phy131h1f - class 15 - department of physics

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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 “elastic”, that means it

conserves

A. Momentum p=mv

B. Kinetic energy E = ½ mv2

C. Both

D. Neither

https://grabcad.com/library/newton-s-cradle-assembly-1

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Quick Quiz 2 of 3:

• Two objects collide. All external forces on the

objects are negligible.

• If the collision is “inelastic”, 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 “totally inelastic”, 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.

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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?

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Class 15 Preclass Quiz Student Comments

“Does the explosive collision always need external forces?”

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.

“when 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...”

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.

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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

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Harlow answer: “What day of the

week are your Practicals on?”


“how do practicals schedule work again?

.................................................................haha just kidding”


“what kind of collision is for car accident?

Harlow answer: Always inelastic. If the cars ‘stick together’

during the collision, then this is totally inelastic.

“I 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: “Mechanical Energy” 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.

“could you please make the second term test easy?”

Harlow answer: No. Sorry.

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• 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:

– Uncertainties Reading

– Wolfson Chapters. 5-9 and Sections 10.1-10.3

– Classes 7-16

– Practicals 3-7

– 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 “Alternative Sitting

Registration Test 2”. The deadline to register for the alternative sitting

is 4:00pm on Thursday Nov. 12, 2015.


“What topics do the test 2 cover?”


“I 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.”

Harlow comment: For future reference, there is a “back

door” to much of my physics content. Just google “Jason

Harlow Teaching” and you’ll get to my teaching page. Click

on PHY131 Fall 2015. Links to all the preclass videos are

just sitting there.

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“we never learned integrals in high school, but were

expected to just know them now for physics?”

Harlow answer: Yes, you should learn them. It’s not too

hard.

Alyssa: “Give me hope by posting this in lecture!”

Paul: “I 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!”

Caroline: “I want to make it to the board too...”

“In Prof Thomson's book, he mentions "the spherical cow" as

a "classic" joke about physicists approaching real world

problems. elaborate.”

Consider the two integrals below. What’s the difference?

𝐽 = 𝐹𝑑𝑡

𝑊 = 𝐹 ∙ 𝑑 𝑟

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]

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Recall from Ch.4: Momentum

Momentum is the product of a particle’s 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.

𝑝 = 𝑚 𝑣

A basketball with mass 0.1 kg is traveling down and

to the right with vxi = +5 m/s, and vyi = –5 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. –0.5 kg m/s

C. +0.1 kg m/s

D. +0.9 kg m/s

E. zero

x

iv

fv

y

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A basketball with mass 0.1 kg is traveling down and

to the right with vxi = +5 m/s, and vyi = –5 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. –0.5 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 particle’s momentum is equal to the impulse on it:

𝐽𝑥 = 𝑡1

𝑡2

𝐹𝑥𝑑𝑡

∆𝑝𝑥= 𝐽𝑥 ∆𝑝𝑦= 𝐽𝑦

𝐽𝑦 = 𝑡1

𝑡2

𝐹𝑦𝑑𝑡

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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 –1.0 N s.

What is the velocity of the cart

immediately after the collision?

Was this collision elastic or

inelastic?

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• 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

http://images.thecarconnection.com/med/25-years-of-the-airbag_100223157_m.jpg

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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 balls’ velocities after the collision are v1f and v2f.

1 2Before:

During:

After:

𝑣1i 𝐾i

1 2𝑣1f 𝑣2f

𝐾f = 𝐾i

1 2During the collision energy is

stored as elastic potential energy.

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There are two equations, and two unknowns: v1f and v2f.

Solving for the unknowns gives:

Eq. 9.15:

Momentum conservation: 𝑚1𝑣1f +𝑚2𝑣2f = 𝑚1𝑣1i

Kinetic energy conservation: 1

2𝑚1𝑣1f

2 +1

2𝑚2𝑣2f

2 =1

2𝑚1𝑣1i

2

𝑣1f =𝑚1 −𝑚2𝑚1 +𝑚2

𝑣1i

𝑣2f =2𝑚1

𝑚1 +𝑚2𝑣1i

(Elastic collision

with ball 2 initially at

rest.)


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

𝑣1f =𝑚1 −𝑚2𝑚1 +𝑚2

𝑣1i

𝑣2f =2𝑚1

𝑚1 +𝑚2𝑣1i

(Elastic collision

with ball 2 initially at

rest.)


initially at rest.

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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.

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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 = ±[2gh]½ = −4.0 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?

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• 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.)

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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 ]

http://s693.photobucket.com/user/rorimac/media/Shire_dg7.jpg.html

phy131h1f - class 15 - department of physics

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