this collision elastic or
inelastic?
11
⢠Consider a car accident in which a
car, initially traveling at 50 km/hr,
collides with a large, massive
bridge support.
⢠The car comes to an abrupt stop.
Image from http://images.thecarconnection.com/med/25-years-of-the-airbag_100223157_m.jpg
⢠Why is it better to hit the airbag as opposed to the hard plastic
steering wheel or dashboard?
⢠ANSWER:
⢠The people must reduce their momentum from mv to zero.
This requires a force applied over some amount of time. If the
time is very short, the force must be very large (ie hitting
steering wheel).
⢠If the person hits the airbag, this squishes during impact,
lengthening the time of the stop. If the stopping process takes
longer, then the maximum force is less.
A. They receive equal impulses.
B. The damp cloth receives a larger impulse.
C. The rubber ball receives a larger impulse.
A 100 g rubber ball and a 100 g damp cloth are
dropped on the floor from the same height.
They both are traveling at the same speed just
before they hit the floor.
The rubber ball bounces, the damp cloth does
not.
Which object receives a larger upward impulse
from the floor?
Clicker Question
12
Elastic Collisions
Elastic Collision in 1 Dimension when ball 2 is
initially at rest.Consider a head-on, perfectly elastic collision of a ball of
mass m1 having initial velocity v1i, with a ball of mass m2 that
is initially at rest.
The ballsâ velocities after the collision are v1f and v2f.
1 2Before:
During:
After:
đŁ1i đži
1 2đŁ1f đŁ2f
đžf = đži
1 2During the collision energy is
stored as elastic potential energy.
13
There are two equations, and two unknowns: v1f and v2f.
Solving for the unknowns gives:
Eq. 9.15:
Momentum conservation: đ1đŁ1f +đ2đŁ2f = đ1đŁ1i
Kinetic energy conservation: 1
2đ1đŁ1f
2 +1
2đ2đŁ2f
2 =1
2đ1đŁ1i
2
đŁ1f =đ1 âđ2đ1 +đ2
đŁ1i
đŁ2f =2đ1
đ1 +đ2đŁ1i
(Elastic collision
with ball 2 initially at
rest.)
Elastic Collision in 1 Dimension when ball 2 is
initially at rest.
These equations come in especially handy, because you can
always switch into an inertial reference frame in which ball 2
is initially at rest!
Eq. 9.15
đŁ1f =đ1 âđ2đ1 +đ2
đŁ1i
đŁ2f =2đ1
đ1 +đ2đŁ1i
(Elastic collision
with ball 2 initially at
rest.)
Elastic Collision in 1 Dimension when ball 2 is
initially at rest.
14
Demonstration and
Example
⢠A 0.50 kg basketball and a
0.05 kg tennis ball are
stacked on top of each other,
and then dropped from a
height of 0.82 m above the
floor.
⢠How high does the tennis ball
bounce?
⢠Assume all perfectly elastic
collisions.
Segment 1:
freefall of both
balls as they
fall, vi = 0.
Segment 2:
Elastic
collision of
basketball with
floor. Tennis
ball continues
downward,
unaffected.
Segment 3:
Elastic
collision of
upward
moving
basketball (1)
with downward
moving tennis
ball (2).
1
2
Segment 4:
freefall of
upward
moving tennis
ball.
15
Demonstration and
Example
⢠Divide motion into segments.
⢠Segment 1: free-fall of both balls
from a height of h = 0.82 m. Use
conservation of energy: Uf + Kf =
Ui + Ki
0 + ½ m vf2 = mgh + 0
vf = Âą[2gh]½ = â4.0 m/s, for both balls.
⢠Segment 2: basketball bounces
elastically with the floor, so its new
velocity is +4.0 m/s.
Demonstration and
Example
⢠Segment 3: A 0.50 kg
basketball moving upward
at 4.0 m/s strikes a 0.05
kg tennis ball, initially
moving downward at 4.0
m/s.
⢠Their collision is perfectly
elastic.
⢠What is the speed of the
tennis ball immediately
after the collision?
16
⢠A 0.50 kg basketball moving
upward at 4.0 m/s strikes a
0.05 kg tennis ball, initially
moving downward at 4.0 m/s.
⢠Their collision is perfectly
elastic. What is the speed of
the tennis ball immediately
after the collision?
Demonstration and
Example
⢠Segment 4: freefall of
tennis ball on the way up.
vi2 = +10.5 m/s.
⢠Use conservation of
energy: Uf + Kf = Ui + Ki
mgh + 0 = 0 + ½ m vi2
h = vi2/(2g) = 5.6 m.
⢠So the balls were dropped
from 0.82 m, but the tennis
ball rebounds up to 5.6 m!
(Assuming no energy losses.)
17
Before Class 16 next Wednesday Nov. 11
Complete Problem Set 7 on Chapters 8 and 9 by Tuesday
Nov.10 at 11:59pm. (Study break homework!!)
Read the first 3 sections of Chapter 10.
⢠Something to think about:
Why is a door easier to
open when the handle is
far from the hinge, and
more difficult to open when
the handle is in the
middle? ?![image from http://s693.photobucket.com/user/rorimac/media/Shire_dg7.jpg.html ]